Pure Math 450/650, Winter 2012

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1 Compact course notes Pure Math 450/650, Winter 202 Lebesgue integration and Fourier analysis Professor: N. Spronk transcribed by: J. Lazovskis University of Waterloo pril, 202 Contents Toward the Lebesgue integral 2. Review of real analysis Outer measure Lebesgue measure lgebras and measurability The Lebesgue integral 9 2. Function measurability Simple function integration L p spaces 9 3. Construction Inclusion properties Operators and functionals Fourier nalysis Foundations Homogeneous subspaces veraging and kernels Localisation Hilbert spaces The inner product Orthonormal bases The Gibbs phenomenon Review 46 Note: Not all presented theorems are proved.

2 Toward the Lebesgue integral. Review of real analysis Proposition... If ϕ : X Y is a continuous bijection for X, Y compact, then ϕ is a homeomorphism. Definition..2. Banach space is a space over R or C equipped with a norm, that is x 0 for all x X non-negativity x = 0 x = 0 non-degeneracy x + y x + y for x, y X sub-additivity λx = λ x for x X and λ R or C homogeneity such that the metric space (X, ) is complete (Cauchy sequences have limits). Definition..3. metric space X is termed separable if it has a subset D which has a countable number of points and is dense in X. Definition..4. The power set of a set is defined as P() := {S S }. Definition..5. set is a set of first category iff it can be realized as a countable union of sets whose closures are nowhere dense. Definition..6. Let a < b in R and f : [a, b] X a function. Then: a partition of [a, b] is a collection of points P = {a = t 0 < t < < t n = b} a Riemann sum for f over P is any sum of the following form: S(f, P ) = f(t i ) (t i t i ) } {{ } } {{ } vector scalar for t i [t i, t i ] and i {, 2,..., n} Definition..7. We say that f : [a, b] X is Riemann integrable if there is a point x X such that for every ε > 0, there is a partition P ε of [a, b] such that for every refinement P P ε and every associated Riemann sum S(f, P ), we have that x S(f, P ) < ε. Definition..8. Note that if x as above exists, it is unique, and is termed the Riemann integral of f over [a, b] and denoted x = b a f = b a f(t)dt Theorem..9. [Cauchy criterion for Riemann integrability] Let a < b in R and X a Banach space with f : [a, b] X. Then equivalently:. f is Riemann integrable on [a, b] 2. for every ε > 0, there is a partition Q ε such that for any pair of refinements P, Q Q ε and any associated Riemann sums, S(f, P ) S(f, Q) < ε. Proof: ( 2) Easy exercise. ( 2) For each n N, let Q n be a partition of [a, b] such that for any refinements P, Q Q n and any associated Riemann sums, we have S(f, Q) S(f, P ) < 2. n Now let P = Q, P 2 = Q Q 2,..., P n = n Q i, and let X n = S n (f, P n ) be a fixed Riemann sum. Note that P n Q n and P P 2 2

3 If n > m, then we have x n x m = x n x n + x n x m+ + x m+ x m x n x n + x n x n x m+ x m = S n (f, P n ) S n (f, P n ) + D m+ (f, P m+ ) S m (f, P m ) s n + 2 n m = ( 2 m 2 n m + + ) 2 < 2 m If ε > 0 is given, choose m such that 2 < ε, and then (x m n ) n= will be Cauchy in X. Since X is a Banach space, there exists a limit x X of (x n ) n=. Now we must show that x = b f, or that f satisfies the condition of Riemann integrability. a Let ε > 0 and n N be such that 2 < ε n 2. If P n is as above and P P n, then for any Riemann sum S(f, P ) we have that S(f, P ) x S(f, P ) x n+ + x n+ x = S(f, P ) S n+ (f, P n+ ) + lim [ x n+ x m ] m < 2 n + 2 n = 2 n < ε.2 Outer measure Definition.2.. For E R, a sequence {I n } n= of open intervals is termed a cover of E if E n= I n. Definition.2.2. The length of an interval E R is defined naively, with the added condition that l((, a)) = l((b, )) =. Definition.2.3. For E R, define the outer measure of E to be { } λ (E) := inf l(i n ) {I n } n= is a cover of E by open intervals n= Proposition.2.4. The outer measure is a function λ : P(R) [0, ] with the properties:. λ ( ) = 0 2. λ (E) 0 for E P (R) non-negativity 3. E F R = λ (E) λ (F ) increasing 4. i. λ ( i E i) i λ (E i ) σ-subadditivity ii. λ ( i E i) = i λ (E i ) Proof: (4.i.) If λ (E n ) =, then we are done. If not, then we can cover E n by intervals so that l(i i,n) < λ (E n ) + ε 2 n. 3

4 Now consider (I i,n ) n= which covers n E n, such that λ (E) = l(i i,n ) n i ( λ ) (E n ) + ε 2 n n = λ (E n ) + ε 2 n = λ (E n ) + ε Remark.2.5. The notation i E i denotes a union of sets that is disjoint, i.e. E i E j = i j. Proposition.2.6. Let a < b R. Then λ (J) is constant for any J {[a, b], [a, b), (a, b], (a, b)}. Proof: Let ε > 0. Then (a ε, b + ε) covers J, with λ (J) l((a ε, b + ε)) = b a + 2ε. Since ε > 0, λ (J) b a. Fix J = [a, b). Let ε > 0 such that ε < b a, so that K = [a, b ε] is compact, and ((c i, d i )) So it also covers K with a finite subcover, up to some n N. rrange the is so that c < a, d n > b ε, and d i < c i+ for all i =,..., n. Then l((c i, d i )) = l((c i, d i )) (d i c i ) > c d n = b a ε cover J by open intervals. The other intervals are done in a similar fashion..3 Lebesgue measure Definition.3.. [Caratheodory] set R is termed (Lebesgue) measurable if for any E R, λ (E) = λ (E ) + λ (E \ ) Note that the statement λ (E) λ (E )+λ (E \) always holds by σ-subadditivity, so only the opposite direction must be shown for measurability. Furthermore, we define Theorem.3.2.., R L(R) 2. (R) = R \ L(R) 3., 2,.... L(R) = i i L(R) L(R) := { R is measurable} 4

5 Proof: (2.) Suppose that L(R), so then Therefore R \ L(R). λ (E (R \ )) + λ (E \ (R \ )) = λ (E \ ) + λ (E ) = λ (E) (3.) Let, 2, L(R) with E R and = i i, so that E = i (E i) Note that by σ-subadditivity, = (E ) (E 2 ) (E 3 ) = (E ) ((E \ ) 2 ) ((E \ ( 2 )) 3 ) λ (E) λ (E ) + λ (E \ ) Since i L(R) for all i, λ ((E \ λ (E) = λ (E ) + λ (E \ ) i k ) i ) + λ (E \ ) = λ (E ) + λ ((E \ ) 2 ) + λ ((E \ ) \ 2 ) (( ) )) ) i n = (λ E \ n i + λ (E \ n (( (λ E \ i n ) i )) + λ (E \ ) This also holds in the limit as n. Combining the above with σ-subadditivity, we get equality, so that λ (E) = λ (E ) + λ (E \ ). Thus L(R). Definition.3.3. Define Lebesgue measure through outer measure, by Theorem.3.4. Lebesgue measure satisfies:. λ( ) = 0 2. λ() 0 for L(R) 3., B L(R) and B = λ() λ(b) 4. i. λ( i i) i λ( i) ii. λ( i i) = i λ(e i) Lemma.3.5. If a < b R, then (a, b) L(R). λ = λ L(R) : L(R) [0, ] Proof: If λ (E) =, we are done. Suppose λ (E) <. Let ε > 0. Then we can find for E a cover (I n ) n= of open intervals such that n= l(i n) < λ (E) + ε 2. For each n N, let J n = I n (a, b) L n = I n (, a) R n = I n (b, ) 5

6 Then (J n ) n= covers E (a, b) and K n = {L n, R n, (a ε 8, a + ε 8 ), (b ε 8, b + ε 8 )} covers E \ (a, b), and By the definition of λ, we have l(k n ) = (l(l n ) + l(r n )) + ε 2 n= n= λ (E (a, b)) + λ (E \ (a, b)) l(j n ) + l(k n ) n= n= = (l(j n ) + l(r n ) + l(l n )) + ε 2 = l(i n ) + ε 2 n= < λ (E) + ε 2 + ε 2 = λ (E) + ε Since ε was arbitrary, λ (E (a, b)) + λ (E \ (a, b)) λ (E). Becauce of σ-additivity, equality holds. Corollary.3.6. ny open set G L(R). Proof: By ssignment, question 4, we can find (a n, b n ) for all n N such that G = pply the above lemma to all these intervals. (a n, b n ). n=.4 lgebras and measurability Definition.4.. Let X be a set. n algebra of subsets of X is any family of subsets M P(X) such that., X M 2. M = X \ M 3., 2,..., n M = n M Further, M is termed a σ-algebra if it satisfies. and 2. above, and 3. for countable unions. Example.4.2. These are some of the most common σ-algebras: Trivial σ-algebra M = {, X} Power set M = P(X) Lebesgue measurable sets M = L(R) For {M β β B} P(X) a family of σ-algebras, M = M β = { X M β β B} Borel σ-algebra B(R) = {M M P(R) is a σ-algebra containing all open sets in X} L(R) Remark.4.3. Heuristically speaking, a Borel set is a subset of R that may be formed by taking countable intersections and/or unions of open subsets of R. Definition.4.4. Let P(X) be a family of sets such that, X. Then we define { } { } σ := n, 2, δ := n, 2, n= β B n= 6

7 Proposition.4.5. If M is a σ-algebra and, 2, M, then n= n M. Proof: Suppose M P(X). Then each X \ n M, and hence n= (X \ n) M, which leads to ( ) ( ) n = X \ X \ n = X \ (X \ n ) n= n= n= M Definition.4.6. Define Then we have that G δ = G and F σ = F. G := {open sets in R} F := {closed sets in R} Definition.4.7. set R is a G δ -set if may be expressed as a countable intersection of open subets of R. Similarly, is an F σ -set if it may be expressed as a countable union of closed subets of R. Proposition.4.8. For the sets as defined above, G F σ and F G δ. Proof: Let G G. By ssignment, question 4, G = n= (a n, b n ) For each k N, let F k = k n= [a n + n, b n n ] for [c, d] = if d < c, and [a, n ] = [a, ]. Then each F k, being a finite union of closed sets, is closed. Moreover, G = F k. Therefore G F σ. If F F, then G = R \ F G. Hence R \ F = F k for each F k F. Then F = R \ (R \ F ) = R \ F k = (R \ F k ) G δ Definition.4.9. Define the Cantor set in the following manner. C 0 = [0, ] C = [0, 3 ] [ 2 3, ] = C 0 \ I C 2 = [0, 9 ] [ 2 9, 3 ] [ 2 3, 7 9 ] [ 8 9, ] = C \ (I 2 I 22 ).. C n = C n= \ (I n C n2 n ) Then C = n= C n is termed the Cantor set. We note that C is compact and non-empty. Proposition.4.0. For C the Cantor set as above,. C is nowhere dense in R 2. λ(c) = 0 3. C = R Note that the Cantor set has Jordan content. 7

8 Proof: (3.) For x [0, ], we may express x in ternary expansion, x = 0.t t 2 t 3 = t i 3 for t i {0,, 2} However, this expansion is not unique, as = We claim that C is the set of points with ternary expansions not containing s. To see this, observe that for I n,k as in the definition of the Cantor set, I = ( 3, 2 3 ) = {x = 0.t 2t 3... t i {0,, 2}, t i 2 for some i 2 and t i 0 for some i 2} I 2 = ( 9, 2 9 ) = {x = 0.0t 3t 4... t i {0,, 2}, t i 2 for some i 3 and t i 0 for some i 3} I 22 = ( 7 9, 8 9 ) = {x = 0.2t 3t 4... t i {0,, 2}, t i 2 for some i 3 and t i 0 for some i 3}. I n,k = {x = 0.t t 2... t n t n+ t n+2... t i {0,, 2}, t i 2 for some i n + and t i 0 for some i n + and t i i {,..., n } Further we note that This induces a bijection C = C n = [0, ] \ n= n= 2 n I n,k } {{ } set of points with t n = ϕ : C {0, 2} N 0.t t 2... (t i ) Since we know that R = {0, } N, this completes the proof. Proposition.4.. If x R and E R, define the translate of E by x by x + E := {x + y y E} The translate has the following properties:. λ (E) = λ (x + E) 2. E L(R) = (x + E) L(R) From these two we conclude that λ(e) = λ(x + E) for all x R. Proof: (.) For G R open, we know G = n= (a n, b n ) with λ(g) = n= λ((a n, b n )) = n= (b n a n ). Then for fixed x R, x + G = n= (x + a n, x + b n ) with λ(x + G) = n= ((x + b n) (x + a n )) = λ(g). Now note that E G iff x + E x + G. Hence λ (E) = inf{λ(g) E G, G open} = inf{λ(g) x + E x + G, G open} = λ (x + E). (2.) Let E L(R) and R. Then λ ( (x + E)) + λ ( \ (x + E)) = λ (x + (( x + ) E)) + λ (x + (( x + ) \ E)) = λ (( x + ) E) + λ (( x + ) \ E) = λ ( x + ) = λ () 8

9 Theorem.4.2. There exists a subset E R that is not Lebesgue measurable. Proof: Fix a > 0. On ( a, a), define a relation x y iff x y Q. For x [ a, a], define the equivalence class [x] = (x + Q) ( a, a). Let E ( a, a) be such that:. if x, y E, then x y implies x y 2. ( a, a) = x E [x] Note that E contains exactly one point from each equivalence class, which requires the axiom of choice. Now enumerate ( 2a, 2a) Q = (r k ) We claim that ( a, a) (r k + E) ( 3a, 3a). Note that (r k + E) (r j + E) = i j, since x = r k + y r j + z implies y z = r j r k Q. To see the first inclusion, note if x ( a, a), then x [y] for some y E. Thus x y Q and x y < 2a, so x y = r k for some k. To see the second inclusion, note that r k + x < 3a for any x ( a, a) and r k ( 2a, 2a). Now we will show that E / L(R). Suppose E L(R), so either λ(e) = 0 or λ(e) = α > 0. If λ(e) = 0, then λ(k + E) = 0 for all k. Then by the increasing and σ-additivity properties, ( ) 2a = λ(( a, a)) λ (E k + r k ) = λ(e + r k ) = 0 Hence its measure must be non-zero. Then by the increasing, σ-additivity, and translation invariance properties for n N, ( ) ( ) n nα = λ(r k + E) = λ (r k + E) λ (r k + E) λ(( 3a, 3a)) = 6a Clearly this cannot hold for n > 6a α. Therefore E / L(R). Remark.4.3. If the axiom of choice is weakened to only allow countable choice, then P(R) = L(R) = B(R). This makes irrelevant most of the previous work, so we acept the axiom of choice. Remark.4.4. For the non-measurable set E above, we have i. 0 < λ (E) 2a ii. λ (E) = 0 We will learn about the Lebesgue inner measure λ in the current ssignment. Definition.4.5. subset N R is termed a (Lebesgue) null set if λ (N) = 0. Note that null sets are measurable, as λ (E } {{ N } ) + λ (E \ N) λ (N) + λ (E) = λ (E) } {{ } N E 2 The Lebesgue integral 2. Function measurability Definition 2... Define the (characteristic) indicator function χ : X {0, } for a set X by { x χ (x) = 0 x / 9

10 Definition function f : R R is termed measurable if f ((α, )) = {x x R, f(x) > α} is measurable for all α R. We say that f is Borel measurable if f ((α, )) B(R) for all α R. Proposition Let f : R R be a function. Then the following are equivalent.. f is measurable 2. f ((, α]) L(R) for all α R 3. f ((, α)) L(R) for all α R 4. f ((α, )) L(R) for all α R Proof: (. 2.) First note that f ((, α]) = {x R f(x) α} = R \ {x R f(x) > α} Now recall that for R, we had that L(R) iff R \ L(R). (2. 3.) Observe that ( ) f ((, α)) = f (, α n ] ( = f ( (, α n ])) n= n= (3. 4.) as (. 3.) (4. 2.) as (2. 3.) Corollary Let f : R R be a function. Then f is measurable iff f (B) L(R) for any B B(R). Proof: ( ) Let G R be open, so that G = n= (a n, b n ), and ( ) f (G) = f (a n, b n ) = f ((, b n )) f ((a n, )) L(R) } {{ } } {{ } n= n= L(R) L(R) ( ) Let M f = {M R f (M) L(R)}. Note that f (R) = R implies R M f. lso, M, M 2, M f implies f ( M i) = f (M i ) L(R). Further, if M M f, then f (R \ M) = R \ f (M) L(R). Therefore M f is a σ-algebra. From above we have that G M f. Since B(R) is the smallest σ-algebra containing G, we must have B(R) M f. Proposition Let f, g : R R be measurable and c R fixed with ϕ : R R continuous. Then are all measurable functions. cf f + g ϕ f fg Proof: (.) Let α R. Then f (( α c, )) c > 0 (cf) R c = 0, α < 0 ((α, )) = c = 0, α 0 f ((, α c )) c > 0 0

11 Since all of the results are in L(R) by the assumption, cf is Lebesgue measurable. (2.) Enumerate the rationals by (r k ), so that (3.) Let α R. Then (f + g) ((α, )) = {x R f(x) + g(x) > α} = {x R f(x) > α g(x)} = {x R f(x) > r k and r k > α g(x)} = = ({x R f(x) > r k } {x R g(x) > α r k }) ( f ((r k, )) g ((α r k, )) ) L(R) (ϕ f) ((α, )) = f (ϕ ((α, ) )) L(R) } {{ } open set } {{ } open set (4.) Merely observe that, for m signifying a measurable function, fg = 4 ((f } {{ + g } (f g } {{ } ) } m {{ } } m {{ } } m {{ m } m ) 2 ) 2 Definition Let f : R R. Then we define non-negative functions f (x) = f(x) f (x) = max{f(x), 0} f + (x) = max{ f(x), 0} Corollary Let f : R R. Then f, f +, f are all measurable. To see this, note that x x is continuous, and f ± = 2 ( f ± f), and use a previous result. Definition Let L(R). Then define M() = {f : R f is measurable}, which is an algebra of functions. We extend f to f : R R by letting f(x) { = f(x) x so then f is measurable iff f is measurable. 0 x / Definition Define the extended real numbers to be R := R {, }. Remark function f : R R is termed extended real valued. Then f is measurable if f (B) L(R) for each B B(R), and f ({± }) L(R). Proposition 2... Let f n : R R be measurable for all n. Then are all measurable functions. sup{f n } inf {f n} n n lim sup{f n } lim inf {f n}

12 Proof: (.) Fix α R. Then ( sup{f n }) ([, α]) = {x R sup{f n } α} = n n (2.) Similarly as above, show that (inf n {f n }) ([, α]) L(R). (3.) Observe that [ ] lim sup[f n ] = lim sup{f k } kn Since sup kn {f k } is measurable by., the result follows. { = inf n fn n= sup{f k } kn ([, α]) } {{ } L(R) (4.) Same as (3.), } Corollary If f n : R R is measurable for all n and lim [f(x)] R for all x, then lim [f(x)] is measurable. Proof: The following sets are all measurable: lim [f n] = lim sup[f n ] = lim inf [f n] 2.2 Simple function integration Definition Let L(R). function f : R is termed simple if f() = {a,..., a n } is finite. Then we express f in standard form as f() = {a 0 < < a n }, and for E i = f ({a i }), then f = a i χ Ei. Proposition simple function f : R is measurable iff when written in standard form as above, each E i is measurable. Definition Define S() := {ϕ : R ϕ is simple and measurable} S + () := {ϕ S() ϕ 0} Then we define the proto-integral for ϕ S + () in standard form as I (ϕ) := a i λ(e i ) which may be infinite, and with the condition that 0 = 0 = 0. Proposition Let ϕ, ψ S + () and c 0. Then. I (cϕ) = ci (ϕ) 2. I (ϕ + ψ) = I (ϕ) + I (ψ) 3. ϕ ψ = I (ϕ) I (ψ) 2

13 Proof: (2.) Let ϕ() = {a < < a n }, ψ() = {b < < b m } and E i = ϕ ({a i }), F j = ψ ({b j }). Consider {a i + b j i n, j m} = {c < < c l }. For k l, let D k = a i+b j=c k (E i F j ), so then On the other hand, we also have ϕ + ψ = = = = I (ϕ) + I (ψ) = = = = m a i χ Ei + b j χ Fj o= a i m j= χ Ei F j + j= j= m (a i + b j )χ Ei F j j= l c k χ Dk m a i λ(e i ) + b j λ(f j ) m a i j= j= j= λ(e i F j ) + n b j χ Ei F j m m (a i + b j )λ(e i F j ) l c k λ(d k ) = I (ϕ + ψ) (3.) Let a i, b j, E i, F j as above. Then E i F j implies a i b j, as ϕ ψ, so then I (ϕ) = j= m a i λ(e i F j ) j= n b j j= λ(e i F j ) m b j λ(e i F j ) = I (ψ) Definition Given L(R), define M + () := {f : [0, ] f is measurable} Definition If L(R) and f M + (), then define S + f () := {ϕ S+ () ϕ f} f := sup{i (ϕ) ϕ S + f ()} The latter is termed the Lebesgue integral of f. 3

14 Proposition Let L(R) \ { } and f, g M + (). Then. f g on = f g 2. B and B L(R) = B f = fχ B 3. ϕ S + () = ϕ = I (ϕ) Proof: (.) If S + f () S+ g (), then f = sup {I (ϕ)} sup {I (ψ)} = g ϕ S + f () ψ S g + () (2.) For ϕ S + f (B), define ϕ on by ϕ(x) = { ϕ(x) x B 0 x \B. We note that ϕ is simple and measurable. So then ϕ S + f() and { ϕ ϕ S + f (B)} = S+ fχ B (), so fχ B = sup {I (ϕ)} = sup {I ( ϕ)} = sup {I B (ϕ)} ϕ S + fχ () ϕ S + B f (B) ϕ S + f (B) (3.) First we note if ψ S ϕ + (), then I (ψ) I (ϕ), and hence ϕ = sup {I (ψ)} I (ϕ) ψ S ϕ + () On the other hand, ϕ S + ϕ (), so I (ϕ) ϕ. Lemma If 2 L(R), then λ ( n= n) = lim [λ( n)]. Proof: Let C = and C n = n \ n for n 2. Since 2, we have C i C j = iff i j. Then ( ) ( ) λ n = λ C n = n= n= [ N ] λ(c n ) = lim λ(c n ) N n= n= = lim N [λ( N )] Theorem [Monotone convergence theorem] Let (f n ) n= M + () with f f 2 pointwise, and let f = lim [f n]. Then In particular, if sup n N [ f ] n <, then f <. [ f = lim f ] n Proof: Since f f 2, we have that f f 2. [ ] [ Hence lim f n = sup f n ]. a n N lso, f M + () by a previous result. Since f n f for each n, we have that f n [ f which implies lim [ ] Thus it remains to establish that lim f n f n. 4 ] f n a a f.

15 Let ϕ S + f () and 0 < η <. Let n = {x f n (x) ηϕ(x)}, so then: 2 i = Now let ηϕ() = {a < a 2 < < a m } and E i = (ηϕ) ({a i }) for all i =,..., m. Then, for each n, m f n f n χ n = f n ηϕ = a i χ (Ei i) n n By a previous lemma, since E i = E i = n= (E i n ), taking the limit as n, we get m a i λ(e i ) = ηϕ = η ϕ [ Therefore lim f ] n η ϕ. [ Since this holds for all 0 < η <, we have that lim f [ n] lim η η ϕ] = ϕ. Thus, since ϕ S + f () [ ] [ ] lim f n sup ϕ = f ϕ S + f () Lemma Let f : [0, ] for L(R) \ { }. Then f M + () there exists a sequence (ϕ n ) n= S + () with lim [ϕ n] = f pointwise Proof: ( ) limit of a sequence of measurable functions is measurable. ( ) For each k N, let F k = f ([k, ]). For each i =, 2,..., k2 k, let E k,i = f ([ i Then we have that 2 k, i 2 k ]). k2 = F k k E k,i ϕ k = kχ Fk + k2 k i 2 k χ E k,i Then clearly ϕ ϕ 2 and lim k [ϕ k] = f. Corollary Let L(R) \ { }. Then. if f, g M + (), c 0, then cf = c f and (f + g) = f + g 2. if (f n ) n= M + (), then n= f n = n= f n 3. if i L(R) for all i N with = i N i, then f = i N Proof: (.) Let (ϕ n ) n=, (ψ n ) n= S + () nondecreasing with lim [ϕ n] = f and lim [ψ n] = g. Then ϕ + ψ ϕ 2 + ψ 2 and lim [ϕ n + ψ n ] = f + g. 5 i f

16 Using MCT, we have that [ (f + g) = lim [ = lim [ = lim = f + g ] (ϕ n + ψ n ) ] ϕ n + ψ n [ ϕ n ] + lim ] ψ n (2.) Let g n = f k M + () with g g 2 such that lim [g n] = f k. Use (.) to see that g n = n f k and MCT for [ n ] [ ] f k = lim f n = lim g n = f k (3.) Let f n = χ n, and note fχ n M + () with f = n= fχ n Use (2.) above and the fact that fχ n = n f to get the result. Definition Define M() := {f : R f is measurable}. Definition Let L(R)\{ }. Then f M() is termed Lebesgue integrable if f +, f <. For such f, we define the Lebesgue integral of f to be f = f + f We introduce new sets: L() := {f : R f M(), f is Lebesgue integrable} L() := {f : R f M(), f is Lebesgue integrable} Lemma If f L(), then λ(f ( ({, })) = 0 2. If f M(), then f = 0 λ(f ([, 0)) f ((0, ])) = 0) Proof: (.) Let E + = f ({ }). Then for any n N, nχ E + f +, and nλ(e + ) = nχ E + f + < Rearranging, λ(e + ) n f +, and so λ(e + ) = 0. Doing a similair approach for the pullback of { }, we see that the statement holds. Corollary Let f L(). Then there exists f 0 L() such that f(x) = f 0 (x) for all x \ N with λ(n) = 0. This relationship is defined as f = f 0 almost everywhere, expressed f = f 0 (a.e.). 6

17 Proof: We know that since f L(), λ(f ( ) f ( )) = 0. Define f 0 : R by { f(x) x / f f 0 (x) = ( ) f ( ) 0 else Theorem [Properties of the integral] Let f, g L() and c R. Then (cf) L() and (f + g) L() and f L() and cf = c f (f + g) = f + g f f Moreover, if f : R, then f L() iff f L() and f M(). Proof: (2.) First note that (f + g) = (f + g) + (f + g), and (f + g) ± f ± + g ±. Hence (f + g)± (f ± + g ± ) = f ± + g± Therefore f + g L(). Claim: For h, k, ϕ, ψ L + () and h k = ϕ ψ, we have that h k = ϕ ψ. By a previous corollary, + ψ = (h + ψ) = (ϕ + k) = ϕ + k for finite integrals. Now we have (f + g) + + (f + g) = f + g = f + f + g + g = (f + + g + ) (f + g ). Since all these functions are Lebesgue integrable, (f + g) = (f + g) + (f + g) = (f + + g + ) (f + g ) = f + + g + f g = f + g (3.) First we recall that f = f + + f, so then f = f + + f f + + f = f + + f = (f + + f ) = f Since all the terms are finite, the result is finite. If f L() and f M(), then we get that f +, f M(). Then the first assumption gives us that f +, f f + + f = (f + + f ) = a f < 7

18 Lemma [Fatou] Let (f n ) n= M + (). Then [ ] lim inf [f n] lim inf f n Proof: Let g n = inf {f k}. kn Then 0 g g 2, and lim [g n] = lim inf [f n] by definition. pplying MCT, we get lim inf [f n] = lim [ g n]. Since g n f k for all k n, we have both g n f k and g n lim inf [ k f k]. Combining the above equations we get the desired result. n example of a sequence where strict inequality holds for the above lemma is f n = nχ (0,/n). Note that both Fatou and MCT still hold if we replace f = lim inf [f n] pointwise by f = lim inf [f n] (a.e.). Theorem [Dominated convergence theorem - Lebesgue] Let L(R) with λ() > 0 and (f n ) n= M() and g L + () such that. there exists f : R such that f = lim [f n] (a.e.) 2. for each n, f n g where g is termed the integrable majorant Then f L() and f = lim [ f n]. Proof: Let N = n= {x f n(x) > g(x)} {x lim [f n(x)] f(x) or DNE}, which is a null set. Note that N f n = 0 = g. Thus let \ N. Note that f = lim [f n] pointwise is measurable and that f = lim [ f n ] g. Since f g <, we have that f is integrable. a Next note that g + f n 0 and g + f = lim [g + f n] = lim inf [g + f n] pointwise. Then Fatou gives us that (g + f) lim inf [ (g + f n)], and so [ ] [ ] [ ] g + f = (g + f) lim inf (g + f n ) = lim inf g + f n = g + lim inf f n This shows that f lim inf [ f n]. Further, note that g f n 0 and g f = lim [g f n] = lim inf [g f n]. s above, from Fatou we get that g [ ] f = (g f) lim inf (g f n ) [ ] [ ] = lim inf g f n = g lim inf f n This shows that f lim inf Combining, we have that lim sup Therefore f = lim [ f n]. [ f n]. [ f n] f n lim inf [ f n]. 8

19 3 L p spaces 3. Construction Proposition 3... For f L(), let f = f. Then cf = c f for c R -homogeneity f + g f + g for g L() subadditivity Definition Define an equivalence relation on L() by: f g iff f = g (a.e.) Definition Define the L -space by L () = L()/. Note that elements in this vector space are sets of equivalence classes. However, we treat them as functions with the proviso that f = f in L () iff f = f (a.e.). Remark We say that f = lim [f n] in L iff lim [ f f n ] = 0. Definition Let < p < and L(R) with λ() > 0. Define the L p -space and its norm by L p () = { f : R ( ) /p f p = f p f p < } / (a.e.) where f L p (). It will be proved below that p is indeed a norm. Definition If < p < is fixed, define the conjugate index q by p + q =, so q = p p. Lemma For p and q as above, if ab 0, then ab ap p + bq q with equality iff ap = b q. Theorem [Holder] Let L(R) with λ() > 0 and < p < with q the conjugate index. If f L p () and g L q (), then with equality holding iff g q f p = f p g q (a.e.). fg L () fg f p g q Theorem [Minkowski] Let L(R) with λ() > 0 and < p <. If f, g L p (), then f + g L p () f + g p f p + g p with equality holding iff there exist c, c 2 R 0 with c + c 2 > 0 such that c f = c 2 g (a.e.). Lemma Let (X, ) be a normed vector space. Then X is complete with respect to iff for every sequence (x n ) n= X with n= x n < we have [ N ] x n = lim x n N n= n= 9

20 Proof: ( ) Here we employ the abstract Weierstrass test. Let (x n ) n= X with n= x n <. Let S n = n x k, so if m < n, then m S m S n = x k k=n+ m k=n+ x k Since n= x n <, we can make S m S n small so that (S n ) n= is Cauchy in X. Since X is complete, (S n ) n= converges in S X for [ n ] S = lim [S n] = lim x k = x k ( ) Let (x n ) n= X be Cauchy. Pick n N such that n, m n implies x n x m < 2 n 2 N such that n 2 n and n, m n 2 implies x n x m < 2 2 n k N such that n k n k and n, m n k implies x n x m < 2 k In this manner we get a sequence (x nk ). For each k, let y k = x nk+ x nk. Then we have that k k y j = x nj x nj < j= j= nd for the infinite sum, N N y j = lim y j lim = N j= j= N k j= 2 j 2 j j= [ j ] By the hypothesis, x = lim j x k exists. We observe that since we are dealing with a telescoping sum, we have. j y k = j (x nk x nk ) = x nk+ x n Therefore x + x n = lim j [x nj+ ] exists. Now we have that (x n ) n= is a Cauchy sequence and has a Cauchy subsequence. Hence (x n ) n= also converges in X. Theorem 3... Let L(R) with λ() > 0. Then L p () is complete. Proof: We use the previous lemma, and let (f n ) n= L p () with M = n= f n p <. We consider each f n as a measurable function on with f p <. Let g n = n f k (so g g 2 ) and for each x, let g(x) = lim [g n(x)] pointwise. Observe that g n p f k p = f k p f k p < 20

21 So by MCT, we have that [ ] g p = lim g p [ n = lim gn p ] p M p < Since g p is integrable, g(x) < (a.e.) on, and so g represents an element of L p (). Then for x (a.e.), n f k(x) = g n (x) g(x), and so f k(x) <. Then there is some f such that f(x) = lim [ n f k(x)], and so [ n ] f p = lim p [( n ) p ] f k lim f k = lim [gp n] = g p (a.e.) Therefore f p gp <, and f represents an element in L p (). Now we have left to show that f n f k p 0. Observe that n f p ( ) p f k f + f k (g + g) p = 2 p g p and Note that from the definition of f, we have lim [ f n f k ] = 0 (a.e.). So by LDCT we have that lim n f p [ f n = lim f p] f k = Hence f k = f is in (L p (), p ). Thus L p () is complete. p 2 p g p < 0 = 0 Definition If we allow p =, then we may construct the L space. Let f M() for L(R) with λ() > 0. Then define f = ess sup{ f(x) } = inf{c > 0 λ({x f(x) > c}) = 0} x Proposition The function is a norm on L (). Proof: If f L (), then f 0. If f = 0, then λ({x f(x) > n }) = 0. So {x f(x) 0} = {x f(x) > 0} = n= {x f(x) > n }. nd a countable union of null sets is still null, so f = 0. Let f, g L (). First note that {x f(x) > f } = n= {x f(x) > f > n }. Thus λ({x f(x) > f }) = 0. Consider {x f(x) + g(x) > f + g } {x f(x) + g(x) > f + g } nd the above union has measure zero. Hence it follows that {x f(x) > f } {x g(x) > g(x) } f + g = inf{c > 0 λ({x f(x) + g(x) > c}) = 0} f + g 2

22 Theorem The metric space (L (), ) is complete and hence a Banach space. Proof: Let (f k ) L (). Suppose that f k <. We need to show that f k defines an element of L (). Let E k = {x f(x) > f k } which is a null set. Hence E = E k is also a null set. So then if x \ E, by absolute convergence f k(x) f k. Therefore f k pointwise defines an element of L (). Remark For (f k ) n= such that f M(), we have notions for several types of convergence, lim k [f k] = f pointwise lim k [f k] = f a.e. lim k [f k] = f in L p 3.2 Inclusion properties Theorem Let [a, b] be a compact interval with a < b. Then for p < r < and f L r ([a, b]), L r ([a, b]) L p ([a, b]) and f p (b a) r p pr f r Proof: Let f L r ([a, b]). Then f p L r/p ([a, b]), i.e. [a,b] ( f p ) r/p = [a,b] f r <. Let q be the conjugate index to r, so q = r r p. Then applying Holder s inequality, we have f p p = f p = [a,b] [a,b] f p ( ) p/r ( ) /q ( f p ) r/p q [a,b] [a,b] ( ) /r p = f r (b a) /q [a,b] = f p r(b a) /q We note some containment relations, for p (, ): C([a, b]) L ([a, b]) L p ([a, b]) L ([a, b]) Remark Observe that L ([a, b]) L p ([a, b]) for p <, and there is k > 0 that is a function of p, a, b such that f p k f. Proposition Let p < r <. Then L p ([a, b]) L r ([a, b]). 22

23 Proof: Let [a, b] = [0, ]. Let f(x) = (a.e.). x p/r Now compute f(x) p dx = [0,] [0,] Thus f L p ([0, ]). It is easy to check that [0,] f r =. [ ] [ ] dx = lim x p/r x p/r dx = lim xp/r a 0 a p/r 0 = r r p < Theorem If a < b in R and p <, then L p ([a, b]) is separable. Proof: By ssignment 4, we know C([a, b]) L p ([a, b]) with f p k f for f C([a, b]) with fixed constant k > 0. lso we know that C([a, b]) is dense in L p ([a, b]). Recall that (C([a, b]), ) is separable. Let R[x] denote then space of polynomial functions in [a, b]. By Stone, Weierstrass, we have that R[x] = C([a, b]). Now note the facts that Q[x] R[x] is countable, call it (d n ) n= for each p R[x] and ε > 0, there is d Q[x] such that p d < ε. So if f L p ([a, b]) and ε > 0, we first find h C([a, b]) such that f h p < ε 2. Then we find g R[x] such that h g p < ε 4k and n N such that g d n < ε 4k. This all gives us f d n p f h p + h d n p < ε 2 + k h d n ε 2 + k ( h g + g d n ) < ε 2 + k ε 2k = ε Theorem L ([0, ]) is not separable. Proof: For each a = {a 0, a,... } {0, } N for a i {0, }, let f a = n= a nχ [ n+, n ]. We observe that if a, b {0, } N, then f a f b = (a n b n )χ [ n+, n ] = sup [ a n b n ] n= Thus if a b, then f a f b =. If there was a dense subset (d n ) n= of L [0, ], then for each a {0, } N there would be a unique n = n(a) such that f a f n(a) < 2. Note that n(a) n(b) for a b, since otherwise we would have f a f b = f a d n(a) + d n(b) f b f a d n(a) + d n(a) f b < This would contradict the above. Then we would have a n(a) : {0, } N N is injective, implying {0, } N N, which is absurd. n N Remark For a < b in R, if f L ([a, b]) then lim p [ f p] = f. 23

24 3.3 Operators and functionals Definition Let X, Y be Banach spaces. linear operator T : X Y is termed bounded iff T = T X = sup{ T x Y x X, x X < } < Proposition Let X, Y be Banach spaces with T : X Y linear. Then equivalently:. T is continuous 2. T is bounded 3. T is Lipschitz If T is Lipschitz, then T x T x Y T x x X where T is the smallest Lipschitz constant. Proof: ( 2) Let B(y) = {y Y y < } which is an open neighborhood of 0y for 0 the zero operator. Since T is continuous and T 0x = 0y, there is some δ > 0 such that x 0 X < δ implies T x 0y y <. Equivalently, the above may be stated as x X < δ = T x Y <. Now suppose x X and x X <. Then δx X = δ x X < δ. Thus δ T x Y = δt x Y <. So T = sup { T x Y } x X δ. x X < (2 3) For x X and ε > 0 we have that x X + ε x = x X + ε x X < Hence ( x X + ε T x Y = T x) Y x X + ε < T So then we have that T x Y T ( x X + ε), which reduces to T x Y T x X. nd if x, x X, then T x T x Y = T (x x ) Y T x x Y. Finally, if 0 < c < T, then there exists x B(x) such that T x Y > c > c x X. That is, T x T 0 Y > c x 0 X, so c is not a Lipschitz estimate. (3 ). Note Lipschitz implies uniform continuity implies continuity. Theorem Let < p < with q the conjugate index. If g L q (), then the functional Γ g : L p () R given by Γ g (f) = gf is a bounded linear functional with Γg = g q. Proof: First, if g L q () and f L p (), then by Holder gf L (), and Γg(f) = gf gf g q f p From above we know that Γg is the smallest c > 0 such that Γg(f) c f P and thus Γg g q. It is easy to verify that Γg is linear. To get the converse equality, observe that fg = f p g q provided f p = c g q. Now define sgn : R {, } by x { x0 which is Borel measurable. x<0 24

25 Further, let f = c g q/p sgn g for c defined below, so that f p =, and f p = c g q/p sgn g p ( = c p g q/p) p sgn g p Thus f p p = c p g q q and so f p = c g q/p Now we let c = to get f q q/p p =. q Finally we compute = c p = c p g q q q. g q Γg = sup{ Γg(f) f L p (), f p } ( Γg g q/p sgn g) g q/p q = g g q/p sgn g g q/p q = g q/p+ g p/q q = g q g p/q q = g q q q/p = g q Therefore Γg = g q. Remark If Γ : L p () R is a bounded linear functional, then there is g L q () such that Γ = Γg. Theorem If ϕ L (), let Γ ϕ : L () R by Γ ϕ (f) = ϕf. Then Γ ϕ(f) is a bounded linear functional with Γ ϕ = ϕ. Proof: First observe that for f L (), we have ϕf ϕ f (a.e.), and ϕf ϕ f = ϕ f = ϕ f Hence ϕf L () and Γ ϕ (f) = ϕf ϕf ϕ f It remains to verify that Γ ϕ ϕ. For ε > 0, let ε = {x ϕ ε ϕ(x) } Then λ( ε ) > 0 by definition, but it may be that the measure is. So we find ε ε such that 0 < λ( ε) <. Let f ε = λ( )χ sgn ϕ so that ε ε f ε = λ( ε) χ ε = λ( ε) 25 χ ε = λ( ε) λ( ε) =

26 Moreover, we have that Γ ϕ Γ ϕ (f ε ) = ϕ λ( ε) χ sgn ϕ ε = λ( ε) ϕ χ ε ϕ ε λ( ε) χ ε = ϕ ε Therefore Γ ϕ = ϕ. Theorem Let a < b in R and for f L [a, b], let Γ f : L [a, b] R be given by Γ f (ϕ) = (a,b) fϕ. i. the functional Γ f is linear and bounded, with Γ f = f ii. for Γ f : C[a, b] R, we have Γ f = sup{ Γ f (h) h C[a, b], h = f } Proof: (i.) Recall from above we have that (a,b) ϕf ϕ f. This tells us that Γ f f (it is clear that Γ f is linear). Let ϕ = sgn f, so that ϕ, and { } Γ f = sup fϕ [a,b] ϕ L [a, b], ϕ fsgn f [a,b] = f [a,b] = f (ii.) From the proof of ssignment, question 4, we know that there exists (h n ) n= C[a, b] such that h n for all n lim [h n] = sgn f (a.e.) Note that fh n f h n f, so f is an integrable majorant of (fh n ) n=. Then fh n fsgn f = f = f [a,b] [a,b] [a,b] Thus as a functional on C[a, b], Γ f sup{ [a,b] fh n } lim [ n N [a,b] fh n ] = f. But also we have that { } { } sup fh n h C[a, b], h sup fϕ ϕ L [a, b], ϕ = f [a,b] [a,b] 4 Fourier nalysis 4. Foundations Definition 4... function f : [a, b] C is termed measurable provided R(f), I(f) : [a, b] R are measurable. If both of these functions are also integrable, then f is integrable, and b a f = b a R(f)+i b a I(f). Definition For n Z, we let e int = e n (t). 26

27 Definition Define the trigonometric polynomials to be members of the set series of the form Trig(T) = span{e n n Z} = n= { N n= N c n e n N N, c n C c n e n for c n C is termed a formal Fourier series, with variants as below. Dirichlet kernel of order n D n := e k k= n π kth Fourier coefficient of f c k (f) := f(s)e iks ds nth Fourier sum of f s n (f, t) := c k (f)e ikt k= n π = } D n (s)f(t s) ds Moreover, we let L (T) = { f : R C f is measurable, -periodic, π f < } / (a.e.) Definition Define the translation function : X {f : X Y } {f : X Y } by 4.2 Homogeneous subspaces (s, f)(x) = (s f)(x) = s f(x) = f(x s) (a.e.) Definition subspace B L (T) is termed a homogeneous Banach space over T if it is equipped with a norm B under which B is Banach and for which. Trig(T) B 2. s f B for s R, f B i. s f B = f B ii. the function R B given by s s f (for fixed f) is continuous for each f in B Example These are some examples of homogeneous Banach spaces: i. C(T) = {f : R C f is -periodic and continuous} ii. L p (T) = {f : R C f is measurable, -periodic, π f p < }/ (a.e.) for p < iii. L (T) is not a homogeneous Banach space Definition Let B be a homogeneous Banach space over T and let h C(T). Define the convolution of h and f by the vector-valued integral h f := π h(s)(s f) ds Observe that our assumptions on h provide that s h(s)(s f) is continuous. 27

28 Note that for t R (a.e.), (h f)(t) = π = π = π = (f h)(t) h(s)f(t s) ds h(t + s)f( s) ds h(t s)f(s) ds Therefore the convolution operator is symmetric in its arguments. Remark We note that s n (f, t) = D n f(t). Proposition If B is a homogeneous Banach space over T with h C(T), then the convolution operator C(h)(f) := h f is linear and bounded, with C(h) B h. Proof: The linearity of C(h) is a consequence of linearity of Riemann integration. s for the inequality, note that for f B, we have C(h)(f) B = h f B = π h(s)(s f) ds = = π π π h f B B h(s)(s f) B ds h(s) s f B ds h(s) f B ds Theorem Let h C(T). Then. C(h) C(T) = h 2. C(h) L(T) = h Proof: (.) Let f C(T). Then if ȟ (s) = h( s), we have h f(0) = π Hence we have that C(h)(f) h f(0) = Γȟ (f). Therefore h(s)f(0 s) ds = π h( s)f(s) ds = Γȟ (f) C(h) C(T) = sup{ C(h)(f) f C(T), f } f sup{ Γȟ (f) f C(T), f } f = Γȟ = ȟ = h 28

29 Then using the previously proved inequality, the desired equality follows. (2.) For n N, let f n = πnχ [0,2/n], so f n =. Now, for t R (a.e.), h f n (t) = π h(s)f n (t s) ds = π h(s + t)f n ( s) ds = n 2 π h(s + t) ds Given ε > 0, there is δ > 0 such that h(t) h(s + t) < ε for s 0 = s < δ. Then for n > δ, we have h (h f n ) = π h(t) (h f(t)) dt = π h(t) n /n h(s + t) ds 2 /n dt Then h(t) = n 2 = ε /n /n π π h(t) ds n 2 n 2 /n /n /n /n Hence h (h f n ) is also bounded above by ε, and This in turn implies (h(t) h(t + s)) ds dt h(t) h(t + s) ds dt lim [ h (h f n ) ] lim [ h (h f n) ] = 0 C(h) Lp(T) = sup{ C(h)(f) f L (T), f } f sup{ C(h)(f n ) } n N lim [ C(h)(f n) ] = h Theorem [Properties of the Dirichlet kernel] The Dirichlet kernel of order n, D n, satisfies the following:. D n is real-valued, -periodic and even π D n = for t [, π], we have D n (t) = { sin((n+ 2)t) sin( t) t 0 2 2n + t = 0 29

30 4. lim [ D n ] = We often call D n = L n the nth Lebesgue constant. Definition Let X be a metric space. set F X is of first category (or meager) if F n= F n for every F n closed and nowhere dense. set U X is of second category (or non-meager) if it is not meager. Recall that the Baire category theorem states that if a metric space X is complete, then it is non-meager. Theorem [Banach, Steinhaus] Let B, X be Banach spaces with F a family of bounded linear maps from B to X. If sup { T f X T F} < for each f in a non-meager set U B, then sup { T } <. f T F Proof: For each n N, let F n := {f B T f X n T F} = {f B T f X n} Note that {f B T f X n} is closed, as for g T (f) = T f X, so is g T ({z C z n}). Then for the specified U non-meager, we have U n= F n with at least some Fn 0. Hence there exists f 0 B and r > 0 such that B r (f 0 ) = {f B f f 0 B < r} F n0. Note that if f B r (f 0 ) F n0, then T f X n 0 for T F. Fix f B with f B, so for such f we have that f 0 + r 2 f and f 0 r 2 f are both in B r(f 0 ). So then for T B r (f 0 ), we have T f X = T ( ( r f0 + r 2 (f 0 r 2 f))) r q ( = r T f0 + r 2 f) T ( f 0 r 2 f) ( X T ( f 0 + r 2 f) X T ( f 0 r 2 f) ) X r 2n0 r < T F Hence T = sup T { T f X f B and f B } < for all T F. Corollary Let B, X be Banach spaces and for n N, let T n : B X be a bounded linear operator. Suppose sup{ T n } =. Then there exists U B with B \ U meager such that sup{ T n f X } = for n N all f U. Proof: Let F = {f B sup{ T n (f) X } < }. n N If F were non-meager, then the Banach-Steinhaus theorem would show that sup{ T } <. This would be a contradiction, so F is meager. Therefole U = B \ F. Theorem The set of f C(T) for which sup { S n (f) } < is a meager subset of C(T) n N 2. The set of f L (T) for which sup { S n (f) } < is a meager subset of L (T) n N Proof: (.) We have seen the following facts: s n (f) = D n f = C(D n )(f) C(D n ) C(T) = D n 30

31 D n = L n Hence by the Banach-Steinhaus theorem, sup{ s n (f) } = for all f C(T) \ F for F meager. n N (2.) Similarly. 4.3 veraging and kernels Definition Let (x n ) n= be a sequence in a Banach space X. Then the expression σ n := n (x + + x n ) is termed the nth Cesaro sum of the given sequence. Further, if f L (T), then the nth Cesaro sum of f is defined to be σ n (f) := n+ (s 0(f) + + n n (f)) = n+ (D 0 f + D n f) = n+ (D D n ) f = K n f With respect to the above, K n is termed the Fejer kernel. Theorem [Properties of the Fejer kernel]. K n is real-valued, -periodic, and even ( sin( (n+)t) ) K n (t) = n+ sin( t) t 0 2 for t [0, ] n + t = 0 3. K n = π K n = 4. if 0 < t, then 0 K n (t) π2 t 2 (n+) Definition summability kernel is a sequence (k n ) n= of -periodic, bounded and piecewise continuous functions such that π. k n = 2. sup{ k n } < n N 3. for any 0 < δ π, lim [ δ k n + ] π δ k n = 0 Proposition The Fejer kernel (K n ) n= is a summability kernel. Theorem [bstract summability kernel theorem] Let B be a homogeneous Banach space on T and (k n ) n= be a summability kernel. Then for f B, lim [ k n f f B ] = 0 k Proof: Fix f B. Let F : R B be given by F (s) = s f for s R (a.e.). The axioms of a homogeneous Banach space tell us that F is continuous and -periodic, and that F (s) B = s f B = f B and 0 f = f = F (0) 3

32 Now we compute (k n f) f B = = = π π π π k n (s)(s f) ds f B k n (s)f (s) ds F (0) B k n (s)(f (s) F (0)) ds k n (s) F (s) F (0) B ds B Given ε > 0, we can find δ > 0 such that F (s) F (0) B < ε 2M Then choose N N such that for all n N, we have ( δ ) π ε + k n < 4 f B δ for 0 s < δ and M = sup{ k n } <. n N We may safely assume that f B > 0. Then for all n N, we have that ( (k n f) f B δ ) π + k n (s) F (s) F (0) B ds + δ k n (s) F (s) F (0) B ds δ δ ( δ ) π + k n (s) (2 f B )ds + δ ( ε ) k n (s) ds δ δ 2M ( = f δ ) π B + k n (s) ds + ε δ k n (s) ds π 4Mπ δ π ε 2 + ε k n (s) ds 4Mπ = ε 2 + ε 2M k n(s) ε 2 + ε 2 = ε δ Corollary For f C(T), we have lim [ σ n(f) f ] = 0 2. If p <, then for f L p (T) we have lim [ σ n(f) f p ] = 0 Corollary Suppose f, g L (T), and c k (f) = c k (g) for each k Z, Then f = g (a.e.). Proof: Recall that σ n (f, t) = n + j=0 s j (f, t) = n + j j=0 k= j c k (f)e ikt 32

33 Since c k (f) = c k (g) for all k, we have that for all n N, f g = f σ n (f) + σ n (g) g f σ n (f) + σ n (g) g 0 Therefore f g = 0, or f = g a.e. Definition Let f L(T) and s R. Then the average value of f at s is defined to be provided that it exists. ω f (s) := 2 lim h 0 + [f(s h) + f(s + h)] Theorem [Fejer]. If f L(T) and x [, π] such that ω f (x) exists, then lim [σ n(f, x)] = ω f (x) 2. If I is an open interval on which f is continuous, then for any closed subinterval J of I, [ ] lim sup { σ n (f, t) f(t) } t J Proof: (.) Suppose that ω f (x) is finite. Given ε > 0, let δ > 0 be such that 0 s δ implies ω f (x) 2 (f(x s) + f(x + s)) < ε. Then we have that for our choice of δ σ n (f, x) ω f (x) = π K n (s)f(x s) ds ω f (x) = π K n (s)f(x s) ds π K n (s)ω f (x) ds = π K n (s)(f(x s) ω f (x)) ds ( δ K n (s)(f(x s) ω f (x)) ds + δ ) π + K n (s)(f(x s) ω f (x)) ds δ Now note that for every n N, since K n is even, δ K n (s)(f(x s) ω f (x)) ds = δ Then we have δ K n (s)(f(x s) ω f (x)) ds δ = 4π = ε ε = ε δ δ δ δ δ δ δ δ π δ δ = 0 π δ K n ( s)(f(x s) ω f (x)) ds K n (s)(f(x s) ω f (x)) ds + δ K n (s)(f(x + s) ω f (x)) ds 4π δ K n (s) ( 2 (f(x s) + f(x + s)) ω f (x) ) ds K n (s) 2 (f(x s) + f(x + s)) ω f (x) ds K n (s) ds K n (s) ds 33

34 s for the other part of the original integral, we have that ( δ ) ( π + K n (s)(f(x s) ω f (x)) ds δ + δ Hence we conclude that lim sup [ σ n (f, x) ω f (x) ] ε. Then we conclude that the limit exists and vanishes. π π 2 2(n + )s 2 π 2 2(n + )δ 2 π 2 2(n + )δ 2 δ ) K n (s) f(x s) ω f (x) ds ( δ + ( δ + π π δ π δ ) ) f(x s) ω f (x) ds ˇf (s x) ω f (x) ds π 2 2(n + )δ 2 (x ˇf ) ω f (x) 0 ˇf (s x) ω f (x) ds (2.) Note that all the estimates performed were done uniformly over x for the choice of δ. Corollary Suppose that f L(T), x [, π] such that ω f (x) exists, as does lim [S n(f, x)]. Then Proof: If lim [S n(f, x)] exists, then lim [S n(f, x)] = ω f (x) lim [σ n(f, x)] = lim n + S j (f) = lim [S n(f, x)] j=0 Since Fejer gives that lim [σ n(f, x)] = ω f (x), the result follows. Definition If f L[a, b], then a point x (a, b) is termed a Lebesgue point of f iff [ n ] lim f(x s) + f(x + s) n 0 + n f(x) 2 ds = 0 Proposition If ω f (x) exists, then x is a Lebesgue point of f. Theorem [Lebesgue, Fejer] If x [, π] is a Lebesgue point for f L(T), then f(x) = lim [σ n(f, x)]. Lemma If f L (T), then for all k Z, c k (f) f. 0 Proof: c k (f) = π f(t)e ikt dt π f(t) e ikt dt = f 34

35 Theorem [Riemann, Lebesgue] If f L (T), then lim [ c k(f) ] = 0. k Proof: Let ε > 0. By the abstract summability kernel theorem, we can find n 0 N such that σ n0 (f) f < ε. We note that σ n0 (f) = n + Thus if k n 0, then we have j j=0 k= j c k (f)e k = n 0 n 0 + j n j= n c k (σ n0 (f) f) = c k (σ n0 (f)) c k (f) = = π n 0 j= n 0 = c k (f) n 0 c j (f)e j = j= n 0 b j e j k c k (f) π b j e }{{} j k c k (f) } {{ } =0 n 0 j= n 0 b j e j We then have that c k (f) = c k (σ n0 (f) f) σ n0 (f) f < ε. Corollary lim [ π ] f(t) cos(nt) dt [ π ] = 0 lim f(t) sin(nt) dt = 0 Definition Define the space C 0 (Z) as { } C 0 (Z) = (c k ) c k C, lim [c k] = 0 k This turns out to be a Banach space. Theorem [Open mapping theorem] If X, Y are Banach spaces and T : X Y is a surjective bounded linear operator, then T is open, i.e. U X is open implies that T (U) Y is open. Corollary [Inverse Mapping theorem] Let X, Y be Banach spaces, with T : X Y a bounded linear operator. If T is bijective, then T : Y X is bounded. Corollary There exist elements of C 0 (Z) which are not arising from Fourier transforms. That is, there exist sequences (c k ) for which there is no f L (T) such that c k = c k (f) for each k. Proof: Let T : L (T) C 0 (Z) be given by T f = (c k (f)) k=. Then T is linear with range(t ) C 0 (Z) (by Riemann-Lebesgue), and T is bounded with T. lso, T is injective by the corollary to the abstract summability kernel theorem. If it were the case that T were bijective, then T : C 0 (Z) L (T) would be bounded. However, consider d n = (..., 0, 2,..., 2, 2, 2,..., 2, 0,... ) C 0(Z) 35

36 So the sequence has nonzero values from the nth to the nth positions only. Then d n = 2, and T (d n ) = D n, the Dirichlet kernel of order n. But this leads us to T sup{ T (d n ) } = sup{ D n } = n N n N This contradicts the inverse mapping theorem. Thus we have a contradiction, and T is not bijective, so not surjective. Remark The map introduced in the proof above is termed the Fourier transform, and is given by 4.4 Localisation Lemma If f L(T) with π T : L (T) C 0 (Z) f (c k (f)) k= f(t) t dt <, then lim [s n(f, 0)] = 0. Proof: Recall that sin(x + y) = sin(x) cos(y) + cos(x) sin(y). Therefore we have that D n (s) = sin (( ) ) ( n + 2 s sin ( sin(ns) cos 2 = s) s) sin ( + cos(ns) s) s n (f, 0) = = π π 2 D n (s)f(s) ds sin(ns) cos ( 2 s) sin ( f(s) ds + 2s) 2 π cos(ns)f(s) ds Note that for 0 < θ < π 2 we have that sin(θ) 2 π θ. Thus we have that ( sin 2 t) π t for t [, π], and π ( cos 2 s) f(s) sin ( 2 s) ds π f(s) π s ds = π f(s) 2 s ds < So the map s cos( 2 s)f(s) sin( 2 s) (for a.e. s [, π] extended periodically) defines an element in L(T). Then we apply the corollary to the Riemann-Lebesgue lemma to see that s n (f, 0) = π sin(ns) cos ( 2 s) sin ( f(s) ds + 2s) π cos(ns)f(s) ds 0 Theorem [Localisation Principle] If f L(T) and f(t) = 0 for a.e. t on an open interval I, then for t I we have lim [s n(f, t)] = 0. Proof: Let g L(T) be given by g(s) = f(t s) = ˇf (s t) so g = t ˇf when t I is fixed. 36

37 Then g(s) = 0 for a.e. s in neighborhood of 0, say for s ( δ, δ), and so π g(s) δ ( s ds = 0 δ ) π s ds + g(s) + s ds = δ δ π π g(s) ds δ t ˇf ds = δ t ˇf δ = δ f < Hence by the lemma, lim [s n(g, 0)] = 0, and by translation and inversion invariance we find that s n (g, 0) = π D n (s)g(s 0) ds = π Therefore we have that lim [s n(f, t)] = lim [s n(g, t)] = 0. D n (s)f(t s) ds = s n (f, t) Corollary If f, g L (T) and f(t) = g(t) for a.e. t on an open interval I, then for t I we have that lim [s n(f, t)] exists iff lim [s n(g, t)] exists, and then two limits coincide. Proof: Observe that lim [s n(f g, t)] = lim [s n(f, t) s n (g, t)] = lim [s n(f, t)] lim [s n(g, t)] = 0 Theorem [Dini] If f L (T) and f is differentiable at t [, π], then for such t we have lim [s n(f, t)] = f(t). Proof: Given ε > 0, there is δ > 0 such that s < δ implies f(t s) f(t) s f (t) < ε. Hence the map s f(t s) f(t) s is bounded on ( δ, δ). Let g = t ˇf f(t), so g(s) = f(t s) f(t) and π g(s) δ s ds = g(s) s ds + g(s) s ds δ δ δ( f (t) + ε) ds + δ [ δ,] [δ,π] π g(s) ds = 2δ( f (t) + ε) + δ t ˇf f(t) < pplying the lemma, we have that lim [s n(g, 0)] = 0. s before, we conclude that s n (g, 0) = s n (t ˇf f(t), 0) = s n (t ˇf, 0) s n (f(t), 0) = s n (f, t) f(t) 37