Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting processes. We hve lernt the limits of sequences of numbers nd functions, continuity of functions, limits of difference quotients (derivtives), nd even integrls re limits of Riemnn sums. As often encountered in pplictions, echngebility of limiting processes is n importnt topic. For emple, we lernt whenever df d d d is integrble; lso lim n f = if {f n } nd {f n} converge uniformly. df, f() =, d d d f n() = d d lim n f n(), Here we consider the following sitution. Let f(, y) be function defined in [, b] [c, d] nd φ(y) = f(, y)d. It is nturl to sk if continuity nd differentibility re preserved under integrtion. Theorem 1. Let f(, y) be continuous in [, b] [c, d]. Then φ defined bove is continuous function on [c, d]. Proof. Since f is continuous in [, b] [c, d], it is bounded nd uniformly continuous. In other words, for ny ε>, δ such tht φ(y) φ(y ) f(, y) f(, y ) d <ε(b ) y, y y <δ, which shows tht φ is uniformly continuous on [c, d].
25-6 Second Term MAT26B 2 Theorem 2. Let f nd nd holds. be continuous in [, b] [c, d]. Then φ is differentible d b dy φ(y) = (, y)d Proof. Fi y (c, d), y + h (c, d) for smll h R, φ(y + h) φ(y) h = 1 h = (f(, y + h) f(, y))d (, z)d where z is point between y nd y + h which depends on. In ny cse, φ(y + h) φ(y) b h (, y)d d. (, z) (, y) Since is uniformly continuous on [, b] [c, d], for ε>, δ such tht (, y ) (, y) <ε, Tking h δ, weget φ(y + h) φ(y) h whence the condition follows. y y <δnd. (, y)d <ε, When y = c or d, the sme proof works with some trivil chnges. In mny pplictions, the rectngle is replced by n unbounded region. When this hppens, we need to consider improper integrls. As typicl cse, let s ssume f is defined in [, ) [c, d] nd set φ(y) = f(, y)d.
25-6 Second Term MAT26B 3 The function φ(y) mkes sense if the improper integrl for ech y. Recll tht this mens f()d is well-defined lim b f(, y)d eists. We introduce the following definition: The improper integrl f(, y)d is uniformly convergent if ε, b > such tht f(, y)d <ε, b,b b. Notice tht in prticulr, this implies tht b f(, y)d eists for every y. Uniform convergence of n improper integrl my be studied prllel to the uniform convergence of infinite series. In fct, if we let φ n (y) = n f(, y)d, it is not hrd to see tht the improper integrl converges uniformly iff the infinite series n=n φ n (y) converges uniformly when f(, y). When f chnges sign, the equivlence does not lwys hold. Nevertheless, techniques in estblishing uniform convergence cn be borrowed nd pplied to the present sitution. As smple, we hve the following version of M-test, whose proof is omitted. Theorem 3. Suppose tht f(, y) h() nd h hs n improper integrl on [, ). Then f(, y)d converges uniformly nd bsolutely. Theorem 4. Let f be continuous in [, ) [c, d]. Then φ is continuous in [c, d] if the improper integrl f(, y)d converges uniformly. Proof. By Theorem 1, the function φ n (y) = n f(, y)d
25-6 Second Term MAT26B 4 is continuous on [c, d] foreveryn. By ssumption, ε >, b such tht m φ n (y) φ m (y) = f(, y)d <ε, n, m b. n Hence {φ n } is Cuchy sequence in sup-norm. Since ny Cuchy sequence in supnorm converges, φ n converges uniformly to some continuous function ψ. As φ n converges pointwisely to φ, φ nd ψ coincide, so φ is continuous. Theorem 5. Let f nd be continuous in [, ) [c, d]. Suppose tht the improper integrls f nd re uniformly convergent. Then φ is differentible, nd dφ dy () = (, y)dy holds. Proof. Applying the men-vlue theorem to φ n φ m, φ n (y) φ m (y) (φ n (y ) φ m (y )) = (y y )(φ n(z) φ m(z)) for some z between y nd y. According to Theorem 2 nd the uniform convergence of, n φ n(z) φ m(z) = m (, y)dy s n, m. This shows tht ε >, b such tht φ n(y) φ n (y ) φ m(y) φ(y ) <ε, n,m b. y y y y Letting m, φ n(y) φ n (y ) φ(y) φ(y ) ε, n b. y y y y By tringle inequlity, φ(y) φ(y ) y y φ(y) φ(y ) + n (, y)d φ n(y) φ n (y ) φ n (y) φ n (y ) + y y y y y y. (, y)d (, y)d n (, y)d
25-6 Second Term MAT26B 5 Fi lrge n b such tht n (, y)d <ε nd, by Theorem 2, we cn lso find δ> such tht φ n(y) φ n (y ) n y y (, y)d <ε y y <δ. Putting things together, we conclude φ(y) φ(y ) y y (, y)d ε + ε + ε<4ε. One my pprecite these results when considering its relevnce in prtil differentil equtions. Consider the Lplce eqution u + u yy = on the disk D = {(, y) : 2 + y 2 < 1}. Epressed in polr coordintes, the eqution becomes u rr + u r r + u θθ =, (r, θ) [, 1) [, 2π]. r2 To solve this eqution mens to find function u = u(r, θ) which stisfies this eqution, nd, moreover, u is periodic in θ for r [, 1). returning to the rectngulr coordintes, u is continuous in D. This is becuse when We observe tht the Lplce eqution is rottionlly invrint. More precisely, for ny solution u(r, θ), the function v(r, θ) =u(r, θ + θ ) is solution for ech θ. From linerity it follows tht n c ju(r, θ + θ j ) is gin solution. In limit form, the function 2π ũ(r, θ) = g(α)u(r, θ + α)dα should lso be solution for ny continuous g. Indeed, define f(r, θ, α) =g(α)u(r, θ+ α). The functions f, θ, 2 f θ, 2 r, 2 f, re continuous in [,d] [, 2π], d<1. r2 It follows from Theorem 2 tht ũ is lso hrmonic. Noting tht g is rbitrry, in
25-6 Second Term MAT26B 6 this wy we hve found mny mny hrmonic functions from single one. In fct, tking the specil hrmonic function to be u(r, θ) = 1 1 r cos θ + r 2, one cn show tht every hrmonic function in D which is continuous in {(, y) : 2 + y 2 1} rises in this wy. We shll prove more sophisticted criterion for uniform convergence. Indeed, recll tht the comprison test is only effective in proving bsolute convergence of infinite series. We need Abel s nd Dirichlet s criteri to hndle the convergence of lternting series. Here the sitution is similr. We shll estblish version of Abel s criterion. The following lemm, which is usully clled the second men vlue theorem, is n integrl nlog of the Abel s lemm. Theorem 6. Let f be integrble nd g be non-negtive, decresing nd continuous on [, b]. Then there eists c [, b] such tht fg = g(c) Proof. Divide [, b] eqully by the prtition = < 1 < < n = b. Wehve f. fg = fg = g( ) f + (g() g( j ))f()d. As g is continuous nd f is bounded on [, b], the second term on RHS of this eqution tends to s n. Writing F () = g( ) f = g( )(F ( j ) F ( )) n+1 = g( j 2 )F ( ) j=2 = g( n 1 )F (b)+ f, the second term g( )F ( )) (g( j 2 ) g( ))F ( )) g()f (). j=2
25-6 Second Term MAT26B 7 As g is decresing, g()m =[g( n 1 )+ [g( n 1 )+ (g( j 2 ) g( ))]m j=2 (g( j 2 ) g( ))]M = g()m j=2 g( ) f for M =supf nd m = inf F. By men-vlue theorem then there eists ξ n [, b] such tht in g() f = g( ) f. Tking n, by pssing to convergent subsequence of {ξ n } we get g(c) f = gf. Now, we cn prove the criterion of Abel. Theorem 7. Suppose tht f(, y)d converges uniformly for y [c, d], nd g(, y) is decresing for ech fied y nd is bounded. Then f(, y)g(, y)d converges uniformly on [c, d]. Proof. By Theorem 6, there eists ξ [b, b ] such tht f(, y)g(, y)d = g(ξ,y) f(, y)d for lrge b nd b, the conclusion follows. b b < (sup g )ε, The following ppliction is of technicl nture. Let s evlute the Dirichlet integrl I = The trick is to consider the integrl ϕ(y) = sin d. e ysin d, y.
25-6 Second Term MAT26B 8 Letting We showed tht By Abel s criterion, given by For ech y δ, conclude tht δ 1, = f(, y) =, y sin e, sin d converges, so it is uniformly convergent in y trivilly. e ysin d converges uniformly. The y-derivtive of f is = e y sin. (, y)d is clerly uniformly convergent. By Theorem 2, we ϕ (y) = = = 1 1+y 2, d dy (e ysin )d e y sin d which holds for y δ>. By integrtion we get ϕ(y) = tn 1 y + C. As ϕ(y) = s y, C =tn 1 = π 2.So e ysin I = lim y + ϕ(y) =π 2. e y = 1 y