3 Lipschitz condition and Lipschitz continuity

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1 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity7 3 Lipschitz condition and Lipschitz continuity 3. Definition Lipschitz-continuous function) Lipschitz-continuity comes in three different flavours. Let f : R m R m. a) Given an open set B R m, we say that f is Lipschitz-continuous on the open subset B if there eists a constant Λ R + called the Lipschitz constant of f on B) such that f fy) Λ y,, y B. 5) b) The function f is called locally Lipschitz-continuous, if for each z R n there eists an L > such that f is Lipschitz-continuous on the open ball of center z and radius L B L z) := {y R m : y z < L}. 6) c) If f is Lipschitz continuous on all of the space R m i.e. B = R m in 5)), then f is called globally Lipschitz-continuous. 3.2 Remark local vs. global) Notice the fundamental difference between the local and global versions of the Lipschitz-continuity. Whereas in the local version the Lipschitz constant Λ and the open set B depend on each point z R m, in the global version the constant Λ is fied and B = R m. In particular, a globally Lipschitz-continuous function is locally Lipschitz-continuous, but the viceversa is not true. 3.3 Remark norms and Lipschitz constants) In 5) the norm can be any norm. However, once a norm has been chosen one should stick to that single norm, as the Lipschitz constant Λ depends on the particular choice of this norm. Unless otherwise stated, we use the Euclidean norm in all our analysis. 3.4 Interpretation To see what these definitions mean, let us consider the situation in one dimension. Suppose f is a Lipschitz function on a neighborhood B of R. This implies that, y B, f fy) Λ y f fy) y Λ f + h) f h Λ by putting y = + h. If we were to let h and if the function f were differentiable, then the result would mean f Λ; that is that the derivative is bounded by the Lipschitz constant. However, there is nothing in the defintion of Lipschitz-continuity that implies that f is differentiable. So in general we can t proceed to this limit, since we don t know if f is differentiable at. But this tells us all we need to know: being Lipschitz just means f can t be too steep, the bound on the difference quotient being Λ.

2 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity8 3.5 Eamples and Countereamples The functions below are pictured in figure ; we will eamine them in turn, with B = [, ]. 3.6 Eample Consider f = 2. We will show that f is locally Lipschitz-continuous but not globally so. This function is continuously differentiable. Pick up any point R, we observe that sup y,+) f y) = sup 2y 2 + 7) y,+) being y + for y, + ) by the triangle inequality. Now picking up two points, y, z, + ) it follows by the mean value theorem that for some ξ between y and z, that f z) f y) = f ξ)z y) sup f θ) z y. 8) θ,+) Using 7), we conclude that f z) f y) 2 + ) z y, z, y, + ). 9) Hence the Lipschitz constant of f on, + ) is Λ = 2 + and the function is locally Lipschitz-continuous on all of R. Observe that the Lipschitz constant depends on and its neighbourhood. In particular, if then Λ. This is an indication that the function may not be globally Lipschitz continuous. Indeed, for any y f y) f ) y = y, as y, 2) which means that there is no Λ that can satisfy the global Lipschitz property for y R. { 3.7 Eample f 2 = 2 sin/ 2 ) if, This function is differentiable everywhere: if =. { f 2 2 sin ) = 2 2 cos ) if, 2 if =. One has to use the definition of derivative to obtain f 2 ).) However, since f 2 as, the derivative is not continuous. Is f 2 Lipschitzcontinuous? Define n = 2nπ + π/2) /2 and y n = 2nπ) /2 for n =, 2,..., and suppose that f 2 is Lipschitz-continuous. Since n, y n B for all n, there must eist Λ such that Λ f 2 n ) f 2 y n ) n y n for all n 2nπ + π 2 = ) 2nπ) /2 2nπ + π 2 ) /2 = 4n 2nπ) /2 + 2nπ + π ) ) /2 as n. 2

3 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity9 Plainly no such Λ can eist. Hence f 2 is not Lipschitz-continuous. In the first eample we used the theorem continuously differentiable = locally Lipschitz. Here we saw an eample where a differentiable function whose derivative is not continuous turned out not to be Lipschitz. However, this is not true of all such functions. For eample, consider the function shown in the handout 2.2 Continuity and Differentiability some facts of life ) { f = 2 sin/) if, if =, which is differentiable, but f is not continuous. It is left as an eercise to show that f is globally Lipschitz continuous. 3.8 Eample This eample shows that differntiability is a stronger concept than Lipschitz continuity. f 3 = This function is not differentiable at =, but { f 3 if <, = if >. Since f 3 f 3 y) = y y, for all, y R, we see that f 3 is globally Lipschitz continuous, with Lipschitz constant Λ =. 3.9 Eample f 4 = /2 This function is not differentiable at =, but { f 4 = 2 ) /2 if <, 2 /2 if >. Is f 4 Lipschitz-continuous? Define n = /n 2 and y n = for n =, 2,..., and suppose that f 4 is Lipschitz-continuous. Since n, y n B for all n, there must eist Λ such that Λ f 4 n ) f 4 y n ) n y n for all n = n as n. Plainly no such Λ can eist. Thus f 4 is not Lipschitz-continuous. 3. Relationships between Lipschitz continuity, continuity and differentiability 3. Theorem Lipschitz C ) Every locally Lipschitz-continuous function is continuous. The proof of this Theorem is left as an eercise. 3.2 Theorem C Lipschitz) Every continuously differentiable function is locally Lipschitz.

4 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity Proof Let f : R m R m be continuously differentiable. Fi any two points z, y R m and define the function f : [, ] R as fθ) := fz + θy z)). It is clear that f) = fz) and f) = fy)). 2) Furthermore, by the chain rule, we know that f is differentiable and that Here Dfw) is the Jacobian matri of f at w: d fθ) = Dfz + θy z)) y z). 22) dθ f f w w) w m w) Dfw) = ) f m f w w) m w m w) where f i w j w) is the partial derivative of the i-th component of f with respect to the j-th coordinate. By the definition of f and then by the fundamental theorem of calculus, we have fy) fz)) = f) f) = f θ) dθ = Df z + θy z) ) ) dθ y z); where the integral in the last line is a matri whose i, j-th component is given by It follows that f i w j z + θy z) ) dθ. fy) fz) Df z + θy z) ) dθ y z, where we use the notation for both the vector norm and its associated matri norm. Notice that, by the triangle inequality for integrals, we have Df z + θv z) ) dθ Df z + θv z) ) dθ sup Dfz + θv z)) θ [,] = sup Dfz + θv z)). θ [,] dθ

5 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity Furthermore by the equivalence of matri norms we have c > : A c A, A R m m, 24) where c depends only on m and is the maimum row sum norm from last handout. Thus, to establish Lipschitz continuity, we fi an arbitraty point R m and we establish a bound for Dfz + θy z)) dθ in an appropriate neighbourhood B of. Let B = B L, with L arbitrary. Since f is continuously differentiable on B, which is closed and bounded, there eists Λ such that sup f i w) w j Λ, ij [ : m]. w B Here we have applied the Weierstrass theorem which says that each continous function wj f i ) is bounded on a closed and bounded set B). Now, given z, y B, it follows that z + θz y) B for all θ [, ] because B is a ball, so It follows that sup θ [,] Df z + θy z) ) c mλ =: Λ. fy) fz) Λ y z, y, z B; which is to say that f is Lipschitz-continuous on B L. Since is arbitrary this means that f is locally Lipschitz-continuous on R m. 3.3 Remark Notice that in the proof above Λ, and therefore Λ, might depend on the point and the constant L. However, if we can find a bound that is independent of and Li.e., a uniform bound), then it means that the function f is actually globally Lipschitz-continuous. 3.4 Corollary If Dfw) is uniformly bounded as w, then f is globally Lipschitz. 3.5 Lipschitz-continuity with respect to some arguments As you may have noticed in the course ntoes we used a slightly different version of Lipschitz continuity: the Lipschitz continuity with respect to the first argument of a function f : R m R R. 25) We give now an adaptation of our earlier definition. The main point is that the only variable that counts is the first one. It is useful though to have the function continuous with respect to both variables. 3.6 Definition A function f such as in 25) is locally Lipschitz continuous with respect to its first argument if is continuous and if for each, t) R m R there eists a L > and a Λ > such that fy, s) fz, s) Λ z y 26)

6 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity 2 for all z, y B L and s t L, t + L). Notice how the right-hand side is independent of the second argument s.) The function f is called globally Lipschitz-continuous with respect to its first argument if it is continuous and if there is a Λ > such that 26) is satisfied for all z, y in R m and s R. Notice again the secondary role played by the time variable.) A fact that we have used in the proof of uniqueness is the following: 3.7 Lemma Characterization of local Lipschitz-continuity) A continuous function f, such as in 25), is locally Lipschitz continuous with respect to its first argument if and only if for each closed and bounded subset K of R m R, there eists an open set A R m such that K A and a constant Λ K > such that f, t) fy, t) Λ K y,, t), y, t) A. 27) Proof For the bold and the knowledgeable.) The if part is easy and is left as an eercise. The only if part can be proved by contradiction. Suppose that the conclusion is false. This means is, that there eists a closed and bounded subset K R m R such that any n Z + there eist n, y n R m, t n R, such that n, t n ), y n, t n ) A n, with A n = {z, s) R m R : y, t) K : z y + s t /n} and for which f n, t n ) fy n, t n ) > n n y n. 28) Notice that A n A which is closed and bounded. It follows that the sequences n, t n )) n Z + and y n, t n ) n Z + are bounded. There eists thus a family of integers X Z + and a point, t ) A the closure of A) such that Likewise, there eists y, t ) A and Y X such that lim n, t n ), t ), n X. 29) n lim y n n, t n ) y, t ), n Y. 3) Notice that, t ), y, t ) n Y A n = K. We show net that = y. Indeed, by the continuity of f, it has to be bounded by a constant M over the closed and bounded set K. So 28) implies that n y n 2M/n, for n Y, which means, by passing to the limit, as n, that y =, i.e., = y. 3) To conclude the proof it is sufficient to observe that the function f fails to be Lipschitzcontinuous in any neighbourhood of, t ). Indeed for any Λ > and for any open ball B, t ), for n Y big enough, we have that n, t n ) y n, t n ) B, and f n, t n ) fy n, t n ) > n n y n Λ n y n, 32) in contrast with the local Lipschitz-continuity with respect to the first argument.

7 ADE G56) Spring 26 Handout 3: Lipschitz condition and Lipschitz continuity 3 f f 2 f 3 f 4 Figure : Eample functions