Lipschitz condition and differentiability result


About lipschitz continuity of differentiable functions the following holds.

Theorem 1.

Let X,Y be Banach spacesMathworldPlanetmath and let A be a convex (see convex set), open subset of X. Let f:A¯Y be a function which is continuousMathworldPlanetmathPlanetmath in A¯ and differentiableMathworldPlanetmath in A. Then f is lipschitz continuous on A¯ if and only if the derivativePlanetmathPlanetmath Df is bounded on A i.e.

supxADf(x)<+.
Proof.

Suppose that f is lipschitz continuous:

f(x)-f(y)Lx-y.

Then given any xA and any vX, for all small h we have

f(x+hv)-f(x)hL.

Hence, passing to the limit h0 it must hold Df(x)L.

On the other hand suppose that Df is bounded on A:

Df(x)L,xA.

Given any two points x,yA¯ and given any αY* consider the function G:[0,1]

G(t)=α,f((1-t)x+ty).

For t(0,1) it holds

G(t)=α,Df((1-t)x+ty)[y-x]

and hence

|G(t)|Lαy-x.

Applying Lagrange mean-value theorem to G we know that there exists ξ(0,1) such that

|α,f(y)-f(x)|=|G(1)-G(0)|=|G(ξ)|αLy-x

and since this is true for all αY* we get

f(y)-f(x)Ly-x

which is the desired claim. ∎

Title Lipschitz condition and differentiability result
Canonical name LipschitzConditionAndDifferentiabilityResult
Date of creation 2013-03-22 13:32:42
Last modified on 2013-03-22 13:32:42
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 5
Author paolini (1187)
Entry type Result
Classification msc 26A16