Lipschitz condition and differentiability result

About lipschitz continuity of differentiable functions the following holds.

Theorem 1.

Let $X\mathrm{,}Y$ be Banach spaces and let $A$ be a convex (see convex set), open subset of $X$. Let $f\mathrm{:}\overline{A}\mathrm{\to }Y$ be a function which is continuous in $\overline{A}$ and differentiable in $A$. Then $f$ is lipschitz continuous on $\overline{A}$ if and only if the derivative $D\mathit{}f$ is bounded on $A$ i.e.

Proof.

Suppose that $f$ is lipschitz continuous:

$$\parallel f(x)-f(y)\parallel \le L\parallel x-y\parallel .$$

Then given any $x\in A$ and any $v\in X$, for all small $h\in ℝ$ we have

$$\parallel \frac{f(x+hv)-f(x)}{h}\parallel \le L.$$

Hence, passing to the limit $h\to 0$ it must hold $\parallel Df(x)\parallel \le L$.

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

$$\parallel Df(x)\parallel \le L,\forall x\in A.$$

Given any two points $x,y\in \overline{A}$ and given any $\alpha \in Y^{*}$ consider the function $G:[0,1]\to ℝ$

$$G(t)=⟨\alpha ,f((1-t)x+ty)⟩.$$

For $t\in (0,1)$ it holds

$$G^{\prime }(t)=⟨\alpha ,Df((1-t)x+ty)[y-x]⟩$$

and hence

$$|G^{\prime }(t)|\le L\parallel \alpha \parallel \parallel y-x\parallel .$$

Applying Lagrange mean-value theorem to $G$ we know that there exists $\xi \in (0,1)$ such that

$$|⟨\alpha ,f(y)-f(x)⟩|=|G(1)-G(0)|=|G^{\prime }(\xi )|\le \parallel \alpha \parallel L\parallel y-x\parallel$$

and since this is true for all $\alpha \in Y^{*}$ we get

$$\parallel f(y)-f(x)\parallel \le L\parallel y-x\parallel$$

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