On “creating” a Lipschitz continuous derivative from a strictly convex function.

convex optimizationconvex-analysisreal-analysis

I believe the following statement to be true:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a strictly convex twice differentiable function, call it's unique global minimum $x^*$ .

Take any differentiable function $S(x) : \mathbb{R} \rightarrow \mathbb{R}$ such that $$ 0<\ell < \frac{ d(S(x) f'(x))}{dx} < L $$

for all $x$, where $\ell, L \in \mathbb{R}$.

Then $g(x):= S(x) f'(x) $ is Lipschitz continuous and the derivative of a strongly convex function with global minimum $x^*$ .

To apply this reasoning one only needs the derivative of the function $f$ and the knowledge of strict convexity.

Is this true or is something incorrect? Is strict convexity necessary or would convexity suffice? I have the impression that if there where multiple critical points we would need to know the minimum a priori.

Best Answer

  • Why $f$ has a global minimum? Take $f(x)=e^x$ on $\mathbb R$, you should add it to your hypothesis.

So suppose $f$ has a minimum point $x^*$ in $\mathbb R$, in general this result holds:

Theorem Let $f:\mathbb R\rightarrow\mathbb R$ a differentiable function and let $M>0$ such that $\lvert f'(x)\rvert \leq M$ for every $x\in\mathbb R$, then $f$ is a Lipschitz continuous function on $\mathbb R$.

Proof For every $x, y\in\mathbb R$ with $x\leq y$ by mean value formula we get $$ \lvert f(x)-f(y)\rvert =\lvert f'(\xi)(x-y)\rvert\leq M\lvert x-y\rvert $$ for some $\xi\in [x, y]$.
$\square$

Then $g(x)=S(x)f'(x)$ is clearly a Lipcschitz function. Remember that $f$ is strictly convex if and only if $f''(t)>0$ up to countable many $t\in\mathbb R$, then set $$ F(x)=\int^x_0S(t)f'(t)\ dt $$ you have $F''(x)=\frac{d}{dx}[S(x)f'(x)]> l>0$ so $F$ is strictly convex.

Finally because $f$ is convex we have $x^*$ is a minimum $\Leftrightarrow f'(x^*)=0$ so $$ F'(x^*)=S(x^*)f'(x^*)=0 $$ and $x^*$ is the unique global minimum of $F$.

If $f$ is only convex we have $$ F''(x)=S(x)f''(x)+S'(x)f'(x)=\frac{d}{dx}[S(x)f'(x)] $$ so if $f$ has multiple minimums then doesn't exists any function $S$ that satisfies your inequality, because $f''(y)=f'(y)=0$ for all these minimums. If $f$ has only one minimum then your assertion is true and $F$ is strictly convex.

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