On "creating" a Lipschitz continuous derivative from a strictly convex function. 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.
 A: *

*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. 
