prove that $||d^2f(x)||\le M \Rightarrow ||df(x)||\le \sqrt{2Mf(x)}$

Question

let E be a banach space , $f : E \to \mathbb R$ a function of $C^2$ / $f>0$ we suppose that $\exists M $ cte and :

$||d^2f(x)||\le M $

prove that :

$||df(x)||\le \sqrt{2Mf(x)}$

Answer 1

Suppose to the contrary that there is $x_0$ such that $\|df(x_0)\|>\sqrt{2Mf(x)}$. Then there is a unit vector $v$ such that the one-dimensional restriction $\varphi(t)=f(x_0+tv)$ satisfies $$\varphi'(0)<-\sqrt{2M\varphi(0)}\tag{why?}$$ Now the Banach space disappears from the picture: we have a problem about a $C^2$ function $\varphi:\mathbb R\to\mathbb R$ which is positive and satisfies $|\varphi''(t)|\le M$ for all $t$.

A couple of hints for what to do next: (i) for all $t\in\mathbb R$

$$ \begin{split} 0 < \varphi(t) &=\varphi(0)+ \varphi'(0)\, t+\int_0^t (t-s)\varphi''(s)\,ds \\ &\le \varphi(0)+ \varphi'(0)\, t+ \int_0^t (t-s)M\,ds \end{split} \tag{why?} $$ (ii) when a quadratic polynomial is always positive, something can be said about its coefficients.