# F is strictly convex and $\lim_{|x|\to \infty}\frac{F(x)}{|x|^2}=C.$ Then F has bounded second derivative [closed]

Can you let me know how to prove the following?

Suppose $F:\mathbb{R}^n\mapsto \mathbb{R}$ is smooth. If $F$ is strictly convex and $$\lim_{|x|\to\infty}\frac{F(x)}{|x|^2}=C$$, then $\xi^{t}D^{2}F(x)\xi< A$ for some constant A.

Thanks

The claim certainly fails for $n=1$: consider a function $f:\Bbb R\to\Bbb R$ such that:

1. $f\ge 0$ and, for all $x\le 0$, $f(x)=0$
2. $f\in C^\infty(\Bbb R)$
3. $\int_0^\infty f(t)\,dt=\alpha<\infty$
4. $\limsup_{x\to\infty} f(x)=+\infty$

Trivia: such a function exists.

Then consider the function $H(x)=\int_0^x dy\int_0^yf(t)\,dt$.

By Lagrange's theorem, for all $x\in\Bbb R$ there is $\xi_x\in\Bbb R$ such that $$\left\lvert\frac{H(x)}x\right\rvert=\left\lvert\int_0^{\xi_x} f(t)\,dt\right\rvert\le \alpha$$

And therefore $$\limsup_{\lvert x\rvert\to\infty} \left\lvert\frac{H(x)}{x^2}\right\rvert\le\limsup_{\lvert x\rvert\to\infty}\frac\alpha{\lvert x\rvert}=0$$

Thus $\lim_{\lvert x\rvert\to\infty} x^{-2}H(x)=0$ and $\lim_{\lvert x\rvert\to\infty} \frac{x^2+H(x)}{x^2}=1$.

However, $$(x^2+H(x))''=2+f(x)$$ which is unbounded. And it is strictly positive, making $x^2+H(x)$ strictly convex.

Personally, I don't see a reason why such a counterexaple couldn't be "radially" adapted to $n>1$.