# Prove boundedness of 2nd derivatives

Let $f \, \colon \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a smooth and convex function. Assume $f$ behaves asymptotically as a cone at infinity, i.e.,

$\lim_{R \rightarrow \infty} \frac{f(R x)}{R} = F (x),$

for every $x \in \mathbb{R}^d$, where $F$ is a homogeneous function of degree one which is smooth away from the origin.

I would like to show that the second derivatives of $f$ are bounded.

So far, I was able to show that $f$ is Lipschitz and therefore I know that the gradient $\nabla f$ is bounded. But I have no idea how to prove boundedness of the second derivatives. I would be really grateful for any suggestions. Thanks!

## 1 Answer

This is not true. Here's a one-dimensional counterexample (that also works in higher dimensions). Begin with the function $g(x) = \sqrt{x^2+1}$ which is smooth, convex, asymptotic to $|x|$, and has $g''(0)=1$. Define $$f(x) = \sum_{n=0}^\infty 3^{-n} g(2^n(x-n))$$ Since $|x|\le g(x)\le |x|+1$, it follows that $$\sum_{n=0}^\infty (2/3)^{-n} |x-n| \le f(x) \le 3+ \sum_{n=0}^\infty (2/3)^{-n} |x-n|$$ hence $3|x|-C_1\le f(x)\le 3|x|+C_2$ for some constants $C_1,C_2$. So, $f$ is asymptotic to a cone.

On the other hand, $$f''(n) \ge 3^{-n} 4^n g''(0) = (4/3)^n$$

• Nice one! When I saw you were claiming it was false, I thought of this: Take any continuous positive unbounded $h$ on $[0,\infty)$ such that $\int_0^\infty h = L <\infty.$ Then $f(y) = \int_0^y\int_0^x h(t)\,dt\,dx$ is a counterexample. – zhw. May 31 '15 at 2:51
• Thank you for this nice example! But I have one question: why is the function f twice differentiable? – Mike Jul 1 '15 at 7:52