Limits along lines for the gradient of a convex function

Question

It is easy to see that if a function $f: \mathbb{R} \to \mathbb{R}$ is strictly convex, $C^1$ and $f'$ has bounded image, then as $t\to \infty$ the limit $$ \lim_{t\to\infty} f'(t) = \lim_{t\to\infty} \frac{f(t)}{t} $$ exists and is finite.

Suppose instead that we have a function $F : \mathbb{R}^n \to \mathbb{R}$ that is strictly convex, $C^1$ and the image of $\nabla F$ is bounded. Then do we know that for each $v \in \mathbb{R}^n$ the following limit exists $$ \lim_{t\to\infty} \nabla F(vt) $$ (and belongs to $\mathbb{R}^n$)? If not, is there a reasonable assumption we can add to $F$ to guarantee that these limits exist?

We know that the curve $t \mapsto F(vt)$ is $C^1$ and strictly convex and so the limit of it's derivatives $$ \lim_{t\to\infty} \langle \nabla F(vt), v \rangle $$ exists in $\mathbb{R}$ but can we say anything about convergence for $\nabla F$ in general?

Thanks in advance for any help/suggestions.

Answer 1

Here is a counterexample to the convergence of $\nabla F$ in the conditions you give.

Let $f(x)=\sqrt{1+x^2}$, let $s(x)=\cos(\ln(\ln(x^2+10)))$ and let $\phi:\mathbb{R}\to[0,1]$ be some $C^\infty$ bump function with supp$(\phi)\subseteq[-1,1]$ and $\phi(0)=1,\phi'(0)=0$. Then, for a small enough constant $\varepsilon>0$, the function $g:\mathbb{R}^2\to\mathbb{R}$;

$$g(x,y)=f(x)+f(y+\varepsilon\phi(y)s(x))$$

has $\frac{\partial}{\partial y}g(x,0)=f'(\varepsilon s(x))$ (which does not converge when $x\to\infty$). But $g$ is convex, as we will see below by just computing its hessian. We just need to check that the Hessian is positive definite for $|y|\leq1$, as otherwise $g(x,y)=f(x)+f(y)$, which is strictly convex. As $|y|\leq1$, we have $y+\varepsilon\phi(y)s(x)\in[-2,2]$ if $\varepsilon\leq1$.

$$\frac{\partial}{\partial x}g(x,y)=f'(x)+f'(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s'(x). $$ $$\frac{\partial}{\partial y}g(x,y)=f'(y+\varepsilon\phi(y)s(x))(1+\varepsilon s(x)\phi'(y)).$$

Note that $\frac{\partial}{\partial x}g(x,y)$ and $\frac{\partial}{\partial y}g(x,y)$ are bounded.

$$\frac{\partial^2}{\partial x^2}g(x,y)=f''(x)+f''(y+\varepsilon\phi(y)s(x))(\varepsilon\phi(y)s'(x))^2+f'(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s''(x)$$ $$\frac{\partial^2}{\partial x\partial y}g(x,y)=f''(y+\varepsilon\phi(y)s(x))\varepsilon\phi(y)s'(x)(1+\varepsilon s(x)\phi'(y))+f'(y+\varepsilon\phi(y)s(x))\varepsilon s'(x)\phi'(y)$$ $$\frac{\partial^2}{\partial y^2}g(x,y)=f''(y+\varepsilon\phi(y)s(x))(1+\varepsilon s(x)\phi'(y))^2+f'(y+\varepsilon\phi(y)s(x))\varepsilon s(x)\phi''(y)$$

Now, for some constant $C$ we have $f''(x)=\frac{1}{(1+x^2)^{3/2}}>C|s''(x)|$ for all $x\in\mathbb{R}$. So by taking $\varepsilon$ to be small enough, we have $$\frac{\partial^2}{\partial x^2}g(x,y)>\frac{1}{2}f''(x)=\frac{1}{2(1+x^2)^{3/2}}.$$ Also, as $|y|\leq1$ implies $f''(y+\varepsilon\phi(y)s(x))>0.01$, taking $\varepsilon$ small enough we have $\frac{\partial^2}{\partial y^2}g(x,y)>0.001$.

Meanwhile, for some constant $K$, the term $\frac{\partial^2}{\partial x\partial y}g(x,y)$ is bounded in norm by $K\varepsilon s'(x)\leq\frac{2K\varepsilon}{\sqrt{x^2+10}}$. From these estimations it follows that the trace and determinant of the Hessian of $g$ are positive, so the Hessian of $g$ is positive definite, so $g$ is convex.