Consider a special type of convex function $g(\cdot):\mathbb{R}^d \to \mathbb{R}_+\cup\{+\infty\}$ such that $g(x)=+\infty$ as $x\to \infty$. Then $g$ is differentiable almost everywhere within its domain. Now consider the quantity $$\frac{\nabla g(x)}{(g(x))^{1+\delta}},$$ where $\delta>0$. Can we get some estimate of type $$\nabla g(x)_2\leq c_g(1+x)^{\gamma} ((g(x))^{1+\delta}$$ for some $\gamma>0$ and all $x>R$, where $c_g$ may depend on some 'local' information of $g$ but not asymptotically as $x \to \infty$, and $R$ is independent of $g$? Or is there a counterexample?

1$\begingroup$ You can find a counterexample for $d=1$ among functions which are linear on each interval $[n,n+1]$. $\endgroup$– Anton PetruninApr 12, 2015 at 6:15

$\begingroup$ @AntonPetrunin For piecewise function the left hand side is always $0$, where is the contradiction? $\endgroup$– Roy HanApr 12, 2015 at 6:41

1$\begingroup$ $g$ is piecewise linear (not piecewise constant). $\endgroup$– Anton PetruninApr 12, 2015 at 17:50
1 Answer
Let me explain first what Anton wrote: we construct piecewise linear function $f$ by on $[0,n]$ by induction in $n$. Suppose that it is already constructed on $[0,n]$. Define the right derivative $D^+f(n)$ so that your inequality is violated. Then extend $f$ linearly on $[n,n+1]$. Then smoothen, if you need a smooth function.
However, inequality $f'\leq f^{1+\epsilon}$ holds on ``most'' of the positive ray: the set where it is violated has finite measure. Indeed, assuming that $f$ is increasing, let $E$ be the set where $f'>f^{1+\epsilon}$, and $a>0$ is the leftmost point of $E$. Then $$E\leq\int_E\frac{f'}{f^{1+\epsilon}}dx=\int_{f(E)}\frac{dw}{w^{1+\epsilon}}<\int_{f(a)}\frac{dw}{w^{1+\epsilon}}<\infty.$$ I do not know what is the appropriate extension of this argument to higher dimensions, but certainly some statements of this sort can be made.