Let $d\in\mathbb N$ and $f:\mathbb R^d\to\mathbb R$ be convex with $$\int e^{-f(x)}\:{\rm d}x<\infty\tag1.$$ How can we show that $$\liminf\_{|x|\to\infty}\frac{x\cdot\nabla f(x)}{|x|}>0?$$ $f$ should only be differentiable outside a countable set. Why this is not an issue?



How can we show that if $f$ is convex, then $\liminf_{|x|\to\infty}\frac{x\cdot\nabla f(x)}{|x|}>0$?
We first show that $\liminf\_{|x|\to \infty} \frac{f(x)}{|x|} > 0$. Indeed, denote $ a = \max\{f(0)+1,1\}$, then the set $A =\{x: f(x) < a\}$ is a convex set and
$$|A| = \int\_A dx \leq \int\_A e^{-f(x) + a}dx \leq e^a \int\_{\mathbb R^n} e^{-f} dx < \infty.$$
The function $f$ is continue at $0$ and $f(0) < a$, then there exists $r >0$ such that $f(x) < a$ for all $x\in B\_r(0)$ with center at $0$ and radius $r$. For any $x \in A$, $x\not=0$, the convex hull of $x$ and the ball $B\_r(0)\cap x^{\perp}$, hence 
$$|A| \geq \frac {\omega\_{n-1}} n |x| r^{n-1},$$
where $\omega\_{n-1}$ is the volume of $n-1$ dimensional unit ball. Therefore $A$ is bounded. Let $R >0$ such that $A \subset B\_R(0)$. Hence $B\_R(0)^c \subset \{f(x) \geq a\}$. For any $x$ with $|x| > R$, we have $\frac R{|x|} x = \frac R{|x|} x + (1 -\frac R{|x|}) 0$ and by using the convexity,we get
$$a \leq f\left(\frac R{|x|} x\right) \leq \frac R{|x|} f(x) + \left(1-\frac R{|x|} \right) f(0).$$
From this we have 
$$\liminf\_{|x|\to \infty} \frac{f(x)}{|x|} \geq \frac{a}{R} >0.$$
Again by the convexity, we have $f(0) \geq f(x) + \langle \nabla f(x), 0-x\rangle$ which implies
$$\liminf\_{|x|\to \infty} \frac{\langle \nabla f(x),x\rangle}{|x|} \geq \liminf\_{|x|\to \infty} \frac{f(x) -f(0)}{|x|} >0.$$


