How can we show that if $f$ is convex, then $\liminf_{|x|\to\infty}\frac{x\cdot\nabla f(x)}{|x|}>0$? 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?
 A: $\newcommand{\R}{\mathbb{R}}
\newcommand{\ep}{\varepsilon}
$
For any $x\in\R^d$, let us understand $\nabla f(x)$ as any one of the subgradients of the convex function $f$ at point $x$ (so, we will not have to worry about differentiability of $f$). 
To obtain a contradiction, suppose that the desired conclusion does not hold. Then for some sequence $(x_n)$ in $\R^d$ such that $t_n:=|x_n|\to\infty$ and for some $\ell\in[-\infty,0]$ we have 
\begin{equation}
u_n\cdot\nabla f(x_n)\to\ell \tag{1} 
\end{equation}
where $u_n:=x_n/t_n$. By the compactness of the unit sphere $S^{d-1}$ in $\R^d$, without loss of generality (wlog) $u_n\to u$ for some $u\in S^{d-1}$. 
For any $w\in\R^d$, consider the convex function $g_w\colon\R\to\R$ defined by the condition that $g_w(t)=f(tw)$ for all real $t$. Let $g'_w$ denote the left derivative of $g_w$. Then $g'_{u_n}(t_n)\le u_n\cdot\nabla f(x_n)$. So, in view of (1), wlog $g'_{u_n}(t_n)\to m$ for some $m\in[-\infty,\ell]\subseteq[-\infty,0]$. 
So, for any $t\in[0,\infty)$ and any real $\ep>0$ there is some natural $n_{t,\ep}$ such that for all natural $n\ge n_{t,\ep}$ we have $t_n\ge t$ and $g'_{u_n}(t_n)\le\ep$, so that (by the convexity of $g_{u_n}$) $g'_{u_n}\le\ep$ on $[0,t_n]$ and hence on $[0,t]$, which yields $g_{u_n}(t)\le g_{u_n}(0)+\ep t$, that is, $f(tu_n)\le f(0)+\ep t$. Since $f$ is convex and real-valued, it is continuous. Therefore, $f(tu)\le f(0)+\ep t$, for all $t\in[0,\infty)$ and all real $\ep>0$. Thus, $f(tu)\le f(0)$ for all $t\in[0,\infty)$; that is, $f\le f(0)$ on $R_u:=\{tu\colon t\in[0,\infty)\}$. 
Also, by the mentioned continuity of $f$, for some real $b$ we have $f\le b$ on $B:=\{x\in\R^d\colon|x|\le1\}$. So, $f\le \max[f(0),b]=b<\infty$ on the convex hull (say $C$) of $R_u\cup B$. The Lebesgue measure $|C|$ of $C$ is $\infty$. So, 
\begin{equation}
 \int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty, 
\end{equation}
which contradicts the first display in the OP. $\Box$
Notes added in response to a comment by the OP:


*

*The OP commented: "I'm not familiar with the concept of a subgradient. From Wikipedia, $\nabla f(x)$ is actually a set." Response: $\nabla f(x)$ was the notation you used, incorrectly. To make sense of such usage, I suggested: "let us understand $\nabla f(x)$ as any one of the subgradients". 

*The OP commented: "And we clearly need that each $\nabla f(x)$ is nonempty." Response: It is a well known fact that, for any real-valued convex function $f$ on $\R^d$ and any $x\in\R^d$, the subdiferential of $f$ at $x$ (which is the set of all subgradients of $f$ at $x$) is nonempty. See e.g. Theorem 23.4 or Proposition 2. Also, any real-valued convex function $f$ on $\R^d$ is continuous (and even locally Lipschitz continuous) -- see e.g. Theorem 10.1 or Theorem 3.3.1.  
A: 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
$$\liminf_{|x|\to \infty} \frac{\langle \nabla f(x),x\rangle}{|x|} \geq \liminf_{|x|\to \infty} \frac{f(x) -f(0)}{|x|} >0.$$
