# 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?

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 $$\liminf_{|x|\to \infty} \frac{\langle \nabla f(x),x\rangle}{|x|} \geq \liminf_{|x|\to \infty} \frac{f(x) -f(0)}{|x|} >0.$$

• I'm not familiar with the concept of a subgradient. From Wikipedia, $\nabla f(x)$ is actually a set. So, how do we need to read $\liminf_{|x|\to \infty} \frac{\langle \nabla f(x),x\rangle}{|x|} >0$? Does it mean, for any choice $g(x)\in\nabla f(x)$, it holds $\liminf_{|x|\to \infty} \frac{\langle g(x),x\rangle}{|x|} >0$? And we clearly need that each $\nabla f(x)$ is nonempty. – 0xbadf00d Jan 27 '19 at 11:33
• @0xbadf00d : Perhaps you meant this comment rather as a comment to my answer, since the above answer does not contain the term "subgradient". Anyhow, I have now added notes to my answer in response to your comment. Please let me know if anything is still not clear enough. – Iosif Pinelis Jan 27 '19 at 15:26
• @IosifPinelis Yes, actually I've intended to comment your answer. However, what is $\nabla f(x)$ in nguyen0610's answer if it's not the "subgradient"? – 0xbadf00d Jan 27 '19 at 16:04
• @0xbadf00d : "However, what is ∇f(x) in nguyen0610's answer if it's not the "subgradient"?" -- I think this question should be addressed to nguyen0610. Is there anything in my answer that is still not quite clear to you? – Iosif Pinelis Jan 27 '19 at 16:09


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 $$$$u_n\cdot\nabla f(x_n)\to\ell \tag{1}$$$$ where $$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, $$$$\int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty,$$$$ which contradicts the first display in the OP. $$\Box$$

Notes added in response to a comment by the OP:

1. 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".

2. 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.

• Thank you very much for your notes. If $r>0$ is small, is it possible that $\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}=-\infty$? – 0xbadf00d Jan 27 '19 at 16:36
• @0xbadf00d : No, this is not possible, because (i) the $\liminf$ is $>0$, (ii) $\langle\nabla f(x),x\rangle\ge f(x)-f(0)$, and (iii) $f$ is continuous and hence locally bounded. If you have no further questions about my answer, then, I think, any further questions should be asked in another post. – Iosif Pinelis Jan 27 '19 at 17:05