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

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – 0xbadf00d Jan 27 '19 at 11:33
  • $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Jan 27 '19 at 15:26
  • $\begingroup$ @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"? $\endgroup$ – 0xbadf00d Jan 27 '19 at 16:04
  • $\begingroup$ @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? $\endgroup$ – Iosif Pinelis Jan 27 '19 at 16:09

$\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:

  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.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for your notes. If $r>0$ is small, is it possible that $\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}=-\infty$? $\endgroup$ – 0xbadf00d Jan 27 '19 at 16:36
  • 1
    $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Jan 27 '19 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.