Let $f:X\to \mathbb{R}\cup\{+\infty\}$ be a proper, convex and l.s.c. function on a Banach space $X$. The mapping $\partial f:X\to 2^{X^*}$ defined by $$\partial f(x)=\{x^*\in X^*: (x^*, v-x)\le f(v)-f(x)\,\,\mbox{for all}\, v\in X\}$$ is called the convex subdifferential of $f$. I was thinking: what if $f(x_0)=\infty$? Since $f$ is proper, then there exists at least one element $v\in X$ such that $f(v)\neq\infty$. It follows that $x^*=-\infty$ for all $x\in X$ if I consider $\partial f(x_{0})$. Does it have any sense? Natural question emerges: is always $\partial f\neq\emptyset$?


It might be easier to consider the real case $X = \mathbb R$ first. Then, if $f(x_0) = \infty$ and $f(v) < \infty$, you never have $$\mathbb R \ni (x^*, v - x_0) \le f(v) - f(x_0) = -\infty,$$ hence $\partial f(x_0) = \emptyset$.

Even if $f(x_0) \in \mathbb R$, the subdifferential might be empty, consider $f(x) = -\sqrt{1-x^2}$ for $x \in [-1,1]$ and $f(x) = +\infty$ otherwise, at the points $x = \pm 1$.

  • $\begingroup$ OP is asking : " is always $\partial f\neq\emptyset$? " the answer is positive for proper lsc , convex functions. $\endgroup$ – Red shoes May 23 '17 at 21:44
  • $\begingroup$ @nonlinearthought The counter-example works exactly the same way for $f(x) = -\sqrt{1-x^2}$, which is convex (or for the even simpler choice $f(x) = -\sqrt{x}$ for $x\geq 0$, and $f(x)=\infty$ else). $\endgroup$ – Christian Clason May 24 '17 at 13:23
  • $\begingroup$ @ChristianClason: Thank you for spotting the missing "$-$". $\endgroup$ – gerw May 24 '17 at 13:25
  • $\begingroup$ @ChristianClason What I see here is that, you and gerw couldn't understand the question yet !? $\partial f$ is a set-valued function. $\partial f = \emptyset$ means $\partial f(x) = \emptyset$ for ALL $x \in X$. $\partial f$ for the function $f(x) = -\sqrt{1-x^2}$ is not EMPTY set valued function since $\partial f(0) =\{0\} $. $\endgroup$ – Red shoes May 24 '17 at 18:55
  • $\begingroup$ @nonlinearthought That's indeed one possible interpretation, but not the only one (and, in my opinion, not the most reasonable one) -- it might also be a simple typo. Why not ask the OP to clarify his question if you think it is being misunderstood? $\endgroup$ – Christian Clason May 24 '17 at 19:02

Let assume $X$ be Banch.

Yes. always $\partial f\neq\emptyset$ !

Since epi$(f)$ is closed, nonempty, convex subset of $X \times R$. Therefore based bishop-phelps theorem supporting point on its boundary.

Note that This is a partial answer you must elaborate it.

  • $\begingroup$ Only on the interior of the effective domain (note that the OP considers $f:X\to \mathbb{R}\cup\{\infty\}$!), otherwise your claim is false as gerw's counter-examples show. $\endgroup$ – Christian Clason May 24 '17 at 11:48
  • $\begingroup$ Here's what's wrong with your argument: You can indeed put a supporting hyperplane at the boundary of $\mathrm{epi}(f)$ at a point $x \in \mathrm{dom}f\setminus (\mathrm{dom}f)^o)$, but it would be (considering $X=\mathbb{R}$ for visualization) vertical. That means the slope is $x^*=\infty$, which is not an element of $X^*$ ($=\mathbb{R}$) and hence not admissible as a subgradient. $\endgroup$ – Christian Clason May 24 '17 at 13:54

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