Is convex subdifferential always non-empty? 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 $f(v) &lt; \infty$, 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 $x = \pm 1$.
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.