Find a subgradient of a real function

Question

Let $f(x)=\chi_{\{-1\}}(x)+I_{[-1,1]}(x)$, where $\chi_{A}(x)=1$ when $x\in A$ and $\chi_{A}(x)=0$ when $x\notin A$, and $I$ is the indicator funcion (which is zero in its domain and infinity otherwise). I want to calculate the subgradient of $f$ at $x=-1$. So, $$x^*\in \partial f(-1) \Leftrightarrow x^*(x+1)\le f(x)-f(-1), \qquad \mbox{for all }x\in \mathbb{R}.$$ The last expression is equivalent to $$x^*\in \partial f(-1) \Leftrightarrow x^*(x+1)\le f(x)-1, \qquad \mbox{for all }x\in \mathbb{R},$$ by the definition of $f$. How can I get this subgradient? It is my intuition that it must be the empty set, but I am not sure how to prove it. Please, any help will be appreciated

Answer 1

Pick $x=-1+1/n$, $n\ge 2$ in the inequality defining the sub gradients. $$ \frac{1}{n}x^*\le-1 $$ and, therefore, $x^*\le -n$ for all natural $n$. No real number $x^*$ satisfies this condition.