Assuming that $f$ is convex and continuous from the right at zero show that $∂f(0)$ is empty using the definition of the subdifferential.

Question

Suppose that $$ f(x)= \begin{cases} +\infty,\;\;\;\;\;\;\;if\; x<0\\ 0,\;\;\;\;\;\;\;\;\;\;if\;x=0\\ x\ln(x)\;\;\;\;if\;x>0 \end{cases} $$ Assuming that $f$ is convex and continuous from the right at zero show that $∂f(0)$ is empty using the definition of the subdifferential.

I am looking for help specifically with the case where $x>0.$ I know that the definition of subdifferential is given by $$v∈∂f(x ̄) ⇔ (∀x∈X)\;\;\;f(x ̄)+⟨v,x−x ̄⟩≤f(x).$$

So applying this to $∂f(0)$ for $x>0$ gives $$v∈∂f(0) ⇔ (∀x∈X)\;\;\;f(0)+⟨v,x−0⟩≤x\ln(x).$$ However, I am not sure ow this will help me. I am looking for suggestions on where to go with this proof next.

Answer 1

You have got this $f(0)+\langle v,x−0\rangle\le x\ln(x)$. Using $f(0)=0$ it means, in particular, that $$ \forall x>0\colon\ vx\le x\ln x\quad\Leftrightarrow\quad\forall x>0\colon\ v\le\ln x. $$ The last inequality is trivially impossible for any $v\in\mathbb{R}$ since $\ln x\to-\infty$ as $x\to 0^+$.