Can L-smooth (L>0) convex function to be non-differentiable?

Question

As we know, a function $f:\mathbb{R}^n\to \mathbb{R}$ is called L-smooth (with a finite $L>0$), if $x\mapsto \frac{L}{2}\|x\|^2 - f(x)$ is convex. This definition does not restrict $f$ to be differentiable. I would like to know: "is there any $\textbf{convex}$ L-smooth function which is not differentiable?" Otherwise, how to prove that any L-smooth convex function is differentiable? Many thanks for any comment.

Answer 1

I think the answer is: any convex L-smooth function $f$ is differentiable. A possible proof is as follows: we try to show that the subdifferential $\partial f(x)$ is always a singleton at any point $x\in \mathbb{R}^n$.

First, the non-emptyness of $\partial f(x)$ is known.

Second, by contradiction, suppose that there exists $x\in \mathbb{R}^n$ such that $\partial f(x)$ contains two different subgradients $v_1$ and $v_2$. Then by the convexity of $f$, we have
$$\forall h\in \mathbb{R}^n,\quad f(x+h)\geq f(x) + \langle v_1, h \rangle.\qquad (1)$$ And by the convexity of $g(x)=\frac{L}{2}\|x\|^2 - f(x)$, we have $$\forall h\in \mathbb{R}^n,\quad g(x+h)\geq g(x) + \langle z, h \rangle.\qquad (2)$$ where $z\in \partial g(x)$. Using the fact $$\partial g(x) = Lx - \partial f(x).$$ Then we take $z = Lx - v_2$ and get from $(2)$ that $$\forall h\in \mathbb{R}^n,\quad g(x+h)\geq g(x) + \langle Lx-v_2, h \rangle.$$ That is (by replacing the expression of $g$): $$\forall h\in \mathbb{R}^n,\quad \frac{L}{2}\|x+h\|^2 - f(x+h)\geq \frac{L}{2}\|x\|^2 - f(x) + \langle Lx-v_2, h \rangle.\qquad (3)$$ Summing $(1)$ and $(3)$. Then $$\forall h\in \mathbb{R}^n,\quad \langle v_1-v_2, h \rangle \leq \frac{L}{2}\|x+h\|^2 - \frac{L}{2}\|x\|^2 -\langle Lx, h \rangle = \frac{L}{2}\|h\|^2.$$ Taking $h=t(v_1-v_2)$ with $t\geq 0$. Then $$t \|v_1-v_2\|^2 \leq \frac{L}{2}t^2\|v_1-v_2\|^2.$$ That is $$\forall t\geq 0,\quad (\frac{L}{2}t^2-t)\|v_1-v_2\|^2\geq 0.$$ Consider the case where $0<t<\frac{2}{L}$, then $\frac{L}{2}t^2-t<0$ and we get $$\|v_1-v_2\|^2\leq 0.$$ This implies (by the definiteness of the norm) that $$v_1=v_2,$$ which contradicts the assumption $v_1\neq v_2$. Therefore, the subdifferential $\partial f(x)$ is a singleton at any point $x\in \mathbb{R}^n$, implying that $f$ is differentiable over $\mathbb{R}^n$.