# Find the subdifferential for $\max\left(x^2,|x|\right)$

We define the function $$f:\mathbb R \rightarrow \mathbb R$$ as the following: $$f\left(x\right)=\max\left(x^2,|x|\right)$$ Find the subdifferential $$\partial f\left(x\right)$$ for all $$x\in \mathbb R$$.

If we let $$S$$ be a subset of $$\mathbb R^n$$. A vector $$\xi\in\mathbb R^n$$ is called a subgradient for the function $$f:S\rightarrow\mathbb R$$ at $$x_0\in S$$ if $$f\left(x\right)\geq f\left(x_0\right)+\xi^t\left(x-x_0\right)$$ for every $$x\in S$$. The set of all subgradients of $$f$$ at $$x_0$$ is called the subdifferential.

I know that the derivitive doesn't exist at $$x_1=-1$$, $$x_2=0$$ and $$x_3=1$$.

By knowing this we would be able to solve the problem.

For $$x_2=0$$, we have that: $$|x| \geq |0|+\xi\left(x-0\right)\Leftrightarrow \xi\in \left[-1, 1\right]$$

For $$x_1=-1$$: $$x^2 \geq \left(-1\right)^2+2x\left(x-\left(-1\right)\right)$$ $$x^2 \geq 1+2x^2+2x$$ $$0 \geq 1+x^2+2x$$ $$0 \geq \left(1+x\right)^2$$ Since $$u^n \leq 0$$, if $$n$$ is even, then $$u=0$$, which gives us the following: $$0 = 1+x \Leftrightarrow x=-1$$ Which is the same for $$x_3=1$$, so we get that $$x=1$$. So the subdifferential is given as: $$\xi=\begin{cases} -1, & x = -1 \\ \left[-1,1\right], & x=0 \\ 1, & x=1 \end{cases}$$ I am not sure if this is the right way of doing it. I will appreciate it if I can get some help.

The subdifferential for $$0$$ is okay. For $$1$$ and $$-1$$ it is not. Consider $$x=1$$ ($$-1$$ ist just mirrored).
Then you need to solve $$1+\theta(x-1) \leq x^2$$ for $$x>1$$ and $$1+\theta(x-1) \leq x$$ for $$0\leq x < 1$$.
The first one is satisfied if $$\theta\leq \frac{\mathrm d}{\mathrm d x}x^2 |_{x=1} = 2$$. The second one is satisfied if $$\theta\geq \frac{\mathrm d}{\mathrm d x}x|_{x=1} = 1$$. So the subdifferential in $$x=1$$ is $$[1,2]$$.
• Note that since $f$ is even, $\partial f$ is odd, so I would say that the subdifferential at $-1$ is a mirrored (not reflected) version of same at $+1$. Commented Sep 25, 2021 at 18:18