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I have to find the subgradients of the following function.

$$F(x) := \max \left\{0, \frac1{2}(x^2 - 1)\right\}$$

Analytically I can see subdifferentials at $x=-1$ is $\nabla f(-1) \in [-1 ,0] $ and at $x =1$ is $\nabla f(1) \in [0,1]$.

I am facing difficulties while obtaining these subdifferentials ($v$) using following inequality, $f(x) - f(\bar x) \ge \langle v,x-\bar x \rangle, x\in R$.

If I apply $\bar x = -1$ what should be my $f(x)$?

Similarly what should be the $f(x)$ at $\bar x =1$?

How can we obtain above-observed subdifferential using the definition of subdifferential? (above inequality)

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  • $\begingroup$ use \max, \langle and \rangle $\endgroup$
    – LinAlg
    Jul 30, 2018 at 15:25

1 Answer 1

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For $\bar{x}=-1$ you get: $$f(x) - f(-1) \geq \langle v,x-(-1) \rangle$$ $$f(x) \geq \langle v,x+1 \rangle$$ Clearly $v=0$ is a subdifferential, because $f(x) \geq 0$ for all $x \in \mathbb{R}$. But also $v=-1$ is a subdifferential because $f(x) \geq -x-1$ for all $x \in \mathbb{R}$ (just draw a plot of $F$ and of $g(x) = -x-1$).

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