How to find the subgradients of $f(x)=\|x\|^2-1$ if $\|x\|\geq 1$, $f(x)=0$ if $\|x\|\leq 1$, $x\in\mathbb{R}^2$.

Asked Sep 26 '15 at 18:44

Active Sep 26 '15 at 18:53

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By definition a subgradient $a$ must satisfy $f(x+y)\geq f(x)+a\cdot y$

I just have problems in the case $\|x\|=1$, when $\|x+y\|\leq 1$.

In this case $a$ must satisfy

$0\geq a\cdot y$, for all $y$ such that $\|x+y||\ \leq 1$ or $\|x+y||\ ^2 \leq 1$.

$0\geq a\cdot y\Rightarrow0\geq \|a\|\|y\|cos\theta_{ay}\Rightarrow0\geq cos\theta_{ay}\space\forall y$ such that $\|x+y||\ \leq 1$

I drew that condition and then the solution is $a=\lambda x,\lambda \geq 0$

But I dont know how to justify it.

edited Sep 26 '15 at 18:53

asked Sep 26 '15 at 18:44

Guadalupe

I guess that $f(x)=0$ if $|x|\leq 1$? // Can you solve the analogous question in dimension $1$, that is, find the subgradients of $g(t)=t^2-1$ if $|t|>1$, $g(t)=0$ if $|t|\le1$? – Did Sep 26 '15 at 18:49
I already did it, and it is exactly $a\leq 0$, in this case, if $t=-1$, $a\geq 0$ if $t=1$ – Guadalupe Sep 26 '15 at 18:56
It is easier since there is not $cos\theta$ – Guadalupe Sep 26 '15 at 18:59

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