Question:
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ a sub linear function and $ a \in \mathbb{R^n}$. We have to prove that: $\partial f(a)= \left \{ z \in \partial f(0) | z\cdot a = f(a) \right \}$
I am really stuck here as I ve tried different ways to prove it. By induction, prove of the type $A \subseteq B$ and $B \subseteq A$ to conclude $A=B$, but i didn't succeed in all of them.
Any details answer will be very helpful as I have no one to ask.
Thank
Thank a lot to @Theo Bendit
1/ Prove that: $\left \{ z \in \partial f(0) | z.a =f(a) \right \} \subseteq \partial f(a)$
By definition $\partial f(0) =\left \{ z \in \mathbb{R}^n | f(y) \geq f(0)+z\cdot (y-0), \forall y \in \mathbb{R}^n \right \}=\left \{ z \in \mathbb{R}^n | f(y) \geq z\cdot y , \forall y \in \mathbb{R}^n \right \} $. As by definition of a sub linear function we have: $f(0)=f(0\cdot x)=0f(x)=0$.
Now if $z \in \partial f(0)$ by definition we have that $f(y) \geq z\cdot y, \forall y \in \mathbb{R}^n$.
And on this set of $z \in \partial f(0)$ by assumption we only take the "$z$" s.t. $f(a)=z \cdot a$.
So we get the following inequation $f(y)-f(a) \geq z \cdot y - z \cdot a, \forall y \in \mathbb{R^n} $
But pay attention to the fact that by definition $\partial f(a) = \left \{ z \in \mathbb{R}^n | f(y) \geq f(a)+z \cdot (y-a), \forall y \in \mathbb{R}^n \right \}=\left \{ z \in \mathbb{R}^n | f(y) - f(a)\geq z \cdot y- z \cdot a, \forall y \in \mathbb{R}^n \right \}$ so in particular when $z \in \partial f(0) \Rightarrow \left \{ z \in \partial f(0) | z.a =f(a) \right \} \subseteq \partial f(a)$
2/ Prove that: $ \partial f(a) \subseteq \left \{ z \in \partial f(0) | z.a =f(a) \right \}$
We know that $z \in \partial f(a)$ iff $f(y)-f(a) \geq z \cdot (y-a) , \forall y \in\mathbb{R}^n $. If it is true $\forall y$ it is true in particular when $y=\lambda a, \lambda >0$. We get: $f(y)-f(a)=f(a)(\lambda -1) \geq z \cdot (y-a)=z \cdot a (\lambda -1)$
When $\lambda >1 \Rightarrow f(a) \geq z \cdot a$ and this is exactly the definition for $z$ in order to belong to $\partial f(0), \forall a$. And it is especially the case when $f(a) = z \cdot a$ and in such a case this $z \in \left \{ z \in \partial f(0) | z.a =f(a) \right \}$.
When $\lambda <1 \Rightarrow f(a) \leq z \cdot a$ and in the special case when $f(a)=z \cdot a$ we have that this $z \in \left \{ z \in \partial f(0) | z.a =f(a) \right \}$
When $\lambda =1 $ we get $f(0)=0$ and after it is trivial.
Q.E.D.