Where $h_A$ is the support function of $A$ and $j_{A^\circ}$ the Minkowski functional of the polar set of $A$
There is a "proof" in my course which I don't understand:
" Let $x \in A$ and $t>0$ such as $y \in t A^\circ $. Then $y/t \in A^\circ$ therefore $\langle x,y \rangle \leq t$. This proves $ h_A(y) \leq j_{A^\circ}(y) $ "
I don't understand why that implication is true.
My reasoning :
$$ \langle x,y \rangle \leq t \implies \sup_x \langle x,y \rangle \leq \sup_x t \leq t $$
Therefore
$$ h_A(y) \leq t $$
Moreover:
$$ y \in t A^\circ \implies j_{A^\circ}(y) = \inf_s \{s , y \in s A^\circ \} \leq t $$
Which obviously doesn't help the slightest as both inequalities are going the same way. I think I'm missing something obvious or made a simple mistake but it is driving me crazy that I can't prove something apparently so trivial.
The second part of the proof is easy and I understand it but I still put it here :
" Reciprocally, because $0 \in A$, we have $h_A (y) \geq 0$. For all $t > h_A(y)$ we have $y \in t A^\circ $ so $j_{A^\circ}(y) \leq t$. By taking the limit $t \to h_A$ we have the desired result "
So as I originally imagined it is indeed really "trivial" and I was just a tad stupid.
$$ \langle x,y \rangle \leq t \implies \sup_x \langle x,y \rangle \leq t \implies \sup_x \langle x,y \rangle \leq \inf_{t, \, y \in t A^\circ } t \implies h_A(y) \leq j_{A^\circ}(y) $$
It's a little embarassing