I'd like to prove a result but I do not know where to start. Any help or little idea is more than welcome (:
If I had to ask my proof issue in therm of an exercise question, It would look like that :
Let $x \in \mathbb{R}^2$, $f(x): \mathbb{R}^2 \to \mathbb{R}$ a convex function with $f(x^*)$ as minimum and $\chi := \{x:||x||_F = r \in \mathbb{R}\}$. If we know that $f(x^*)\leq 0$ and $\text{inf}\{f(x):x\in\chi\} > 0$, then, is it possible to prove that $||x^*||_F \leq r$ ?
Here $||x||_F$ is the Frobenius, i.e., the usual Euclidean, norm of $x$ in $\mathbb R^2$.
Thanks in advance,
Deramake
I would start by proving that the sublevel sets $C_t:=\{f\leq t\}$ for all real $t$ form a nested family of convex closed sets. What you're after is $C_{t^*}=:C_*$ where $t^*:=\inf_D f$ where $D$ is the domain of $f$. Now there are two cases $C_*$ is a single point, or not.
$C_*$ single point case (which you called $x_*$): in this case $x_*\in C_*\subseteq C_{t_\flat}=:C_\flat\subseteq B\subseteq C_{t_\sharp}=:C_\sharp$ where $t_\flat:=\int_S f$, $t_\sharp:=\sup_\chi f$ and $\chi=\partial B$ is the sphere or radius $r$, boundary of the ball of radius $r$, $B$. In other words $|x_*|\leq r$, but if $t_*<t_\flat$ then $C_*\subsetneq C_\flat$ and (because $C_*$ has only one point) then $x_*\in B\smallsetminus \chi$; so in fact you have more $|x_*|<r$.
$C_*$ is not a single point: it has to be convex and the same argument as above applies for any point $x_*$, but you need to be careful when the two sublevel sets are tangent: this cannot occur because it would make $f$ discontinuous, which is false because it's convex.
If the function is strictly convex, only the first case applies.