We are given a function $f: \mathbb{R}^n \to \mathbb{R}$. We know that $f$ is convex and continuously differentiable on $\mathbb R^n \setminus\{0\}$. Also $f$ is homogeneous of degree $1$, which means that $\forall t \geq 0$, $f(tx)=tf(x).$
We are given $2$ spaces: $S$ is unit the unit sphere of $\mathbb{R}^n$ and $\Sigma$ that is the image of $S$ under the gradient of $f$. So $$ \Sigma =\{ u \in \mathbb{R}^n \mid \exists x \in S, u = \nabla f(x) \}\, . $$
What I want to prove is that $$ \forall x \in \mathbb{R}^n, f(x) = \max \{ u \cdot x \mid u \in \Sigma \} \, . $$
To do so I tried to use one the properties of the function $f$: $\forall (x,x') \in \mathbb{R}^n \times \mathbb{R}^n $ is $f(x') \geq \nabla f(x) \cdot x' \implies f(x) \geq f(x) \cdot x$,$ \forall x $.
Which means that $\forall x$, $f(x) = \max\{ \nabla f(x) \cdot x \}$.
That's what I found on my own but there was a hint given in the exercise : We can consider the point $ \epsilon = \frac{1}{\Vert x \Vert }x $ when $ x \neq 0$, and then we can show that the equality is true when $ x = 0 $. I don't understand how to use this hint.
Please tell me if the demonstration I did is correct, and maybe tell me how to use the hint to resolve the exercise.
We want to show that for all $x \in \Bbb R^n$ $$ \tag{*} f(x) = \max \{ \nabla f(y) \cdot x \mid y \in S \} \, . $$
Your solution is a bit unclear. You probably meant $f(x) \ge \nabla f(x) \cdot x$ instead of $f(x) \ge f(x) \cdot x$. That would be correct, but does not imply $(*)$.
$f(x) = \max\{ \nabla f(x) \cdot x \}$ is unclear as well: Is the $x$ on the right-hand side the same as the $x$ on the left-hand side?
I would proceed as follows:
As a homogeneous function, $f$ satisfies $f(0) = 0$, so that both sides of $(*)$ are zero for $x=0$.
From now on, we consider a fixed $x \in \Bbb R^n$, $x \ne 0$. In Prove that $f(x') \geq \nabla f(x) \cdot x' $ for $x \in \mathbb{R}^n$ and $ x' \in B(x,r) \subset \mathbb{R}^n \setminus \{0\}$ it was demonstrated that for all $y \in \Bbb R^n$ with $y \ne 0$ $$ \tag{**} f(x) \ge \nabla f(y) \cdot x \, , $$ with equality for $y=x$, which implies $$ f(x) \ge \max \{ \nabla f(y) \cdot x \mid y \in S \} \, . $$ It remains to show that equality holds in $(**)$ for some $y \in S$. Here we use the hint and set $y = x/\Vert x \Vert$. Then $\Vert y \Vert = 1$ and $$ f(x) = \nabla f(x) \cdot x = \nabla f(y) \cdot x \, . $$