The support function of a set $A \in \mathbb{R}^n$ is defined as the following
$$ S_A=\sup_{y \in A} x^Ty $$ where $x \in \mathbb{R}^n$.
Let $A=B_1=\|x\|_2<1$.
Show that the support function of the Euclidean unit ball is $|x|$.
I tried the following by defining the Lagrangian function
$$ L(y,\lambda)=x^Ty+\lambda(\|y\|_2^2-1) $$ Because $\|y\|_2^2 <1$ and $\|y\|_2 <1$ are the same.
I do not know how to proceed?
On
Let $A=B_1$. We can assume that $x \ne 0$.
For $y \in A$ we have , by Cauchy-Schwarz: $x^Ty \le ||x||_2 ||y||_2 \le ||x||_2$, hence
$(1) \quad S_A \le ||x||_2$.
Now let $t>1$ and put $y_t:= \frac{1}{t||x||_2}x$, then $y_t \in A$ and $x^Ty_t=\frac{1}{t}||x||_2 \le S_A$ With $t \to1$ we get
$(2) \quad ||x||_2 \le S_A$.
The result follows now from $(1)$ and $(2)$.
If $x=0$, the solution is obvious. Suppose $x \ne 0$,
$$ L(y,\lambda)=x^Ty+\lambda(\|y\|_2^2-1) $$
We want to solve for $y$, hence differentiating with respect to $y$,
$$x+2\lambda y = 0$$
If $\lambda =0$, then $x=0$. Hence, $\lambda \ne 0$.
$$y = -\frac{1}{2\lambda}x$$
That is we know that $y$ must be parallel to $x$, let's write $y = \frac{kx}{\|x\|}$, $|k|\le 1$.
The problem is now $$\sup_{|k| \le 1} kx^T\frac{x}{\|x\|}=\sup_{|k|\le 1}k\|x\|=\|x\|.$$
Remark:
Cauchy-Schwarz seems to be the natural approach to me.