Prove that $(\mathbb{R}^n, ||.||_2)$ is strictly convex. Prove that $(\mathbb{R}^n, \|\cdot\|_2)$ is strictly convex.
I want to use the subsequentialized negation, but I don´t see the contradiction.
The definition of a norm strictly convex:
$\forall x,y\in X$ such that $x\neq y$ and $\|x\|=1=\|y\|$, we got that $\left\|\frac{x+y}{2}\right\|<1$
It´s negation:
$\exists  x,y\in X$ such that $x\neq y$ and $\|x\|=1=\|y\|$, we got that $\|\frac{x+y}{2}\|\geq1$
Now the subsucsequentialized negation:
$\exists  (x_n),(y)_n\in X$ and $n\in \mathbb{N}$ such that $(x_n)\neq (y_n)$ and $\|x\|=1=\|y\|$, we got that $\left\|\frac{(x_n)+(y_n)}{2}\right\|\geq1$
Any suggestions would be great!
 A: Suposse that $$\|x+y\|_2 =2$$ and $\|x\|_2 =1 , \|y\|_2 =1 , x\neq y.$ Then $$4=\|x+y\|^2_2 $$
and from the pararellogram identity we get $$4+\|x-y\|^2_2 =2 +2$$ therefore $$x=y$$ and we get that the norm $\|\cdot \|_2$ is strictly convex on $\mathbb{R}^n .$
A: For all $\;x=(x_1,x_2,\ldots,x_n)\in\mathbb{R}^n,$ $\;y=(y_1,y_2,\ldots,y_n)\in\mathbb{R}^n\;$ such that $\;x\ne y\;$ and $\;\Vert x\Vert_2=\Vert y\Vert_2=1\;,$ it results that
$\left(\dfrac{x_i+y_i}{2}\right)^2\le\dfrac{x_i^2+y_i^2}{2}\;,\;$ for any $\;i\in\big\{1,2,\ldots,n\big\}\;.$
Since $\;x\ne y\;,\;$ there exists $\;i^*\in\big\{1,2,\ldots,n\big\}\;$ such that
$x_{i^*}\ne y_{i^*}\;$ and $\;\left(\dfrac{x_{i^*}+y_{i^*}}{2}\right)^2<\dfrac{x_{i^*}^2+y_{i^*}^2}{2}\;,\;$ hence
$\bigg\Vert\dfrac{x+y}{2}\bigg\Vert_2=\sqrt{\sum\limits_{i=1}^n\left(\dfrac{x_i+y_i}{2}\right)^2}<\sqrt{\sum\limits_{i=1}^n\left(\dfrac{x_i^2+y_i^2}{2}\right)}=$
$=\sqrt{\dfrac{1}{2}\left(\sum\limits_{i=1}^n x_i^2+\sum\limits_{i=1}^n y_i^2\right)}=\sqrt{\dfrac{1}{2}\bigg(\Vert x\Vert_2^2+\Vert y\Vert_2^2\bigg)}=1\;.$
A: Here is a nifty trick combining the polarization identity with Cauchy-Schartz.  You have
$$ \langle x, y \rangle = {1\over 4}(\|x + y\|^2
 + \|x - y\|^2)$$
for any $x, y  \in H$.  Now suppose that $\|x\| = \|y\| = 1$ and that $ \|(x + y)/2\| = 1.$
Then
$$   {1\over 4}(\|x + y\|^2
 + \|x - y\|^2) = \langle x, y\rangle \le \|x\| \|y\| = 1.$$
Since $ \|(x + y)/2\| = 1,$, you have $\|(x - y)/2\| = 0,$
so $x = y$.
