TITLE: a subset of $l_2$.
QUESTION [6 upvotes]: the exercise is: show that the set $A=\{x=(x_n)\subset l_2: \sum(1+\frac{1}{i})x_i^2)\leq 1\}$ doesn't contain an element with norm equal to $\sup\{\|x\|_2,x\in A\}$.
My attempt: i showed that $\forall x\in A, \|x\|_2^2=\sum x_i^2< \sum x_i^2+\frac{x_i^2}{i}\leq 1$. And $\sup_{x\in A} \|x\|=1$. Using that $l_2$ is a Hilbert space, is there a way to show that, supposing that exists an element that there is norm equal to $\sup\{\|x\|_2,x\in A\}$, find some contradiction?  This exercise i found at functional analysis and infinite-dimensional geometry- Marián Fabian.

REPLY [5 votes]: Consider $x\in A$. Since for all $i\ge 1$ we have $x_i^2\le (i+1/i)x_i^2$ we conclude by adding these inequalities that $\Vert x\Vert_2^2\le 1$, so
$$\sup_{x\in A}\Vert x\Vert_2\le 1.$$
Now, if we define $x^{(k)}\in\ell_2$ by $x^{(k)}_k=1/\sqrt{1+1/k}$ and  $x^{(k)}_i=0$ for $i\ne k$, then $x^{(k)}\in A$ and
$$\Vert x^{(k)}\Vert_2=\frac{1}{\sqrt{1+1/k}}\le \sup_{x\in A}\Vert x\Vert_2\le 1$$
But since $k$ is arbitrary we conclude by letting $k$ tend to $\infty$ that
$$ \sup_{x\in A}\Vert x\Vert_2=1.$$
Now, suppose that there exists $y\in A$ such that $\Vert y\Vert_2=1$ then
$$\sum_{i\ge1}\frac1i y_i^2=
\sum_{i\ge1}\left(1+\frac1i \right)y_i^2
-\sum_{i\ge1}y_i^2\le1-\Vert y\Vert_2^2=0$$
Thus
$$\sum_{i\ge1}\frac1i y_i^2=0$$
So $y_i=0$ for all $i$ and this contradicts the fact that $\Vert y\Vert_2=1$. This contradiction proves that there is no $y\in A$ such that $\Vert y\Vert_2=1$. $\qquad\square$