A closed and convex subset of an Hilbert space on which the norm does not attain its maximum I want to prove that the norm $\lvert \lvert \cdot \rvert \rvert _2$ has no maximum on  $X=\{x\in \ell^2(k):\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2 \le 1\}$, where $k=\mathbb{R}$ or $\mathbb{C}$.
I have managed to see that this is close and bounded, I can't figure out if this is also convex and I don't know if this would be helpful.
I've tried to do some contradiction argument, I have observed is that since $2^{1/n}>1$, if there is $x \in X$ which attains the maximum for $\lvert \lvert \cdot \rvert \rvert _2$, then this maximum, say $M$, must be strictly smaller than $1$, otherwise $x$ would be outside of $X$. Now my idea would be that in this case we could add a small $\epsilon$ to some term $x_n$ of $x$ in such a way that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2$ stays below one, but I can't carry out the computations, even because if $\lvert \lvert x \rvert \rvert _2<1,$ then it does not follow that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2<1$, as we can see taking the element $(2^{-1/4}, 0, 0, \ldots)$.
 A: TOO LONG FOR A COMMENT - NOT AN ANSWER.
The set $X$ is indeed convex, and this is interesting; see the remark at the bottom of this post.
Proof of convexity: Let $$
Lx=(2^{\frac1{2n}}x_n)_{n\in \mathbb N}, $$ and consider $x_1, x_2\in X$, which by definition means that $\langle Lx_j|x_j\rangle \le 1$ for $j=1, 2$.
For $\alpha, \beta>0,\alpha+\beta=1$, let $$x' = \alpha x_1+\beta x_2.$$
Then  $$\langle Lx'|x'\rangle\le \alpha^2+\beta^2+2\alpha\beta \langle Lx_1|x_2\rangle.$$
Now since $L$ is a positive definite operator, we have the Cauchy-Schwarz inequality
$$
\lvert \langle Lx_1|x_2\rangle \rvert\le \sqrt{\langle Lx_1|x_1\rangle\langle Lx_2|x_2\rangle}\le 1.$$
We conclude that
$$\langle Lx'|x'\rangle\le (\alpha+\beta)^2=1,$$
that is, $x'\in X$.

Remark. As I mentioned in a comment to deb's answer, the fact that $X$ is convex is interesting. Indeed, it is a standard theorem that the norm attains its minimum on closed convex sets. Here we see that the norm can fail to attain its maximum.
A: Take $x_n=(y_k)_k$, where $y_k=0$, if $k\neq n$ and $y_n=2^{-1/2n}$. Then $x_n\in X$ and $\|x_n\|_2 \to 1$. So the sup of the norm over the set $X$ is one.
Now, if $x\in X$ and $\|x\|_2=1$, let $k$ be the smallest number such that $\sum_{i=1}^k|x_k|^2\geq 1/2$. Write $a:=\sum_{i=1}^k|x_k|^2$ and $b:=\sum_{i>k}|x_k|^2$. So $$\sum_{n\in\mathbb{N}}2^{1/n}|x_k|^2\geq 2^{1/k}a+b>a+b=1,$$ which is a contradiction. So the sup of the norm is unattained.
