The convex subbed of Hilbert space with no maximal norm I wanted to show that  if  $\{e_n \mid n\in \mathbb{N}\}$ is an orthonormal basis for Hilbert space $H$ and put 
$$C= \left\{x\in H \mid \sum_{n \in \mathbb{N}} \left(1 +\frac{1}{n} \right)^2 | \langle x | e_n \rangle |^2 \le 1 \right\}.$$
Show that $C$ is closed bounded,convex set but that it contains no vector with maximal norm

Notice that given an orthonormal basis $\{e_n,n\in\mathbb N^*\}$, the dot product is given by
$$
\langle x,y\rangle=\sum_{n\in\mathbb N^*}\langle x,e_n\rangle\overline{\langle y,e_n\rangle}.
$$
Define the bilinear function
$$
(x,y):=\sum_{n\in\mathbb N^*}(1+1/n)^2\langle x,e_n\rangle\overline{\langle y,e_n\rangle}.
$$
It is easy to check that $(.,.)$ is a dot product on $H$, and its associated norm $\Vert \Vert_*$ is equivalent to the norm $\Vert.\Vert$ associated to $\langle .,.\rangle$, since $1\leq(1+\frac{1}{n})^2\leq 4$ for any $n$.
Then, $C=\{x\in H,\Vert x\Vert_*\leq 1\}$ is the (closed) unit ball with respect to $\Vert.\Vert_*$, and as such is convex. The two norms being equivalent gives that $C$ is bounded and closed. 
We're now trying to find if $s:=\sup_{x\in C}\Vert x\Vert=\sup_{\Vert x\Vert_*\leq 1}\Vert x\Vert=\sup_{\Vert x\Vert_*= 1}\Vert x\Vert$ is reached. Once again, using the fact that $\Vert x\Vert\leq\Vert x\Vert_*$, we deduce that $s\leq 1$. Next, note that $\Vert e_n/(1+1/n)\Vert_*=1$ and $\Vert e_n/(1+1/n)\Vert\to 1$ as $n\to\infty$, so $s=1$.
Suppose that $s$ is reached: there exists $x\in H$ such that $\Vert x\Vert^2_*=\Vert x\Vert^2=1$, thus 
$$
\sum_{n\in\mathbb N^*}(1+1/n)^2|\langle x,e_n\rangle|^2=\sum_{n\in\mathbb N^*}|\langle x,e_n\rangle|^2,
$$
and necessarily $\langle x,e_n\rangle=0$ for all $n$, so $x\in H^\perp=\{0\}$; this is a contradiction.
