He had studied functional analysis in the book "Lectures on Functional Analysis and the Lebesgue Integral" by V. Komorkik. On page 9 there is an example of properties in finite-dimensional spaces that do not hold in infinite-dimensional spaces.
For example, the diameter of a bounded and closed set is finite, but maximizers do not exist.
How can I calculate the diameter of $A$ given by:
\begin{equation*}A:=\left\{x\in\ell^{2} : \sum_{n\in\mathbb{N}} \left(1+\frac{1}{n}\right)^{2}|x_{n}|^{2} \leq 1\right\}.\end{equation*}
My idea is, given $x\in A$, we note that $-x\in A$ (since the constraint defines a set symmetric about the origin). So it turns out that\begin{align*} \operatorname{diam}(A) &:= \sup _{x,y\in A} \|x-y\|_{2} \\&= \sup _{x\in A} \|x-(-x)\|_{2} \\&= 2\sup _{x\in A}\|x\|_{2} \\&= 2\sup _{x\in A} \left[ \sum _{n\in\mathbb{N}} \left(1+\frac{1}{n}\right)^{2}x_{n}^{2 } \right]^{1/2} = 2. \end{align*}However, we note that there does not exist $x\in A$ such that its norm allows reaching the supreme, that is, given that for all $x\in A$:\begin{eqnarray*} \|x\|_{2} < \left[ \sum_{n\in\mathbb{N}} \left(1+\frac{1}{n}\right)^{2}x_{n}^{2} \right]^{1/2} \leq 1, \end{eqnarray*}Therefore the supreme one that is studied in the calculation of the diameter is not reached. But, I'm not sure of the reasoning.
Regards!