Let $(\ell_2,\|\cdot\|)$ be the usual Banach space of the real sequences with summable squared, and $B$ its closed unit ball. Define $T:B\longrightarrow B$ as
$$ T(x):=(\sqrt{1-\|x\|^{2}},x_{1},x_{2},\ldots), $$ for all $x:=(x_{n})_{n\geq 1}\in B$. Then, it is easy to check that $T$ has not fixed points. However, I have read in some papers that the following holds: $$ \inf\{\|x-T(x)\|:x\in B\}=0 $$
How can we prove this equality?
Many thanks in advance for your comments.
I will denote the $m$th point in a sequence $x \in \ell^2$ by $x(m)$.
For each $m, n \in \Bbb{N}$, let $$x_n(m) = \begin{cases} \dfrac{1}{\sqrt{n}} & \text{if } 1 \le m \le n \\ 0 & \text{if } m > n. \end{cases}$$ Then $\|x_n\| = 1$, hence $x_n \in B$. Further, $$(T(x_n) - x_n)(m) = \begin{cases} -\dfrac{1}{\sqrt{n}} & \text{if } m = 1 \\ \dfrac{1}{\sqrt{n}} & \text{if } m = n + 1 \\ 0 & \text{otherwise.} \end{cases}$$ Hence, $$\|T(x_n) - x_n\| = \frac{\sqrt{2}}{\sqrt{n}} \to 0.$$