Let $E$ be a normed space and $T \in L(E)$ with $\|Tx\|\lt\|x\|$ for all $x\ne0$ and $\|T\|=1$.
I want to prove the following:
$A=\{x\in E: \|Tx\|\ge1\}$ is closed.
There is no $x\in A$ with $$\inf_{y\in A} \|y\|=\|x\|.$$
Let $(y_n)_{n \in \mathbb{N}} \subset \mathbb{R}^+$ be convergent to $1$ from below. Then $$T: \ell_2 \rightarrow \ell_2, \\ (x_n)_{n\in \mathbb{N}} \mapsto (x_n \cdot y_n)_{n\in \mathbb{N}}$$ has the properties stated above.
I was able to show 1., but I'm stuck with 2. and 3.
We have $$1=\|T\|=\sup_{x\in E,\ x \ne 0} \frac{\|Tx\|}{\|x\|}$$ and $$1 \le \|Tx\| \lt \|x\|$$ for $x \in A$.
But I don't see how to derive 2. from that - towards a contradiction, I suppose.
For 3., I want to show that $$1=\|T\|=\sup_{\|(x_n)\|=1}\|T(x_n)\|=\sup_{\|(x_n)\|=1}\|(y_n\cdot x_n)\|=\sup_{\|(x_n)\|=1}\left(\sum |y_n\cdot x_n|^2\right)^{1/2}.$$ Why does this hold?
I also don't see why we have $\|(y_n\cdot x_n)\| \lt \|(x_n)\|$.
Can anyone help me to understand this?
[edit] I forgot to mention that in c), $(y_n)$ is a sequence of positive real numbers.