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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:

  1. $A=\{x\in E: \|Tx\|\ge1\}$ is closed.

  2. There is no $x\in A$ with $$\inf_{y\in A} \|y\|=\|x\|.$$

  3. 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.

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up vote 1 down vote accepted

For 2:

As $\|T\|=1$, there exists a sequence $\{z_n\}$ with $\|z_n\|=1$ and $\|Tz_n\|\to1$. Putting $z_n'=z_n/\|Tz_n\|$ we get a sequence in $A$ (because $\|Tz_n'\|=1$) with $\|z_n\|\to1$. So $$ \inf\{\|y\|:\ y\in A\}=1. $$

But if $x\in A$, then $\|x\|>1$, so the desired infimum cannot exist.

For 3:

Take any $x\neq0$. Then there exists $n_0$ with $x_{n_0}\ne0$. $$ \|Tx\|^2=\sum_n|x_n|^2\,|y_n|^2=|x_{n_0}|^2|y_{n_0}|^2+\sum_{n\ne n_0}|x_n|^2\,|y_n|^2 <\sum_n|x_n|^2=\|x\|^2. $$ So $\|Tx\|<\|x\|$ for all nonzero $x$.

Finally, $\|T\|=1$. We can see this by considering the vector $x$ that has $1$ in the $n^{\rm th}$ coordinate and zero elsewhere. Such $x$ satisfies $\|x\|=1$ and $$ \|Tx\|=|y_n|. $$ As $y_n\to1$, this shows that $\|T\|\geq1$. And we already had $\|T\|\leq1$ in the previous paragraph, so $\|T\|=1$.

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