1
$\begingroup$

I am reading the book A short Course on Banach Space Theory by Carothers, and I am trying to solve exercise 2 in chapter 2.

Suppose $Y$ is a subspace of a Banach space $X$ and let $T\in B(Y, \mathbb{R}^n)$ the space of bounded linear operators from $Y$ to $\mathbb{R}^n$. I am asked to show that there exists an extension $\tilde{T} \in B(X, \mathbb{R}^n)$ such that the operator norm does not increase, i.e. $\|\tilde{T}\| = \|T\|$.

My initial idea was that if $e_1,\ldots, e_n$ is the usual basis for $\mathbb{R}^n$ and $e^i: \mathbb{R}^n \rightarrow \mathbb{R}$ by $e^i (e_1,\ldots, e_n)=e_i$ then $Tx = \sum_{i=1}^n e_i T_ix$, where $T_i= e^i\circ T$ is a linear functional.

Hahn Banach tells us that there exists a norm preserving extension $\tilde{T_i}$ of $T_i$, and we can therefore define $\tilde{T}= \sum_{i=1}^n e_i\tilde{T}_i \in B(X, \mathbb{R}^n)$, which will be an extension of $T$. As $\|T_i\| \leq \|T\|$ we can conclude that $\|\tilde{T}\| \leq \sum_{i=1}^n \|T_i\| = n\|T\|$.

However, it is not obvious to me how to go about the harder problem of making this extension norm preserving. Any hints to get me started would be appreciated!

$\endgroup$
1
$\begingroup$

If we use the max-norm on $\mathbb R^n$, then this is pretty easy. We need to compute the operator norm of $T$ in terms of the $T_i$'s. Let $x\in X$. Then $$ \|Tx\|_\infty = \max_i |T_i(x)| \le \max_i \|T_i\|_{Y^*}\cdot \|x\|_X. $$ Now let $i$ be such that $T_i$ is a functional with maximal norm, $\|T_i\| = \max_j\|T_j\|$. Let $\epsilon>0$ be such that $\|T_i\|> \|T_j\|+\epsilon$ for all $j$ such that $\|T_j\| < \|T_i\|$. Let $x$ with $\|x\|\le 1$ such that $T_i(x) \ge \|T_i\|-\epsilon$. Then $$ \|Tx\|_\infty = \max_j |T_j(x)| = T_i(x) \ge \|T_i\|-\epsilon = \max_j \|T_j\|_{Y^*} -\epsilon. $$ This proves $$ \|T\| = \max_j \|T_j\|_{Y^*}. $$ Now, $\tilde T_j$ is norm-preserving extension of $T_j$, building the extension $\tilde T$. Using the above identity, we see $\|T\|= \|\tilde T\|$.

This construction heavily uses the particular norm on $\mathbb R^n$. I do not see, how one can prove norm-equality for an arbitrary norm on $\mathbb R^n$.

$\endgroup$
  • $\begingroup$ Thank you. I guess I also should start thinking about a possible counter example as well then :-) $\endgroup$ – Mrtny Mar 1 '19 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.