Extension of operator $T$ from a subspace $Y$ of $X$ to $\Bbb R^n$ without increasing the norm

Question

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!

Answer 1

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 $\mathbb R^n$.