Extension of operator $T$ from a subspace $Y$ of $X$ to $\mathbb{R^n}$ (incomplete answer)

Question

I came across the same question in Carothers' A Short Course on Banach Space Theory as another poster (Extension of operator $T$ from a subspace $Y$ of $X$ to $\Bbb R^n$ without increasing the norm) and was stumped.

The question is as follows:

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. $\|\tilde{T}\| = \|T\|$.

But I do not find the posted solution to be satisfactory - the accepted answer proves that we can proceed if we use the supremum norm. This is all well and good, but can we solve the problem for any norm on $\mathbb{R^n}$? The phrasing of the question seems to indicate that the answer is "yes", but I am not so sure. As the commenter for the accepted solution points out, it's not clear that this will hold for any other norm on $\mathbb{R^n}$.

Further, looking at related posts, such as this one (Bounding the norm of an extension of a linear function from a subspace of a normed space to a finite dimensional normed space) there is a comment about extending bounded linear operators "norm preservingly" (only?) when taking the supremum norm.

In another related post (Extension of linear operator), the accepted response states without proof that if the finite-dimensional space of interest is over $\mathbb{C}$ then the result does not hold for any norms other than the supremum norm.

However, I have not been able to come up with a counter example myself. Any further insight is appreciated.

Answer 1

In order not to emphasize any norm on ${\mathbb R}^n$, let us replace it by a given finite dimensional normed space $Z$, so the question becomes:

Question 1. Let $X$, $Y$ and $Z$ be normed spaces, with $Y\subseteq X$, and $\text{dim}(Z)<\infty $. Given a bounded linear transformation $T:Y\to Z$, can one find an extension $\tilde T:X\to Z$, with $\|\tilde T\| = \|T\|$?

In the special case in which $Z=Y$, and $T$ is the identity operator, any extension of $T$ to $X$ is effectively a projection from $X$ to $Y$. So, should the above question have an affirmative answer, the follow would also be answered affirmatively:

Question 2. Let $X$ be a normed space and let $Y\subseteq X$ be a finite dimensional subspace. Does there exist a projection from $X$ to $Y$ with norm one?

Unfortunately the answer to (2) is negative, and hence the same goes for (1).

A counter example is $X={\mathbb R}^3$, equipped with the sup norm, namely $$ \|(x,y,z)\|_\infty = \max\{|x|, |y|, |z| \}, $$ and $Y$ being the two dimensional subspace given by $$ Y=\{(x, y, z) : x+y+z=0\}. $$ The best way to convince oneself that there is no projection from $X$ to $Y$ with norm one, is to make a cardboard model of this cube

cut it along the red line, place one of the two halves on top of the table with the red hexagon down, and attempt to shine a flashlight so that the shadow is restricted to within the hexagon. It is impossible!

Answer 2

As you have found in many places around here, the result is true when the target space is equipped with the supremum norm. Let $W\subset X$ be an inclusion of subspaces and let $Y$ be a finite dimensional space equipped with an arbitrary norm $\|\cdot\|$. Let $T:W\to (Y,\|\cdot\|)$ be a bounded linear operator.

Now it is well-known that in a finite dimensional space all norms are equivalent. So if $\|\cdot\|$ is any norm on $Y$ there exist constants $c_1,c_2>0$ so that $\|\cdot\|\leq c_1\|\cdot\|_\infty$ and $\|\cdot\|_\infty\leq c_2\|\cdot\|$. Actually, the best such constants are interpreted as follows: Take the identity operator $I:(Y,\|\cdot\|_\infty)\to(Y,\|\cdot\|)$. Then $c_1=\|I\|$ and $c_2=\|I^{-1}\|$.

Now consider the operator $S:W\to (Y,\|\cdot\|_\infty)$ defined by $S=I^{-1}\circ T$. Then $S$ is a bounded operator and as we explained this extends to an operator $S':X\to(Y,\|\cdot\|_\infty)$ s.t. $\|S'\|=\|S\|$. Take $T'=I\circ S'$. Then $T':X\to(Y,\|\cdot\|)$ extends $T$. Now $$\|T\|\leq \|T'\|\leq c_1\|S'\|=c_1\|S\|\leq c_1c_2\|T\|$$ So if $c_1c_2=1$, the extension is isometric (note that $1\leq\|I\|\cdot\|I^{-1}\|=c_1c_2$). I think that this is the best one can say "for free" without getting more specific about the norm that $Y$ has.

Anyway, I believe that one should not really worry about this; the main point that the exercise is trying to make is that bounded operators towards finite dimensional spaces can be extended from subspaces to the entire space. Now usually a finite dimensional space appears in an auxilliary way, so the norm we consider on it can be chosen freely, because all norms are equivalent, so we simply choose the supremum norm if we want to make our extension isometric. As one of the linked answers state, the result fails for the $\ell^p$ norms on $Y$, so one cannot expect to prove that the extension is isometric in general.

Sure though, it would be more appropriate to explicitly state that $\mathbb{R}^n$ is equipped with the supremum norm.