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.
Best Answer
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!