Suppose $Y$ is a closed subspace of Banach space $X$ and $T:Y\to X$ is a bounded finite rank operator. Can we extend $T$ to $\tilde{T}:X\to X$, in the sense that:
- $T=\tilde{T}$ on $Y$
- Range($T$)$=$Range($\tilde{T}$)
- $||T||=||\tilde{T}||$
If $T$ is rank $1$, this follows pretty quickly from Hahn-Banach, but if the rank of $T$ is greater than $1$, I am having trouble with the last bullet, I cannot show the equality of norms.
No. Kakutani proved that a Banach space $X$ is isometric to a Hilbert space if and only if each two-dimensional subspace $Y$ is complemented by a norm-one projection.
Now suppose that $X$ is not isometric to a Hilbert space and take a two-dimensional subspace $Y\subset X$ such that each projection $P\colon X\to X$ with ${\rm im}\,P = Y$ has norm $>1$. Now take $T$ to be the inclusion map $Tx=x\;(x\in Y)$.
Thus, you can find such extension if and only if $X$ is a Hilbert space.