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Suppose $Y$ is a Banach space and $X$ is a finite-dimensional subspace of $Y$. Further assume $T:X \rightarrow X$ is a linear operator which is power bounded from above and below, in other words there is $0 < c \leq 1 \leq C < \infty$ such that $c \cdot \|x\| \leq \|T^n(x)\| \leq C \cdot\|x\|$ for all $n \in \mathbb{N}$ and all $x \in X$.

Can $T$ be extended to a linear operator $\tilde{T}:Y \rightarrow Y$ which satisfies the same power bounds?

If it is not true in general, are there nice conditions which make it true? (It is important to my application that there is a lower bound.)

I apologize in advance if this is too trivial for Mathoverflow. I tried to look it up but couldn't find the answer. The impression I got is that extending linear operators is not as well known as extending linear functionals (Hahn-Banach). Also, while this is for an analysis paper, I am not an analyst. Hence I am not always sure what is common knowledge and what isn't. Thanks!

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up vote 3 down vote accepted

A finite-dimensional subspace is complemented, so we can write $Y = X \oplus Z$ for some closed subspace $Z$. There are constants $m$ and $M$ such that for all $x \in X$ and $z \in Z$, $m (\|x\| + \|z\|) \le \|x + z\| \le M (\|x\| + \|z\|)$. Extend $T$ to $\overline{T}$ so that $\overline{T} = I$ on $Z$. Then $\overline{T}$ is power-bounded above and below (though not necessarily with the same constants as $T$): $$ \|\overline{T}^n (x + z)\| = \|T^n x + z\| \le M (\|T^n x\| + \|z\|) \le C M (\|x\| + \|z\|) \le \frac{c M}{m} \|x + z\| $$ and similarly in the other direction.

It can't always be done with the same constants. Consider $Y = {\mathbb R}^3$ with the norm $\|(x,y,z)\| = \max(\sqrt{x^2 + y^2}, |x+z|)$, and $X = \{(x,y,0): x,y \in {\mathbb R}\}$. Note that $\|(x,y,0)\| = \sqrt{x^2 + y^2}$. Let $T: X \to X$ be a rotation $(x,y,0) \to (\cos(\theta) x + \sin(\theta) y, -\sin(\theta) x + \cos(\theta) y, 0)$ where $\theta$ is not an integer multiple of $\pi$. This is an isometry (so $c = C = 1$), but has no extension to an isometry of $Y$ because $(\pm 1, 0, 0)$ are the only points of $X$ that are extreme points of the unit ball of $Y$.

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Thank you! What determines the bounds $m$ and $M$? I don't really need the same bounds for $T$ and $\tilde{T}$, but I do need some uniformity in deriving the new bounds from the previous ones (and the subspace $X$ can change). – Jason Rute May 2 '13 at 6:51
I did some more thinking. While $M$ can be 1, it seems that without more information $m$ can be arbitrary small depending on the choice of $Y$ and $X$. Oh well... – Jason Rute May 2 '13 at 8:15
I don't know, how good estimate you need, but one can always take $m$ to be equal to $C\sqrt{dimX}^{-1}$ (Kadec-Snobar theorem, $C$ is universal). – Mateusz Wasilewski May 2 '13 at 8:56
Mateusz, my $X$'s grow in dimension so I don't think this will help. Also, if it matters, the $X$'s are all isomorphic (informally so) to $\ell^n_p$ where $n$ grows in size. But thank you for the reference! – Jason Rute May 2 '13 at 10:16
In general you cannot do better than the projection constant of $\ell_p^n$, which is of order $n^{1/p}$ when $p \ge 2$ and $n^{1/2}$ otherwise. To see that, look at $Y= \ell_p^n \oplus_p C(K)$ and take an isometric copy $E$ of $\ell_p^n$ in $C(K)$. On $\ell_p^n \oplus_p E$ let $T$ be an isometry that maps $\{0\} \oplus E$ to $\ell_p^n \oplus \{0\}$ and $\ell_p^n \oplus \{0\}$ to $\{0\} \oplus E$. – Bill Johnson May 2 '13 at 13:06

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