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 $x \in X$.



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




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*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!*



Extension of power bounded operators over a finite subspace
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 $Y$.


