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I've been pondering about the following Lemma for a while now, but can't think of a proof. In fact, I can't even think of a way to prove it.

Let $E$, $F$ and $G$ be Banach spaces, $T \in \mathcal{K}(E,F)$ (e.g. a compact operator) and $S \in \mathcal{L}(F,G)$ injective. Then for each $\epsilon \gt 0$ there is a $c_\epsilon \gt 0$ such that for all $x \in E$ holds $$\|Tx\| \le \epsilon \|x\| + c_\epsilon \|STx\|.$$

I have no idea what properties to use here. How can I see that this holds?

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Are you assuming that $S$ is bounded? – Martin Argerami Jul 10 '13 at 23:17
Yes, $\mathcal{L}(F,G)$ for me is the set of all bounded linear mappings from F to G. – Amarus Jul 10 '13 at 23:54
up vote 7 down vote accepted

Suppose it weren't so. Then there'd be an $\varepsilon > 0$ such that for every $n \in \mathbb{N}$ there is an $x_n \in E$ with

$$\lVert T x_n\rVert > \varepsilon \lVert x_n\rVert + n\cdot \lVert ST x_n\rVert.$$

$x_n$ cannot be $0$, hence we may without loss of generality assume that $\lVert x_n\rVert = 1$.

$T$ is compact, hence $T x_n$ has a convergent subsequence, say $T x_{n_k} \to y \in F$. Then $\lVert y\rVert \geqslant \varepsilon$ since $\lVert T x_{n_k}\rVert > \varepsilon$. In particular, $y \neq 0$, hence also $S y \neq 0$. But then

$$\lVert S y\rVert = \lim_{k \to \infty} \lVert ST x_{n_k}\rVert \leqslant \limsup_{k\to\infty} \frac{1}{n_k}\lVert T x_{n_k}\rVert = \lim_{k \to \infty} \frac{1}{n_k} \lVert y\rVert = 0,$$

which contradicts the injectivity of $S$.

It might be good to point out that if $S$ has closed range (in particular if it is surjective), you have the estimate $\lVert Tx \rVert \leqslant \lVert S^{-1}\rVert \cdot \lVert STx\rVert$ regardless of whether $T$ is compact.

And that in general, when the range of $S$ is not (necessarily) closed, you need both parts of the right hand side in the estimate for $\varepsilon < \lVert T\rVert$. As an example, consider $E = F = G = \ell^2$ and $S = T$ the operator $(x_k) \mapsto \left(\frac1k x_k\right)$.

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Very nice. +1 $ $ – Martin Argerami Jul 10 '13 at 23:52

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