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?
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 $(x_k) \mapsto \left(\frac1k x_k\right)$.