Let $E,F$ and $G$ be normed spaces, $f\in \mathcal{L}(E,F)$ and $g\in \mathcal{L}(F,G)$. Suppose that $g$ is injective and $f$ is compact. Show that, $\forall \varepsilon > 0$, there exists $M>0$ such that: $$\|f(x)\| \leq \varepsilon \|x\| + M \|g(f(x))\|, \quad \forall x \in E.$$

$$\|f(x_n)\| > \varepsilon \|x_n\| + M \|g(f(x_n))\|$$

My attempt:

Assume that the conclusion does not hold. Then there exists $\varepsilon > 0$ and $x_n$ such that $$\|f(x_n)\| > \varepsilon \|x_n\| + n \|g(f(x_n))\| \iff \|f(x_n)\| > \varepsilon + n \|g(f(x_n))\|, \quad \|x_n\|=1, \quad \forall n \in \mathbb{N}.$$

Since $f$ is compact and $x_n$ is bounded, there exists a subsequence $(x_{n_k}) $ such that $f(x_{n_k}) \to y \in F$.


$$\|f(x_{n_k})\| > \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad \forall k \in \mathbb{N}$$

If $g(f(x_{n_k}))$ is bounded, it's done. However, I couldn't prove that $g(f(x_{n_k}))$ is bounded and I'm not sure if it holds.

Any hints? New ideas are welcome.


Divide by $n_k$ in the inequality $$\tag{*}\|f(x_{n_k})\| > \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad k \in \mathbb{N}$$ and letting $k$ going to infinity, we get, by boundedness of $\left(f\left(x_{n_k}\right)\right)_{k\geqslant 1}$, that $\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=0$. By continuity of $g$ we also have that $\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=g(y)$ and using injectivity of $g$ we get $y=0$. Since $f$ is bounded, we have for $k$ large enough that $\left\lVert f\left(x_{n_k}\right)\right\rVert\leqslant \varepsilon $ hence by (*), we get a contradiction.


$\|f(x_{n_k})\|\leq \|x_{n_k}\| \|f\|=\|f\|$. This implie that from your inequality,

$\|f\|\geq \|f(x_{n_k})\|>\epsilon +n_k\|g(f(x_{n_k}))\|$

since $f(x_{n_k})$ converges towards $y$, $lim_n\|g(f(x_{n_k}))\|=\|g(y)\|$. We deduce:

$\|f\|\geq \epsilon+n_k\|g(y)\|$. This implies that $g(y)=0$ (since $n_k\|g(f(x_{n_k}))\|$ has to be bounded) and $y=0$ since $g$ is injective. The equality again implies:

$lim_k \|f(x_{n_k})\|>\epsilon+ lim_{k}(n_k\|g(f(x_{n_k}))\|)$ which implies that $0>\epsilon$. Contradiction.


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