Let $X$ and $Y$ Banach spaces and $F: X \to Y$ a linear, continuous and surjective mapping. Show that if $K$ is a compact subset of $Y$ then there exists an $L$, a compact subset of $X$ such that $F(L)= K$.

I know by the Open Mapping Theorem that $F$ is open. What else can I do? Thank yoU!


I suspect there must be a simpler proof for Banach spaces, but as I don't see one, here is what I came up with, using a faint recollection of the proof of Schwartz' lemma (simplified, since we're dealing with Banach spaces, not Fréchet spaces):

By the open mapping theorem, we know that there is a $C > 1$ such that

$$\bigl( \forall y \in Y\bigr)\bigl(\exists x \in X\bigr)\bigl(\lVert x\rVert \leqslant C\cdot \lVert y\rVert\land y = F(x)\bigr).\tag{1}$$

Without loss of generality, assume that $K$ is contained in the (closed, not important) unit ball of $Y$ and nonempty.

Since $K$ is compact, there is a finite set $M_1 = \{ y_{1,\nu} : 1 \leqslant \nu \leqslant n_1\}$ such that

$$K \subset \bigcup_{\nu=1}^{n_1} B_{1/C}(y_{1,\nu}).$$

For every $1\leqslant \nu \leqslant n_1$, choose an $x_{1,\nu} \in X$ with $\lVert x_{1,\nu}\rVert \leqslant C$ and $F(x_{1,\nu}) = y_{1,\nu}$. Let $L_1 = \{ x_{1,\nu} : 1 \leqslant \nu \leqslant n_1\}$.

For $k \geqslant 2$, if $M_1,\dotsc,M_{k-1}$ and $L_1,\dotsc,L_{k-1}$ have already been constructed, let $M_k = \{y_{k,\nu} : 1 \leqslant \nu \leqslant n_k\}$ a finite set such that

$$K \subset \bigcup_{\nu = 1}^{n_k} B_{1/C^k}(y_{k,\nu}).$$

For each $\nu$, choose a $\mu_k(\nu) \in \{ 1,\dotsc,n_{k-1}\}$ such that $\lVert y_{k,\nu} - y_{k-1,\mu_k(\nu)}\rVert < C^{1-k}$, and by $(1)$ a $z_{k,\nu} \in X$ with $\lVert z_{k,\nu}\rVert \leqslant C\cdot \lVert y_{k,\nu} - y_{k-1,\mu_k(\nu)}\rVert$ and $F(z_{k,\nu}) = y_{k,\nu} - y_{k-1,\mu_k(\nu)}$. Let $x_{k,\nu} = x_{k-1,\mu_k(\nu)} + z_{k,\nu}$ and $L_k = \{x_{k,\nu} : 1 \leqslant \nu \leqslant n_k\}$.

By construction, $F(L_k) = M_k$, and hence, setting $L_\ast = \bigcup\limits_{k=1}^\infty L_k$, we see that

$$F(L_\ast) = F\left(\bigcup_{k=1}^\infty L_k\right) = \bigcup_{k=1}^\infty F(L_k) = \bigcup_{k=1}^\infty M_k$$

is dense in $K$.

$L_\ast$ is also totally bounded: Let $\varepsilon > 0$ arbitrary. Choose a $k \geqslant 1$ such that $C^{-k} < \varepsilon(C-1)$. Then

$$L_\ast \subset \bigcup_{j=1}^{k+2}\bigcup_{\nu=1}^{n_j} B_{\varepsilon}(x_{j,\nu}).\tag{2}$$

That is clear for $x \in L_m$ where $m \leqslant k+2$, so let $x'_m\in L_m$ where $m > k+2$. By construction, there is an $x'_{m-1} \in L_{m-1}$ with $\lVert x'_m-x'_{m-1}\rVert \leqslant C\cdot \lVert F(x'_m) - F(x'_{m-1})\rVert$ and $\lVert F(x'_m) - F(x'_{m-1})\rVert \leqslant C^{1-m}$, so $\lVert x'_m-x'_{m-1}\rVert \leqslant C^{2-m}$. If $m-1 > k+2$, again by construction, there is an $x'_{m-2} \in L_{m-2}$ with $\lVert x'_{m-1} -x'_{m-2}\rVert \leqslant C^{3-m}$. We thus obtain a finite sequence $(x'_{r})_{k+2 \leqslant r \leqslant m}$ with $\lVert x'_r - x'_{r+1}\rVert \leqslant C^{1-r}$ and $x'_r \in L_r$, whence

$$\lVert x'_m - x'_{k+2}\rVert \leqslant \sum_{r=k+2}^{m-1} \lVert x'_{r+1} - x'_r\rVert \leqslant \sum_{r=k+2}^{m-1} C^{1-r} < \frac{1}{C^{k}}\frac{1}{C-1} < \varepsilon,$$

showing $(2)$. Since $\varepsilon > 0$ was arbitrary, $L_\ast$ is totally bounded.

Hence $L := \overline{L_\ast}$ is compact, and therefore

$$F(L) \subset \overline{F(L_\ast)} = K$$

is a dense compact subset of $K$, i.e. $F(L) = K$.


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