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Let $X,Y$ be Banach spaces and $f:X \rightarrow Y$ be a linear, continuous and surjective map. Show that $\forall (y_n)\subset Y, \:\: y_n \rightarrow0 \:\:\: \exists (x_n) \subset X \:\:x_n\rightarrow0:f(x_n)=y_n, \:\: n=1,2,\dots$

I know that all facts are needed. So we need to somehow use completness, surjectiveness, continuity and linearity. But I don't have any idea how to start.

luka5z
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    Use the open mapping theorem. – Etienne Jun 08 '14 at 07:53
  • I should also use the fact that $f(0)=0$? – luka5z Jun 08 '14 at 08:06
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    Not really. The key phrase is "open mapping theorem". Do you know this result? – Etienne Jun 08 '14 at 08:10
  • Yes, if $f$ is linear, continuous and surjective map between Banach spaces, then for any open $U$ from $X$, $f(U)$ is open in $Y$ – luka5z Jun 08 '14 at 08:36
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    OK, so you should solve your problem in one line. By the open mapping theorem, you know that there exists some constant $C$ such that, for any $y\in Y$, one can find $x\in X$ with $\Vert x\Vert\leq C,\Vert y\Vert$ such that $f(x)=y$. (Actually, you do need to use the fact that $f(0)=0$ to get this). – Etienne Jun 08 '14 at 08:43
  • How can I solve it in one line? – luka5z Jun 08 '14 at 09:40

2 Answers2

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By the open mapping theorem, $B_\varepsilon(0) \subseteq f(B_1(0))$ for some $\varepsilon > 0$. It follows more generally that $B_r(0) \subseteq f(B_{r/\varepsilon}(0))$ for all $r>0$. Hence, for every $y \in Y$ there is some $x \in X$ with $y=f(x)$ and $||x||\leq ||y||/\varepsilon$ (if $y \neq 0$ take $r = ||y||$, and otherwise choose $x=0$).

If $y_n \to 0$ in $Y$, we therefore find $x_n \in X$ with $y_n = f(x_n)$ and $||x_n|| \leq ||y_n||/\varepsilon$, so that clearly $x_n \to 0$.

Martin Brandenburg
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Claim: For every $m\in\mathbb N$, there exists some $N_m\in\mathbb N$ such that for every $n>N_m$, $\|\widehat{x}_n^m\|<1/m$ and $y_n=f(\widehat{x}_n^m)$ hold for some $\widehat{x}_n^m\in X$.

Proof: Let $B_X(1/m,0)\equiv\{x\in X\,|\,\|x\|<1/m\}$ denote the open sphere around $0$ of radius $1/m$. By the open mapping theorem, $f(B_X(1/m,0))$ is open in $Y$ and contains $0$ (because $f(0)=0$), so there exists some $\varepsilon_m>0$ such that $B_Y(\varepsilon_m,0)\subseteq f(B_X(1/m,0))$. By the convergence hypothesis, in turn, there exists some $N_m$ such that $n>N_m$ implies that $\|y_n\|<\varepsilon_m$. That is, $$n>N_m\Longrightarrow y_n\in B_Y(\varepsilon_m,0)\Longrightarrow y_n\in f(B_X(1/m,0)),$$ which is the claim. $\blacksquare$

Now, suppose, without loss of generality, that $N_1<N_2<N_3<\ldots$ (if it is not true, then you can successively increase the $N_m$ without affecting the claim). If $1\leq n\leq N_1$, find some arbitrary $x_n\in X$ such that $y_n=f(x_n)$ (this is possible because $f$ is surjective). For $N_1<n\leq N_2$, define $x_n\equiv \widehat{x}_n^1$. For $N_2<n\leq N_3$, define $x_n\equiv \widehat{x}_n^2$. For $N_m<n\leq N_{m+1}$, define $x_n\equiv \widehat{x}_n^m$, and so on. It is easy to see that the sequence $(x_n)_{n\in\mathbb N}$ has the properties you need, i.e., it converges to zero and $y_n=f(x_n)$ for all $n$.

triple_sec
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