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Let $X$ and $Y$ be Banach spaces and let $T:X\to Y$ be a bounded linear operator such that $T(X)$ is dense in $Y$ but not equal to $Y$. Show that there exists some $y\in Y$ such that each sequence $(x_k)\subset X$ with $Tx_k\to y$ has the property that $||x_k||\to\infty$.

Example: Let $X=Y=\ell^2$ and define $T(x_k):=(x_k/2^k)$. Then the set $F:=\{(x_1,\ldots,x_m,0,0,0,\ldots)\ |\ m\in\mathbb N\}$ of finite sequences belongs to $T(X)$ and is dense in $\ell^2$, so $T(X)$ is dense as well. But $(x_k)=(1/k)_k\in\ell^2$ does not have a preimage.

This is what I did so far: Since $T(X)$ is dense, we find for each $y\in Y$ a sequence $(x_k)\subset X$ such that $Tx_k\to y$. Since $T$ is bounded there is some $c>0$ such that $c||Tx_k||\leq ||x_k||$. From this it follows that $$ \liminf_{k\to\infty}||x_k||\geq c||y||, $$ which is not so bad. However, I haven't used the fact that $T$ is not surjective. I know that this implies that $T$ cannot be open, but I don't know yet how to work this in. I also have the feeling that the assertion is only true for some $y\in Y\backslash T(X)$.

Since this is homework, please do not provide a full solution. Light hints, however, are highly appreciated. Thank you in advance.

sranthrop
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  • By the way, it seems obvious that for some $y$, when $Tx_k \to y$, we don't have $x_k \to x$. Otherwise you'd have $Tx = y$ and $T(X) = Y$, contrary to the given conditions. – Karolis Juodelė Nov 24 '16 at 10:21
  • Exactly, so if $y\notin T(X)$, then its corresponding $x_k$ does not have a convergent subsequence. But this does not necessarily imply that $||x_k||\to\infty$... – sranthrop Nov 24 '16 at 10:42
  • Can you give an example of $X, Y, T$ ? – Karolis Juodelė Nov 24 '16 at 10:52
  • I added an example. – sranthrop Nov 24 '16 at 11:24
  • I may be wrong, but I think that a "constructive" (not in the technical sense) existence proof would at least be very difficult. So I suggest proving it by contradiction.

    Suppose there didn't exist a $y\in Y$ with the given property. So for every $y\in Y$ there exists a sequence $(x_k)$ in $X$ with $Tx_k \to y$ and $\liminf \lVert x_k\rVert < +\infty$.

    Then look at the open mapping theorem to conclude that $T$ would then be surjective. Depending on the version of the open mapping theorem, it may suffice to look at the statement, or one may need to look at the proof.

     – Daniel Fischer Nov 24 '16 at 11:55
  • And if, by the time the homework is due, nobody has posted a full answer (and you don't want to write one yourself), you can ping me and I could write one. – Daniel Fischer Nov 24 '16 at 12:02
  • @Daniel Fischer: Thank you. Your hint tells me the way to go. I will think about it and come back here later. Thank you again. – sranthrop Nov 24 '16 at 12:41
  • If it turns out you need a further nudge, feel free to ask. – Daniel Fischer Nov 24 '16 at 12:58
  • Hm ok, our version of the open mapping theorem states that in my case $T$ being open and $T$ being surjective is equivalent (since both $X$ and $Y$ are Banach spaces). So assuming that for each $y\in Y$ there exists a bounded sequence $(x_k)$ such that $Tx_k\to y$, do I then have to prove that $T$ is open in order to get (by the open mapping theorem) that $T$ is surjective, which would lead to a contradiction? – sranthrop Nov 25 '16 at 19:56

1 Answers1

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Suppose that $T$ is not bounded below. Then there exists a sequence $(x_n)$ such that $\|x_n\|\to\infty$ and $\|T_n x_n\|\to 0$. Considering $y=0$, we are done for that case.

We are left considering the case where $T$ is bounded below. But this case implies that the range of $T$ is closed, so the hypothesis doesn't hold and we are done.

Jonas Meyer
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