I want to show the following:

Let $Y$ be a normed linear space. Show $Y$ is a Banach space if and only if there exists a Banach space $X$ and a linear, continuous, open, and onto mapping $T: X \rightarrow Y$.

One direction is straightforward: If $Y $ is Banach, take $X = Y, T = Id$, which is linear, open, continuous and onto.

My proof for the other direction has one line that I believe is false, which I mark with (***). I'm using a theorem that states that if T(X) is closed, then $$ \exists M > 0 \text{ s.t. } \forall y \in T(X), \exists x \in X \text{ s.t. } Tx = y \text{ and } \|x \| \le M \| y \|. $$ Suppose $\{y_k\}$ is a Cauchy sequence in Y. Define $\tilde{y}_1 = y_1, \tilde{y}_k = y_k - y_{k-1}$. By open mapping theorem, $T(X)=Y$ is closed. Now find $\tilde{x}_k$ s.t. $T\tilde{x}_k = \tilde{y}_k \text{ and } \|\tilde{x}_k \| \le M \| \tilde{y}_k \|$. Since $y_k$ is Cauchy, $\| \tilde{y}_k \| = \|y_k - y_{k-1}\|$ can be made arbitrarily small for large enough k. Hence $\tilde{x}_k \rightarrow 0$ in $X$. Define $x_k = \sum_{i=1}^k \tilde{x}_i$.

Now note that $y_k = \sum_{i=1}^k \tilde{y}_i = \sum_{i=1}^k T\tilde{x}_i = T(\sum_{i=1}^k \tilde{x}_i) = Tx_k$.

(***) $\|x_m - x_n\| = \|\sum_{i=n+1}^m \tilde{x}_i \|= \sum_{i=n+1}^m\| \tilde{x}_i \|$. This shows that $\{x_k\}$ is Cauchy, hence converges to some $x\in X$. Since $T$ is onto $y:=Tx$ is in $Y$.

By continuity, $y_k = T x_k \rightarrow T x = y$. Hence Y is complete.

  • $\begingroup$ I believe (***) is wrong because m - n is finite but can be arbitrary large, so we need more than that to show that $\|x_m - x_n\| \rightarrow 0$. $\endgroup$ – ertl Jul 28 '18 at 3:58

The issue you are concerned about is indeed a problem. Here's a trick you can use to get around it. By passing to a subsequence, you can assume that $(y_k)$ is Cauchy "as fast as you want". For instance, you can assume $\|y_m-y_n\|<2^{-N}$ whenever $m,n>N$. The advantage of this is that then $\tilde{y}_k$ will not just converge to $0$ but will be absolutely summable, i.e. $\sum_{k=1}^\infty\|\tilde{y}_k\|$ converges. This will then give you the bound on $\sum_{i=n+1}^m\| \tilde{x}_i \|$ you need at the end of your argument. (So, this proves that some subsequence of your original sequence converges to some $y\in Y$. This actually automatically implies that the original sequence converges: if a subsequence of a Cauchy sequence converges, then the original sequence converges. It's a good exercise to prove this if you haven't seen it before).

There are various other significant errors in your proof, though all of them can be easily patched. The theorem which you cite at the start does not apply, since that theorem (which is essentially the same as the open mapping theorem) is only valid assuming that the codomain of $T$ is complete (which is what you are trying to prove here!). Instead you can just get the same conclusion from the stated assumption that $T$ is open. It also does not make any sense to invoke the open mapping theorem as you have done; the open mapping theorem only applies when the codomain is complete and in any case would not give you the conclusion that $T(X)$ is closed. Finally, at the end it does not make sense to say that $T$ is onto; the conclusion that $Tx=y\in Y$ is just a consequence of $T$ being a function from $X$ to $Y$. What you need to prove instead is that this $y$ is actually a limit of the sequence $(y_k)$.

  • $\begingroup$ I worked out the proof, and realized how much harder I had made it for myself than it needed to be - thank you so much for your answer! $\endgroup$ – ertl Jul 28 '18 at 22:12

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