Question in Folland's Proof of Open Mapping Theorem

Question

The Open Mapping Theorem states:

Let $X$ and $Y$ be Banach spaces. If $T \in L(X, Y)$ is surjective, then $T$ is open.

Below is a proof from Folland's text. I am able to follow up his proof until a point, after which things start to become unclear.

I have included his proof as quotes, with my questions included as annotations. I hope my presentation is clear.

Let $B_r$ denote the (open) ball of radius $r$ about $0$ in $X$. It will suffice to show that $T(B_1)$ contains a ball about $0$ in $Y$. Since $X = \bigcup_1^\infty B_n$ and $T$ is surjective, we have $Y = \bigcup_1^n T(B_n)$. But $Y$ is complete and the map $y \mapsto ny$ is a homeomorphism of $Y$ that maps $T(B_1)$ to $T(B_n)$, so Baire's theorem implies that $T(B_1)$ cannot be nowhere dense.

1. How does Folland conclude this from Baire's Theorem? The theorem states that $Y$ cannot be a countable union of nowhere dense sets, so what if the union contains one dense subset, in which case how can we be sure that this subset is $T(B_1)$? Does this follow from the homeomorphism he defines?

That is, there exist $y_0 \in Y$ and $r > 0$ such that the ball $B(4r, y_0)$ is contained in $\overline{T(B_1)}$. Pick $y_1 = Tx_1 \in T(B_1)$ such that $\|y_1 - y_0\| < 2r$; then $B(2r, y_1) \subset B(4r, y_0) \subset \overline{T(B_1)}$. So if $\|y\| < 2r$, $$y = Tx_1 + (y-y_1) \in \overline{T(x_1 + B_1)} \subset \overline{T(B_2)}.$$

2. I am able to follow everything until he considers $\|y\| < 2r$ and the line beneath it. It seems to me he is doing a sort of translation and dilation to recenter the ball at $0$ and have radius $2$, is this correct? If so, I do not see how he is doing this.

Divind both sides by $2$, we conclude that there exists $r > 0$ such that if $\|y\| <r$ then $y \in \overline{T(B_1)}$. Since $T$ commutes with dilations, it follows that if $\|y\| < r2^{-n}$, then $y \in \overline{T(B_{2^{-n}})}$. Suppose $\|y\| < r/2$; we can find $x_1 \in B_{1/2}$ such that $\|y - Tx_1\| < r/4$, and proceed inductively, we can find $x_n \in B_{2^{-n}}$ such that $\|y-\sum_1^n Tx_j\| < r2^{-n-1}$. Since $X$ is complete, the series $\sum_1^\infty x_n$ converges, say to $x$. But then $\|x\| < \sum_1^\infty 2^{-n} = 1$ and $y = Tx$. In other words, $T(B_1)$ contains all $y$ with $\|y \| < r/2$, so we are done.

3. I understand all the technicalities in this part of the proof, but I am having trouble conceptually seeing how it shows that we can replace $\overline{T(B_1)}$ with $T(B_1)$.

Answer 1

This is a standard corollary/rephrasing of Baire’s theorem, and is an extremely powerful tool that I have seen a few times. If all the $T(B_n)$ were nowhere dense, they could not form $Y$ (which is trivially dense in $Y$) as their union - so at least one $T(B_k)$ is not nowhere dense. The key point is that if $T(B_k)$ is not nowhere dense, then, for any other $m$, we have $T(B_m)=\phi(T(B_k))$ where $\phi(y)=\frac{m}{k}y$ is a homeomorphism and the linearity of $T$ is used. Homeomorphy here implies that $T(B_m)$ is also not nowhere dense. This actually holds for all $m,k\gt0$ not just integers, and in particular we get $T(B_1)$ is not nowhere dense. Another way to think about this is, as Folland hints: if $T(B_1)$ were nowhere dense, then the homeomorphism $y\mapsto ny$ (and linearity) would give $T(B_n)$ nowhere dense for all $n$, a contradiction.

He considers $\|y\|\lt2r$ for two reasons. Reason 1: the structure of this proof is to demonstrate that $T(B_1)$ is open about $0$, as this then implies (by aforementioned homeomorphy) that $T$ maps all open balls about the origin to sets open at the origin. By homeomorphy of translation, and linearity, $T(B(x,r))$ is open about $x$, for all pairs $(x,r)$. This is sufficient as an open set is open about every point and thus will map under $T$ to a set which is open about all its points, hence itself open.

Reason 2: he wants to pull of the division by $2$ trick at the end, and he also needs to squeeze $y$ into the ball. Note that $\|y_1-y_0\|\lt2r$ and $\|y\|\lt2r$ are sufficient to say $\|(y_1-y)-y_0\|\le\|y\|+\|y_1-y_0\|\lt4r$, and then $(y_1-y)\in\overline{T(B_1)}$ is iff. $y-y_1\in\overline{T(B_1)}$ because $T$ is linear and $B_1$ is symmetric. This decomposition is clearly important for the proof, and it was convenient to consider it about the origin.

Conceptually, we want to get rid of the closure sign. We can do this here by saying: if $y$ is in the closure of $T(B_\rho)$ I can claim that $y$ is very close to $T(B_\rho)$, for instance it is within $\rho/2$ of the set. That’s all that is going on. This proximity is all that is necessary for the proof via a completeness argument, so we are lucky here and can ditch the $\overline{}$ bar. Taking a sequence due to closure like this is a common theme in these sorts of proofs.

Answer 2

1. The property of being nowhere dense is purely topological and so preserved under homeomorphisms. Thus if $T(B_1)$ is nowhere dense, then so must $T(B_n)$ for every $n$ since it is the image of this set under the given homeomorphism (which is the case since $T$ is linear). Therefore, if $any$ of these subsets were nowhere dense, then they $all$ must be, and we would be able to write $Y$ as a countable union of such sets, a contradiction.

2. The end goal is to show that $T$ is an open mapping, which as said in the proof amounts to showing that $T(B_1)$ contains an open ball centered at the origin. This is equivalent to saying that there exists some radius $\epsilon$ small enough so that if $||y|| < \epsilon$, then $y = Tx$ for some $x$ in the unit ball $B_1 \subset X$. From Step 1, since $T(B_1)$ is $not$ nowhere dense there exists some ball of radius $4r$ centered at $y_0$ contained in $\overline{T(B_1)}$. Notice this is not the actual set we care about but the closure of it, so the center $y_0$ is not guaranteed to be $Tx$ for some $x \in B_1$. Therefore, in Step 2 we look at a slightly smaller ball of radius $2r$ contained in the previous one but with a center $y_1 = Tx_1 $ for some $x_1 \in B_1$. We know this point exists by the definition of the closure; since $y_0 \in \overline{T(B_1)}$, we can always find a point in $T(B_1)$ arbitrarily close to $y_0$. Conceptually, this step allows us to then say that points close to $y_1$ are in some sense equal to or well-approximated by points $Tx$ for $x$ not too far from $x_1$. To be very concrete: if $y_2$ is any other point in this smaller ball, which by construction is contained in $\overline{T(B_1)}$ then by the definition of the closure $y_2 = \lim_{k\to\infty} y_k$ for some sequence $\{y_k\} \subset T(B_1)$, i.e. $y_k = Tx_k \in B_1$ for every $k$. Then $$ y_2 - y_1 = \lim_{k\to\infty}y_k-y_1 = \lim_{k\to\infty}T(x_k - x_1) $$ where $x_k - x_1 \in B_2$ for every $k$ since both $x_1$ and $x_k$ are in $B_1$ (basically the furthest distance between two points in the unit ball is 2). Therefore $y_2 - y_1 \in \overline{T(B_2)}$. Translating this smaller ball by $y_1$ so that it is centered at the origin yields the statement given in the proof; alternatively, you can think of the $y$ in the line you specified as simply the displacement vector $y_2 - y_1$.

3. Here we have a similar idea in that initially, we just have that points in $Y$ close to the origin are in the closure of $T(B_1)$, $not$ $T(B_1)$ itself as desired. However, being in the closure means that it can be approximated as close as we want by elements in that space. Therefore, the trick is combine these two ideas iteratively as follows: If $y$ is close to the origin, we can find an initial point $x_1 \in B_{1/2}$ so that $Tx_1$ approximates $y$ very well, so much so that the error $y - x_1$ is even $closer$ to the origin than $y$ was originally, say by half. Therefore, we can now find another point $x_2 \in B_{1/4}$ so that $Tx_2$ approximates the error even better than our initial approximation of $y$. It should hopefully be clear that successively approximating the errors in this way allows us to reach $y$ in the limit, but the key trick here is that the size of the $x_n$ being used decreases exponentially fast and so their infinite sum absolutely converges. Since in a complete normed space absolutely convergent series converge to a point, we have that at the end that $y = Tx$ for some $x \in B_1$ exactly as desired. It is only in passing to the limit and using the completeness of $X$ that we know such an $x$ exists, otherwise the results of Step 2 are not sufficient.