The Open Mapping Theorem, the Bounded Inverse Theorem, and the Closed Graph Theorem are equivalent theorems in that any can be easily obtained from any other. The Closed Graph Theorem also easily implies the Uniform Boundedness Theorem. But is there a simple way to obtain any of the other three results from Uniform Boundedness, or is Uniform Boundedness really a "lowerlevel" result than the others?

4$\begingroup$ You could see Theorem 27.26 of Schechter's Handbook of Analysis and its Foundations (google books) where it is proved that every space that satisfies UBT also satisfies CGT. I expect one could extract a direct proof from this. I don't have a copy of the book handy right now but I will look when I have a chance. $\endgroup$– Nate EldredgeDec 12, 2014 at 20:06

2$\begingroup$ See also math.stackexchange.com/questions/146910/…. $\endgroup$– Nate EldredgeDec 12, 2014 at 20:08

2$\begingroup$ Thank you. The question of whether the UBT and CGT (say, for Banach spaces) are equivalent in ZF is interesting, but not exactly what I was after. My question is a little more vague: if one has the UBT, is there a proof of the other theorems which is simpler than the usual proof directly from Baire Category? $\endgroup$– Bruce BlackadarDec 12, 2014 at 23:28

$\begingroup$ If you don't care about using AC, here's a simple argument. We prove that a continuous bijection $T\colon X\to Y$ is a topological isomorphism. Let $\Omega=\{ y^* \in Y^* : \ y^*\circ T\_{X^*}\le1\}$. Then, it's bounded at every $y\in Y$. So by UBT, $\Omega$ is uniformly bounded. By HahnBanach, every $x^*\in X^*$ of norm one can be written as $y^*\circ T$ with $y^*\in\Omega$. Given $x\in X$, one applies the above to the norming linear functional $x^*\in X$ and bounds $T(x)$ from below. $\endgroup$– Narutaka OZAWADec 16, 2014 at 0:58

$\begingroup$ @NarutakaOZAWA: Can you explain the "by HahnBanach" part in more detail? I'm worried because I don't see where you used the completeness of $X$. $\endgroup$– Nate EldredgeDec 16, 2014 at 1:29
1 Answer
I don't know whether you'll consider this "simple", but here is a proof. I distilled it from Eric Schechter's Handbook of Analysis and its Foundations, which has a proof of a more general statement at 27.35. The last part is from Folland's Real Analysis, Theorem 5.10.
Suppose $X,Y$ are Banach spaces and $T : X \to Y$ is surjective. We wish to show $T$ is open. Let $B$ be the open unit ball of $X$; it suffices to show $T(B)$ contains a neighborhood of $0 \in Y$.
The first step is to show that the closure $\overline{T(B)}$ contains a neighborhood of 0. The usual method is to use the Baire category theorem: if not, then $Y = \bigcup_{n=1}^\infty n \overline{T(B)}$ meaning that $Y$ is meager. We will use the uniform boundedness principle instead.
For each $n$, construct a new norm $\\cdot\_n$ on $Y$ defined by $$\y\_n := \inf\{\u\_X+n\v\_Y : u \in X, v \in Y, v+Tu=y\}.\tag{*}$$ It is straightforward to verify this is a norm. Now let $Z$ be a countable direct sum of copies of $Y$, i.e., $Z$ is the vector space of all finitely supported functions $f : \mathbb{N} \to Y$, with the pointwise addition and scalar multiplication. Equip $Z$ with the norm $$\f\_Z := \sup_n \f(n)\_n.$$ Then for each $n$, define a linear operator $S_n : Y \to Z$ by $(S_n y)(n) = y$ and $(S_n y)(k) = 0$ for $k \ne n$. Note that $\S_n y\_Z = \y\_n$.
Now by taking $u=0$, $v=y$ in (*), we obtain $\y\_n \le n \y\_Y$, so each $S_n$ is bounded. Moreover, by the surjectivity of $T$, for each $y \in Y$ there exists $x \in X$ with $Tx=y$. Taking $u=x$ and $v=0$ in (*) we see that $\ y\_n \le \x\_X$ independent of $n$; hence $\{S_n\}$ is pointwise bounded. By the uniform boundedness theorem, there is a constant $C < \infty$ such that $\S_n\_{Y \to Z} \le C$ for all $n$.
Fix $\delta < 1/C$. I claim that $\overline{T(B)}$ contains a ball of radius $\delta$ centered at 0. For suppose $\y\_{Y} < \delta$; then $$\y\_n = \S_n y\_Z \le \S_n\_{Y \to Z} \y\_Y \le C \delta < 1$$ for every $n$. Hence for each $n$ there exists $u_n \in X$, $v_n \in Y$ with $y = v_n + T u_n$ and $\u_n\_X + n\v_n\_Y < 1$. In particular, $\v_n\_Y < 1/n$, so we have $T u_n \to y$ where $u_n \in B$. Thus $y \in \overline{T(B)}$ and the proof of this step is complete.
The rest of the proof proceeds as usual. We can show $T(B)$ contains a ball of radius $\delta/2$ centered at 0. Suppose $\y\_Y < \delta/2$, so that by the first step and scaling, $y \in \overline{T(B_{1/2})}$. Hence there is $x_1$ with $\x_1\_{X} < \frac{1}{2}$ and $\y  Tx_1\_Y < \delta/4$. Repeating this process inductively, we construct $x_n$ with $\x_n\_X < 2^{n}$ and $\left\ y  \sum_{k=1}^n Tx_k\right\_Y < \delta 2^{(n+1)}$. Thus $\sum_{k=1}^\infty T x_k$ converges in $Y$ to $y$. Summing a geometric series, $\sum_{k=1}^\infty \x_k\ < 1$, so by completeness of $X$, $\sum_{k=1}^\infty x_k$ converges in $X$ to some $x$ with $\x\_X < 1$. And by continuity of $T$, $Tx=y$. So we have shown $y \in T(B)$.
Note that this proof works even if $Y$ is not Banach, so long as the uniform boundedness principle holds on $Y$. This happens iff $Y$ is barreled, and indeed the proof in Schecheter shows that both uniform boundedness and open mapping are equivalent to being barreled.

$\begingroup$ Well, well,... I consider "the rest of the proof" as the main point of the OMT. In Rudin's book, this is even part of the statement. $\endgroup$ Dec 17, 2014 at 16:32


1$\begingroup$ I believe math.stackexchange.com/questions/146910/… claims that this proof uses the axiom of dependent choice? Is it just the inductive choice of the $x_k$ in the 2nd half, i.e. the paragraph starting with "The rest of the proof"? If it is used in the uniform boundedness principle section, I'm not seeing it. $\endgroup$– D.R.Jan 25, 2022 at 3:57