Direct approach to the Closed Graph Theorem

In the context of Banach spaces, the Closed Graph Theorem and the Open Mapping Theorem are equivalent. It seems that usually one proves the Open Mapping Theorem using the Baire Category Theorem, and then, from this theorem, proves the Closed Graph Theorem.

I was wondering about a more direct approach to the Closed Graph Theorem. What I want to show, possibly using the Baire Category Theorem, but without the Closed Graph Theorem or any of its equivalent theorems, that if $T: X \to Y$ is not continuous, then, there is a convergent sequence $x_n \rightarrow x$ such that $T x_n \rightarrow y \neq Tx$. Of course, $X$ and $Y$ are Banach Spaces.

Because $T$ is not bounded, I know that there is a sequence $a_n$ of unitary vectors such that $T a_n \rightarrow \infty$. Now, taking $b_n = \frac{a_n}{\|T a_n\|}$, we have that $b_n \rightarrow 0$, and $\|T b_n\| = 1$.

I wonder, if there is a simple argument for constructing a $x_n \rightarrow x$, based on $a_n$ or $b_n$, such that $T x_n$ is a Cauchy Sequence, but such that $\|T x_n\|$ is "far from" $\|T x\|$.

Of course, $x_n$'s construction would have to use the fact that $X$ is complete. For example, $x$ could be the limit of an absolutely convergent sequence, pretty much in the same fashion as the construction in the proof of the Open Mapping Theorem.

• What does it mean for two theorems to be equivalent? Commented Apr 27, 2012 at 14:47
• @joriki: I understand what you mean. You are totally right. I myself, would love if someone could explain to me what I MEANT by two theorems being equivalent. :-) However, what I really want to know is that if you can prove a theorem without using the other. Of course, "without the other" is not a well-defined mathematical term, so I guess people will have to appeal to the human inside them in order to answer the question properly. (by the way, your comments are disappearing...) Commented Apr 27, 2012 at 15:02
• @André: I'm thinking that perhaps you mean something like this: In proving the theorems, you can use the axioms of the theory (in this case, the axiomatic properties of Banach spaces). If you can then prove each theorem form the other without making further use of the axioms, i.e. treating the word "Banach space" as a black box, you could call them equivalent, in the sense that the both encapsulate exactly the same information about Banach spaces. Yes, sorry about the disappearing comment; it was worded a bit too categorically so I deleted and rephrased it. Commented Apr 27, 2012 at 15:06
• @AndréCaldas: I usually see two theorems called "equivalent" if there is a proof of each from the other that is substantially shorter than all known proofs which don't assume the other. I seem to recall a discussion of this on MO, but can't find it. Commented Apr 27, 2012 at 15:07
• You can, if you want to, try to formalize equivalence of theorems by saying that two theorems are equivalent if each can be proved from the other over some (specified) very weak base theory, much weaker than needed to prove either theorem from scratch. This is what reverse mathematics is all about. Also, I unupvoted: hope you're happy. Commented Apr 27, 2012 at 15:55

Zabreiko's lemma (from P. P. Zabreiko, A theorem for semiadditive functionals, Functional analysis and its applications 3 (1), 1969, 70–72) is not as well known as it deserves to be and I think it fits the bill to some extent, so let me state that result first:

Lemma (Zabreiko, 1969) Let $X$ be a Banach space and let $p: X \to [0,\infty)$ be a seminorm. If for all absolutely convergent series $\sum_{n=0}^\infty x_n$ in $X$ we have $$p\left(\sum_{n=0}^\infty x_n\right) \leq \sum_{n=0}^\infty p(x_n) \in [0,\infty]$$ then $p$ is continuous. That is to say, there exists a constant $C \geq 0$ such that $p(x) \leq C\|x\|$ for all $x \in X$.

Assuming this lemma, let $T: X \to Y$ be a discontinuous linear map between Banach spaces, consider the seminorm $p(x) = \|Tx\|$ and observe that there must exist an absolutely summable sequence $(x_n)_{n=0}^\infty$ and $\varepsilon \gt 0$ such that $$p\left(\sum_{n=0}^\infty x_n\right) \geq \sum_{n=0}^\infty p(x_n) + \varepsilon.$$ Since the left hand side is finite, both $a_{N} = \sum_{n=1}^N x_n$ and $b_N = Ta_N$ are Cauchy sequences with limits $a$ and $b$, respectively. We have $\|b_N\| \leq \|T(a)\| - \varepsilon$, so $\|T(a) - b_N\| \geq \varepsilon$ for all $N$ and thus $\|T(a) - b\| \geq \varepsilon$. In other words $(a_N,T(a_N)) \to (a,b)$ but $b \neq T(a)$, so the graph of $T$ is not closed.

Of course, I should make the disclaimer that Zabreiko's lemma is actually stronger than the usual consequences of the Baire category theorem in basic functional analysis and thus it does not answer the question as asked. As you mention in your question, as soon as the closed graph theorem is established, the inverse mapping theorem and the open mapping theorem follow easily, as I also explain in this thread. Moreover, the uniform boundedness principle is a straightforward consequence, too:

Exercises:

Use Zabreiko's lemma to prove:

1. the uniform boundedness principle;
Hint: set $p(x) = \sup_{n \in \mathbb{N}} \|T_n x\|$.
2. the inverse mapping theorem.
Hint: set $p(x) = \|T^{-1}x\|$.

The proof of Zabreiko's lemma is very similar to the usual proof by Banach–Schauder of the open mapping theorem:

Proof. Let $A_n = \{x \in X\,:\,p(x) \leq n\}$ and $F_n = \overline{A_n}$. Note that $A_n$ and $F_n$ are symmetric and convex because $p$ is a seminorm. We have $X = \bigcup_{n=1}^\infty F_n$ and Baire's theorem implies that there is $N$ such that the interior of $F_N$ is nonempty.

Therefore there are $x_0 \in X$ and $R \gt 0$ such that $B_R(x_0) \subset F_N$. By symmetry of $F_N$ we have $B_{R}(-x_0) = -B_{R}(x_0) \subset F_n$, too. If $\|x\| \lt R$ then $x+x_0 \in B_{R}(x_0)$ and $x-x_0 \in B_{R}(-x_0)$, so $x \pm x_0 \in F_{N}$. By convexity of $F_N$ it follows that $$x = \frac{1}{2}(x-x_0) + \frac{1}{2}(x+x_0) \in F_N,$$ so $B_R(0) \subset F_N$.

Our goal is to establish that $$$$\tag{\ast} B_{R}(0) \subset A_N$$$$ because then for $x \neq 0$ we have with $\lambda = \frac{R}{\|x\|(1+\varepsilon)}$ that $\lambda x \in B_{R}(0) \subset A_N$, so $p(\lambda x) \leq N$ and thus $p(x) \leq \frac{N(1+\varepsilon)}{R} \|x\|$, as desired.

Proof of $(\ast)$. Suppose $\|x\| \lt R$ and choose $r$ such that $\|x\| \lt r \lt R$. Fix $0 \lt q \lt 1-\frac{r}{R}$, so $\frac{1}{1-q} \frac{r}{R} \lt 1$. Then $y = \frac{R}{r}x \in B_{R}(0) \subset F_N = \overline{A_N}$, so there is $y_{0} \in A_N$ such that $\|y-y_0\| \lt qR$, so $q^{-1}(y-y_0) \in B_R$. Now choose $y_1 \in A_N$ with $\|q^{-1}(y-y_0) - y_1\| \lt q R$, so $\|(y-y_0 - qy_1)\| \lt q^2 R$. By induction we obtain a sequence $(y_k)_{k=0}^\infty \subset A_N$ such that $$\left\| y - \sum_{k=0}^n q^k y_k\right\| \lt q^n R \quad \text{for all }n \geq 0,$$ hence $y = \sum_{k=0}^\infty q^k y_k$. Observe that by construction $\|y_n\| \leq R + qR$ for all $n$, so the series $y = \sum_{k=0}^\infty q^k y_k$ is absolutely convergent. But then the countable subadditivity hypothesis on $p$ implies that $$p(y) = p\left(\sum_{k=0}^\infty q^k y_k\right) \leq \sum_{k=0}^\infty q^k p(y_k) \leq \frac{1}{1-q} N$$ and thus $p(x) \leq \frac{r}{R} \frac{1}{1-q} N \lt N$ which means $x \in A_N$, as we wanted.

Added: A version of this answer appeared on the Mathematics Community Blog. Thanks to Norbert and others for their efforts.

• Thank you very much for your time, t.b.! Your answer looks really interesting. I will study it on Friday. Commented May 2, 2012 at 3:03
• A similar approach is used in "Functional Analysis: an introduction" by Eidelman, Milman, Tsolomitis: books.google.it/… The authors employ the notion of "perfect convexity" and a theorem by Livshic, which states that for perfectly convex sets the two notions of algebraic interior (or centrum) and topological interior coincide. Zabreiko's lemma is a corollary of this: books.google.it/… Commented May 2, 2012 at 17:57
• Of course, owing to the principe de la conservation de l'emmerdement (I hope I have written that right!), use of Baire's lemma is not avoided. Indeed, Livshic's theorem depends heavily on it. Commented May 2, 2012 at 18:00
• @Giuseppe: Thanks a lot for this. Never heard of that principle, but there certainly is some truth to it... I still hope to find a gliding hump argument, but this involves some emmerdement, too :)
– t.b.
Commented May 2, 2012 at 18:02
• @t.b.: Can you make precise what exactly you mean by "stronger than the usual consequences of the Baire category theorem"? Commented Oct 6, 2012 at 22:07