The closed graph theorem, the open mapping theorem, and the bounded inverse theorem are equivalent. I have used the open mapping theorem and the bounded inverse theorem when solving problems, however, I have never used the closed graph theorem. It seems like a less intuitive theorem to me (mainly because it involves a product space). So when is it 'better' to used the closed graph theorem rather than the other two equivalent theorems?
Example of use of closed graph theorem over open mapping theorem and bounded inverse theorem
functional-analysisreal-analysis
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The proof linked to by David is the standard one, but the write-up strikes me as somewhat clumsy.
Here's my take on this argument:
Suppose $T: X \to Y$ is continuous and onto.
The map $\pi: X \to \bar{X} = X/ \operatorname{Ker}{T}$ is open, onto and continuous. The map $T$ factors over $\pi$ via a continuous linear bijection $\bar{T}: \bar{X} \to Y$ by definition of the quotient topology.
By continuity of $\bar{T}$ the graph of $\bar{T}$ is closed. Switching coordinates $(\bar{x},y) \mapsto (y,\bar{x})$ is a homeomorphism $\bar{X} \times Y \to Y \times \bar{X}$ and it maps the graph of $\bar{T}$ to the graph of $\bar{T}^{-1}$, so the graph of the linear map $\bar{T}^{-1}$ is closed, too. By the closed graph theorem $\bar{T}^{-1}$ is continuous.
For all open $U \subset X$ the set $T(U) \subset Y$ is open because it is the pre-image of the open set $\pi(U)$ under the continuous map $\bar{T}^{-1}$, hence $T$ is open.
Note: Step 2. is a proof of the inverse mapping theorem from the closed graph theorem.
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:
- the uniform boundedness principle;
Hint: set $p(x) = \sup_{n \in \mathbb{N}} \|T_n x\|$. - 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 $$ \begin{equation}\tag{$\ast$} B_{R}(0) \subset A_N \end{equation} $$ 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.
Best Answer
I like this application of the closed graph theorem:
Let $X$ be some measure space and $p,q\in[1,\infty]$. If $a$ is a measurable Function with the property that $af\in L^q$ for each $f\in L^p$, then the induced Operator $$M_a:L^p(X)\to L^q(X), f\mapsto af$$ which by assumption is just well defined and nothing more, is automatically bounded.
Proof: Let $f_n$ be a sequence in $L^p(X)$ with $f_n\to f$ in $L^p$ and $M_af_n = af_n\to g$ in $L^q$, then $f_n$ has a subsequence which converges a.e. to $f$ hence $af_{n_k}\to af$ a.e. for some subsequence. On the other hand, also $af_{n_k}\to g$ in $L^q$, hence also convergence for a subsubsequence almost everywhere, hence $af = g$ in $L^q$.
The closed graph theorem is really powerful here because it allows us to only care about pointwise convergence wich is much easier to deal with. As pointed out in the comments, usually when showing continuity with sequences one have to show that for convergent $(f_n)$ with limit $f$, the sequence $M_af_n$ converges to $M_af$. With the GCT we can additionally assume that $M_af_n$ already is convergent to some limit and only need to show equality of limits. This can be very powerful, as seen in this proof.