# Example of use of closed graph theorem over open mapping theorem and bounded inverse theorem

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?

• I don't think there's any need to be dogmatic here. CGT is "better" when its conclusion fits the statement you are proving, or when you find that you can give a nicer / simpler / clearer / shorter / easier proof using it than with the alternatives. With practice you will start to encounter instances where this happens. Aug 2 at 5:53
• Btw you can formulate CGT without product spaces. An equivalent definition of "closed operator" is "for every convergent sequence $x_n$ such that $A x_n$ converges, we have that $\lim A x_n = A \lim x_n$". Then CGT again says that every closed operator whose domain is a Banach space, is a bounded operator. So in other words, when trying to prove that an operator is bounded, you can help yourself to the stronger assumption that $A x_n$ converges. Aug 2 at 5:57

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.