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I read that showing $A$ is bounded by showing its graph is closed is 'more efficient' or 'easier' than showing it is bounded directly. For a specific example, consider the Hellinger-Toeplitz Theorem states that if $H$ is a Hilbert space and $A:H\to H$ is a symmetric linear operator, then $A$ is bounded.

By the Closed Graph Theorem, we can show $A$ is bounded by showing its graph is closed, and the proof is very simple. We take $(x_n)\in H$ such that $x_n \to x$ and $Ax_n \to y$, and then $$ \langle y,z \rangle = \lim_{n\to \infty} \langle Ax_n,z \rangle = \lim_{n\to \infty} \langle x_n, Az \rangle = \langle x, Az \rangle = \langle Ax,z \rangle, $$ for all $z\in H$, and hence $Ax = y$.

Why was proving this theorem using the Closed Graph Theorem easier, than showing it directly?

ManUtdBloke
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1 Answers1

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To prove directly that $A$ is bounded, you have to show that if $x_n\to x$, then $Ax_n\to Ax$. On the other hand, to prove that the graph of $A$ is closed, you only have to show that if $x_n\to x$ and $Ax_n\to y$, then $y=Ax$. The difference is that in the latter case, you get to assume that $(Ax_n)$ does converge to something (namely $y$). The proof you mentioned would not work without this assumption, because you cannot refer to $y$ at all if you don't yet know that the sequence $(Ax_n)$ converges.

Eric Wofsey
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