I'm trying to use the closed graph theorem, i.e, I'm trying to prove that $Graph(T) = \{ (x,Tx) ; ~x \in E \}$ is closed in $E \times $F, but I'm a little bit confused. So, i'd like some help in proving it. Here, given $f$ function, $\langle f,x \rangle = f(x)$. Thanks in advance.

  • 3
    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ Oct 3, 2015 at 15:31
  • $\begingroup$ Is $E$ assumed to be a Banach space? Or are you looking at a more general situation? $\endgroup$ Oct 3, 2015 at 15:45
  • $\begingroup$ Yes, E is banach, Sorry for not saying it. :) $\endgroup$
    – Bohrer
    Oct 3, 2015 at 16:01

1 Answer 1


Let $(x_n)$ be sequence tending to $0$ in $E$ and such that $(Tx_n)$ has a limit $L\in E^*$. One has to show that $L=0$.

Fix any $y\in E$. For any $n\in\mathbb N$ and every $\lambda\in\mathbb R$ we have $\langle T(x_n+\lambda y),x_n+\lambda y\rangle\geq 0$, i.e. $$\langle Tx_n,x_n\rangle+\lambda \,\bigl( \langle Ty, x_n\rangle +\langle Tx_n,y\rangle\bigr)+\lambda^2\,\langle Ty,y\rangle\geq 0\, .$$

Now, $\langle Tx_n,x_n\rangle\to 0$ since $x_n\to 0$ and $(Tx_n)$ is bounded; and likewise $\langle Ty,x_n\rangle\to 0$. So we get $$\lambda\, \langle L,y\rangle+\lambda^2\,\langle Ty,y\rangle\geq0\qquad\hbox{for every $\lambda\in\mathbb R$}\, ; $$ and hence $$ \langle L,y\rangle+\lambda\, \langle Ty,y\rangle\geq 0\qquad\hbox{for every $\lambda>0$}\, . $$ Letting $\lambda\to 0^+$, it follows that $\langle L,y\rangle\geq 0$ for every $y\in E$; and finally $\langle L,y\rangle =0$ by applying this to $-y$. So $L=0$ as required, and hence $T$ is continuous by the closed graph theorem.

  • $\begingroup$ You're welcome! $\endgroup$
    – Etienne
    Oct 4, 2015 at 17:09
  • $\begingroup$ This is a very nice proof. $\endgroup$ Feb 6, 2021 at 2:03

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