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Let $E$ be a Banach space and let $T:E\mapsto E'$ be a linear operato satisfying $\langle Tx,x\rangle\geq0$ for all $x\in E$. How to prove that the graph of $T$ is closed?

Thom Kostova
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1 Answers1

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Suppose that $x_n\to x$ in $E$ and $Tx_n\to f$ in $E'$. Beucase $T$ is Monotone, we have that $$\langle Tx_n-Ty,x_n-y\rangle\ge 0,\ \forall \ y\in E\tag{1}$$

If we pass the limit in $(1)$ we get that $$\langle f-Ty,x-y\rangle\ge 0,\ \forall\ y\in E\tag{2}$$

Now take $y=x+tv$ where $t\in \mathbb{R}$ and $v\in X$. We have that $$\langle f-Tx-tTv,-tv\rangle\geq 0,\ \forall\ t\in\mathbb{R},\ v\in E\tag{3}$$

We get from $(3)$ that $-t\langle f-Tx, v\rangle\geq-t^2\langle Tv,v\rangle$ or equivalently $$\langle f-Tx, v\rangle\leq t\langle Tv,v\rangle,\ \forall\ t> 0,\ v\in E \tag{4}$$

Now it is straightforward to conclude from $(4)$.

Tomás
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  • Hi Tomas, what is $E'$ and the pairing in the question? –  Nov 03 '13 at 00:00
  • Hi @Sanchez, $E'$ is the dual of $E$ and $\langle\cdot,\cdot\rangle$ denotes duality. – Tomás Nov 03 '13 at 00:00
  • thanks! I was confused when I first read the question. –  Nov 03 '13 at 00:02
  • I don't follow. I thought maybe you were trying to show something like symmetry $\langle Tx,y \rangle = \langle Ty,x \rangle$ first? Is that what you have in mind? Though I guess that might not even hold in the case of real scalars... – Mike F Nov 04 '13 at 17:59
  • Dear @Mike, My goal is to show that $Tx=f$. For this purpose we can choose a suitable $y\in E$. – Tomás Nov 04 '13 at 18:27
  • I see. Well, if at some point in the future you feel inclined to post the details, I'd be interested in seeing them. :) – Mike F Nov 06 '13 at 00:32
  • @Mike, tomorrow I post the details. – Tomás Nov 06 '13 at 00:36
  • @Mike, I have added additional details. – Tomás Nov 06 '13 at 12:20