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Let $H$ be an Hilbert space and $T: H \to H$ linear and bounded operator. Suppose $$ \langle Tx, x \rangle \ge \|x\|^2 \quad \quad \text{ for all } x \in H $$

I can prove this implies injectivity of $T$ and closeness of its range. Is it true that $T$ is surjective? In other words, can I prove its range is also dense?

Bremen000
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1 Answers1

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Your claim will follow from the following Lemma

Let $H$ be an Hilbert space, $T:H\rightarrow H$ be a continuous map. Then, if $T^\star$ is injective with closed range then $T$ is surjective

See proof here A proof that $T^*$ injective with closed range implies $T$ is surjective .

You have $\left<x,T^\star x\right> \geq \|x\|^2$ and so your conclusions on $T$ are also true for $T^\star$ (same proof). Therefore, $T^\star$ is injective and has a closed range. Thus, by the lemma $T$ is surjective.

Yanko
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    Thank you very much. I thought the following thing while you were writing: I prove $T(H)^{\perp} = { 0 }$, indeed if $y \in T(H)^{\perp}$ we have $(y,Tx)=0$ for every $x \in H$, in particular $0=(y,Ty) \ge |y|^2$ and this implies the claim. – Bremen000 Apr 18 '19 at 20:39
  • @Bremen000 wow that's so much simpler. I guess this is the direct way without going through $T^\star$. – Yanko Apr 18 '19 at 20:41