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I'm in a Hilbert space $H$ and for $z,v, h \in H$ and $t \in \mathbb C$ I have

$$ \|z\|^2 \leq \|h−(tv+y)\|^2 = \|z−tv\|^2 =\|z\|^2 −2\Re(t⟨v,z⟩)+|t|^2\|v\|^2$$

According to my notes it follows from this that $\Re(t⟨v,z⟩) = 0$ for all $t$. How does that follow? I can't seem to show it. Thanks for your help.

1 Answers 1

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We get that for all $t\in\Bbb C$, $$2\Re(t\langle v,z\rangle)\leq |t|^2\lVert v\rVert^2.$$ For an integer $n$, replacing $t$ by $\frac tn$, we get: $$\frac 2n\Re(t\langle v,z\rangle)\leq \frac 1{n^2}|t|^2\lVert v\rVert^2,$$ hence $$2\Re(t\langle v,z\rangle)\leq \frac 1n|t|^2\lVert v\rVert^2.$$ Letting $n\to +\infty$, we show the result.

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    This is awesome! That's a cool trick. Thank you very much!2012-07-31
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    @Matt Yes, you are right. I will edit it.2012-07-31
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    Ok. : ) (I removed my now obsolete comment.)2012-07-31
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    Couldn't you also take $t$ real, cancel a factor of $t$, and take the limit as $t$ tends to zero?2012-07-31