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Let $H$ be inner product space with inner product $\langle\cdot,\cdot\rangle$ and norm $\lVert \cdot\rVert$. Let $x,y \in H$. Would you help me to prove that $\langle x,y\rangle=0$ if and only if $\lVert x+\alpha y\rVert \geq\lVert x\rVert$ for all scalar $\alpha$?

I have proved that $\langle x,y\rangle=0$ implies $\lVert x+\alpha y\rVert\geq \lVert x\rVert$, but don't know how to prove the converse. I just try to show that $\lVert x+\alpha y\rVert\geq \lVert x\rVert$ implies $\lVert x+\alpha y\rVert=\lVert x-\alpha y\rVert$ (but still not success) since this equality equivalent with $\langle x,y\rangle=0$.

Is there any solution of this problem using this information: ∥x+αy∥=∥x−αy∥ if and only if =0 ? Thanks.

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  • $\begingroup$ The norm $|| \cdot||$ is generated by inner product $<.,.>$? $\endgroup$ – M. Strochyk Sep 17 '12 at 20:14
  • $\begingroup$ Yes, ||⋅|| is generated by inner product <.,.> $\endgroup$ – beginner Sep 17 '12 at 20:31
  • $\begingroup$ Is there any solution of this problem using this information: ∥x+αy∥=∥x−αy∥ if and only if <x,y>=0 ? $\endgroup$ – beginner Sep 17 '12 at 20:37
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Taking the squares and expanding, we get for all real number $\alpha>0$: $$\lVert x\rVert^2+\alpha\langle x,y\rangle+\alpha\overline{\langle x,y\rangle}+\alpha^2\lVert y\rVert^2\geq \lVert x\rVert^2,$$ hence $$\langle x,y\rangle+\overline{\langle x,y\rangle}+\alpha\lVert y\rVert^2\geq 0.$$ Taking $\alpha\to 0$, we get $2\Re\langle x,y\rangle\geq 0$. Working with $-x$ and $-\alpha$, we get $2\Re\langle x,y\rangle\leq 0$. For the imaginary part, work with $e^{i\theta}\beta=:\alpha$ where $\beta\in \Bbb R$ and $\theta$ such that $e^{i\theta}\langle x,y\rangle$ is a real number.

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  • $\begingroup$ Is there any solution of this problem using this information: $\lVert x+\alpha y\rVert=\lVert x-\alpha y\rVert$ if and only if $<x,y>=0$ ? $\endgroup$ – beginner Sep 17 '12 at 20:29
  • $\begingroup$ I don't know. I'm not sure this approach will make the problem easier, as the equality seems hard to show only using the inequality. $\endgroup$ – Davide Giraudo Sep 17 '12 at 20:45
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The implication you mention is not true: take any nonzero $x$, $\alpha=1$, $y=x$. Then $\|x+\alpha y\|=2\|x\|>\|x\|$. But $\|x-\alpha y\|=0\ne 2\|x\|=\|x+\alpha y\|$.

The inequality $\|x+\alpha y\|\geq\|x\|$ forces $\langle x,y\rangle =0$ when it holds for every $\alpha$, and then the canonical way to obtain the result is what Davide did in his solution. But no conclusion can be drawn from the fact that it holds for a single concrete $\alpha$.

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