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Let $x$ and $y$ be two vectors in a Hilbert space $H$.Prove that $\left\|x+cy\right\|\geq\left\|x\right\|$ for all complex number $c$ if and only if $x$ and $y$ are orthogonal.

It's easy to show that if $x$ and $y$ are orthogonal,then the inequality is valid.For the converse conclusion, I think we can choose some special value of $c$ to obtain that $<x,y>=0$. But I don't know how to choose some proper $c$ to get result.

Jack
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1 Answers1

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To elaborate on the suggestion in the comments:

Assume $A=\langle x, y \rangle \ne 0$. After replacing $y$ with $\alpha\, y$, with non-zero $\alpha \in \mathbb C$, you can assume that $A$ is (non-zero) real. At that point, square the inequality, and expand the lhs - you may take $c$ to be real to find your desired contradiction.

Hope this helps.

peter a g
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  • When I expand lhs, I get that $2c<x,y>+c^{2}<y,y>\geq 0$.But I can't find the contradiction. Can you help me?Thank you in advance. – Jack Jul 21 '16 at 00:59
  • under the assumptions of the above hint, the lhs of what you've just written is a real quadratic in $c$, with two distinct real roots (since $\langle x, y\rangle$ and therefore also $\langle y,y\rangle$, are non-zero) - one of roots is $c=0$: hence the quadratic must be negative on one side of $c =0$ (viewing $c$ as a real parameter) - you can see this by factoring - contradiction – peter a g Jul 21 '16 at 01:27
  • other version: for small $c$, the lower order term (here, linear) dominates, and it can be both positive and negative, and thus the whole expression can be both positive and negative. Okay? – peter a g Jul 21 '16 at 01:29
  • clearer version of my first comment, perhaps: - with $c$ real: for large $|c|$, the quadratic is indeed positive, but between its two (distinct) roots, the quadratic is negative (which is your desired contradiction). This we can by factoring the quadratic. Okay? – peter a g Jul 21 '16 at 01:45
  • I got it. Thank you for your beautiful explanation! – Jack Jul 21 '16 at 02:30