Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The norm $|| \cdot||$ is generated by inner product $<.,.>$? – M. Strochyk Sep 17 '12 at 20:14
Yes, ||⋅|| is generated by inner product <.,.> – beginner Sep 17 '12 at 20:31
Is there any solution of this problem using this information: ∥x+αy∥=∥x−αy∥ if and only if <x,y>=0 ? – beginner Sep 17 '12 at 20:37
up vote 3 down vote accepted

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.

share|cite|improve this answer
Is there any solution of this problem using this information: $\lVert x+\alpha y\rVert=\lVert x-\alpha y\rVert$ if and only if $<x,y>=0$ ? – beginner Sep 17 '12 at 20:29
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. – Davide Giraudo Sep 17 '12 at 20:45

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 $\alpha$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.