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Definition of the problem

I have to prove the following statement:

Let $\left(E,\left\langle \cdot,\cdot\right\rangle \right)$ be an inner product space over $\mathbb{R}$. prove that for all $x,y\in E$ we have $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert . $

My efforts

I tried two different ways to prove that, both unsuccessfull..

First:

First, by squaring the whole inequality:

$ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $ We have from Cauchy-Schwarz that $ \left|\left\langle x,y\right\rangle \right|\leq\left\Vert x\right\Vert \cdot\left\Vert y\right\Vert $ So we obtain $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}=\left(\left\Vert x\right\Vert ^{2}+\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $ By Pythagorean theorem, we obtain $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x+y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}. $ We're almost there, except an extra term very annoying: $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert ^{3}\left\Vert y\right\Vert ^{3}. $

Second

I tried after to use only the Cauchy-Schwarz inequality, not squared: $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert \left\Vert y\right\Vert . $

My question

Could you give me a hint/idea on how to solve this problem? which Lemma/Theorem should I use?

Thank you

Franck

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    However, ||x+y|| || <= (||x|| + ||y||) ||x|| ||y|| does hold due to Cauchy-Schwarz, a$n$d the triangle inequality (for normed spaces)2012-06-12

1 Answers 1

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Then, let me remove this question from the dead list of "unanswered questions" by answering it.

The statement is false. A counter-example is as follows. Let $E$ be $\mathbb{R}$ itself, and the inner product be the ordinary multiplication of real numbers. Let $x = 1$ and $y = -1$. Then the left hand side is $(||x||+||y||) \cdot | \langle x, y \rangle| = (1 + 1) 1 = 2.$ The right hand side is $||x + y|| \cdot ||x|| \cdot ||y|| = ||0|| \cdot 1 \cdot 1 = 0$.

  • 0
    For the ones interested, I posted a new question with the corrected statement: $ \left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert. $ [http://math.stackexchange.com/questions/158901/norms-on-inner-product-space-over-mathbbr]2012-06-16