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Let $V$ be a real inner product space, and let $v_1,v_2, \dots ,v_k$ be an orthonormal set of vectors. How do you prove that

$\sum_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot\|y\|?$

When does the equality hold?

I've been trying to do this with the Bessel and Cauchy-Schwarz inequalities, but I can't make it work yet. Any help would be greatly appreciated.

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    I have used cardinal's hint in my answer (although I don't mention it explicitly). I hope that that clears things up.2011-11-20

1 Answers 1

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We don't necessarily know that $\{v_i:1\le i\le k\}$ is a basis of $V$ since we don't know the dimension of $V$, but we are given that $\{v_i\}$ are orthonormal; that is $\left=0$ when $i\not=j$ and $\left=1$.

Consider the vector $ x^\perp=x-\sum_{i=1}^k\leftv_i\tag{1} $ $x^\perp$ is perpendicular to $\{v_i\}$: $ \begin{align} \left &=\leftv_i,v_j\right>\\ &=\left-\left\left\\ &=0\tag{2} \end{align} $ Therefore, $x^\perp$ is perpendicular to $x-x^\perp=\sum\limits_{i=1}^k\leftv_i$.

Next consider $ \begin{align} \left &=\left<\sum_{i=1}^k\leftv_i,\sum_{j=1}^k\leftv_j\right>\\ &=\sum_{i=1}^k\left\left\left\\ &=\sum_{i=1}^k\left\left\tag{3} \end{align} $ Note that since $x^\perp$ is perpendicular to $x-x^\perp$, $ \|x-x^\perp\|^2+\|x^\perp\|^2=\|x\|^2\tag{4} $ which implies that $\|x-x^\perp\|\le\|x\|$.

Now $(3)$, Cauchy-Schwarz, and $(4)$ yield $ \begin{align} \left|\sum_{i=1}^k\left\left\right| &=\left|\left\right|\\ &\le\|x-x^\perp\|\|y-y^\perp\|\\ &\le\|x\|\|y\|\tag{5} \end{align} $ To finish off the proof (thanks to cardinal), consider the vector $ x^+=\sum_{i=1}^k\left|\left\right|v_i\tag{6} $ Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left^2=\|x-x^\perp\|^2$.

Plugging $x^+$ and $y^+$ into $(5)$ gives $ \begin{align} \sum_{i=1}^k\left|\left\left\right| &=\left|\sum_{i=1}^k\left\left\right|\\ &\le\|x^+\|\|y^+\|\\ &\le\|x\|\|y\|\tag{7} \end{align} $

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    This is a very nice answer.2011-11-20