# Proving $\sum\limits_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot \|y\|$

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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Hint: $\langle x, v_i\rangle$ are coordinates. – cardinal Nov 20 '11 at 15:51
Use cardinal's excellent hint, and look again at Cauchy-Schwarz‌​. – robjohn Nov 20 '11 at 15:57
@robjohn: I'm sorry, but I am still very confused! – Freeman Nov 20 '11 at 17:08
@cardinal: How do you calculate the coefficient of $v_i$ in the unique linear combination of $v_1,\dotsc,v_k$ which equals $v$? – user3533 Nov 20 '11 at 18:01
I have used cardinal's hint in my answer (although I don't mention it explicitly). I hope that that clears things up. – robjohn Nov 20 '11 at 18:05

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 $\left<v_i,v_j\right>=0$ when $i\not=j$ and $\left<v_i,v_i\right>=1$.

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

Next consider \begin{align} \left<x-x^\perp,y-y^\perp\right> &=\left<\sum_{i=1}^k\left<x,v_i\right>v_i,\sum_{j=1}^k\left<y,v_j\right>v_j\right>\\ &=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\left<v_i,v_i\right>\\ &=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\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<x,v_i\right>\left<y,v_i\right>\right| &=\left|\left<x-x^\perp,y-y^\perp\right>\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 $$x^+=\sum_{i=1}^k\left|\left<x,v_i\right>\right|v_i\tag{6}$$ Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left<x,v_i\right>^2=\|x-x^\perp\|^2$.

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

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...and, to conclude, note that this holds for arbitrary $x$ and $y$, hence, it holds for $x^+ = \sum_{i=1}^k |\langle x, v_i \rangle | v_i$ and similarly defined $y^+$. Noting that $\|x^+\| \leq \|x\|$ and $\|y^+\| \leq \|y\|$ completes the proof. (+1) – cardinal Nov 20 '11 at 18:10
@cardinal: isn't $(5)$ exactly what was asked? In fact, $x^+=x-x^\perp$ so I think your concerns are already handled. – robjohn Nov 20 '11 at 18:13
The modulus is on the inside of the sum and $x^+ \neq x - x^\perp$ in general. – cardinal Nov 20 '11 at 18:15
@LHS: Pythagoras. (Or, just check $\langle x^\perp, x-x^\perp \rangle = 0$ by the definitions.) – cardinal Nov 20 '11 at 18:27
This is a very nice answer. – cardinal Nov 20 '11 at 18:39