1. ## Bessel's Inequality

I'm having trouble with this problem. I've tried rewriting things in about every way I know how to but I haven't arrived at an answer. I'd appreciate some help.

"Let $\displaystyle V$ be an inner product space and let $\displaystyle S=\{v_1, ..., v_n\}$ be an orthonormal subset of $\displaystyle V$. Prove that for any $\displaystyle x \in V$ we have $\displaystyle ||x||^2 \ge \displaystyle\sum^n_{i=1} |\langle x,v_i \rangle |^2.$"

As a hint the book says to use the fact that $\displaystyle x \in V$ can be written uniquely as $\displaystyle w+w'$ where $\displaystyle w \in W=\mbox{span}(S)$ and $\displaystyle w' \in W^\perp$, the orthogonal complement of W, and use the fact that for $\displaystyle x$,$\displaystyle y$ orthogonal, $\displaystyle ||x+y||^2 = ||x||^2 + ||y||^2$.

Thanks for any help.

2. ## Re: Bessel's Inequality

Originally Posted by AlexP
I'm having trouble with this problem. I've tried rewriting things in about every way I know how to but I haven't arrived at an answer. I'd appreciate some help.

"Let $\displaystyle V$ be an inner product space and let $\displaystyle S=\{v_1, ..., v_n\}$ be an orthonormal subset of $\displaystyle V$. Prove that for any $\displaystyle x \in V$ we have $\displaystyle ||x||^2 \ge \displaystyle\sum^n_{i=1} |\langle x,v_i \rangle |^2.$"

As a hint the book says to use the fact that $\displaystyle x \in V$ can be written uniquely as $\displaystyle w+w'$ where $\displaystyle w \in W=\mbox{span}(S)$ and $\displaystyle w' \in W^\perp$, the orthogonal complement of W, and use the fact that for $\displaystyle x$,$\displaystyle y$ orthogonal, $\displaystyle ||x+y||^2 = ||x||^2 + ||y||^2$.

Thanks for any help.
As the hint gives

$\displaystyle \mathbf{x}=\mathbf{x}_{v}+\mathbf{x}_{v\perp}$

Now

$\displaystyle ||\mathbf{x}||^2=<\mathbf{x},\mathbf{x}>=<\mathbf{ x}_{v}+\mathbf{x}_{v\perp},\mathbf{x}_{v}+\mathbf{ x}_{v\perp}>=||\mathbf{x}_v||^2+||\mathbf{x}_{v \perp}||^2$

Remember that

$\displaystyle \mathbf{x}_{v}=\sum_{i=1}^{n}<\mathbf{x},\mathbf{v }_i>\mathbf{v}_i \implies ||\mathbf{x}_v||^2=\sum_{i=1}^{n}|<\mathbf{x}, \mathbf{v}_i >|^2$

Just put these two facts together and remember that the modulus of a vector is always positive to finish.

3. ## Re: Bessel's Inequality

Wow. I was so close to completing it that it's painful that I didn't see it. Thanks.