Proof involving inner products and norms

Question

For sequence of orthonormal functions $(x_n)$ on [a,b] with $f$ Riemann Integrable on [a,b], prove that $\sum\limits_n(<f,x_n>)^2 \leq ||f||_2^2$

Reviewing problems and I'm stuck. Thanks for any help!

Answer 1

Since you are talking about orthonormal functions $\lbrace x_n\rbrace$ and referring to the norm $\|\cdot\|_2$, I have to guess that your Hilbertspace is $L^2(a,b)$ and with $\langle \cdot , \cdot \rangle$ you mean the natural $L^2$-scalar product.

If this is the case, we are talking about Bessel's inequality (check Wikipedia). A proof should be found in any functional analysis book, but for convenience I'll sketch it for you.

Let $\lambda_1,\dots,\lambda_n$ be a system of numbers. We obviously have
\begin{align*} \|x - \sum_{k=1}^n\lambda_k x_k\|^2 \geq 0 \end{align*}

If we calculate the norm with the fact that $\|x\|^2 = (x,x)$ we obtain \begin{align*} 0 \leq\|x - \sum_{k=1}^n\lambda_k x_k\|^2 = \|x\| - \sum_{k=1}^n |(x_k,x)|^2 + \sum_{k=1}^n |\lambda_k - (x_k,x)|^2. \end{align*}

With $\lambda_k = (x_k,x)$ and $n\to \infty$ Bessel's inequality follows.