How to show that $\sum_{i=1}^n | \langle f, f_i\rangle |^2 \leq \Vert f \Vert^2$

Question

If the set $\{f_1, ..., f_n\}$ is an orthonormal subset of inner product space $E$ and $f\in E$ then how can I show that:

$$\sum_{i=1}^n | \langle f, f_i\rangle |^2 \leq \Vert f \Vert^2.$$

How can I prove this inequality? What is the key step to notice?

Answer 1

Quoting from the wikipedia page for Bessel's inequality, with $f=x$ and $f_i=e_i$, the key point is:

$$0 \le \left\| x - \sum_{k=1}^n \langle x, e_k \rangle e_k\right\|^2 = \|x\|^2 - 2 \sum_{k=1}^n |\langle x, e_k \rangle |^2 + \sum_{k=1}^n | \langle x, e_k \rangle |^2 = \|x\|^2 - \sum_{k=1}^n | \langle x, e_k \rangle |^2$$