# Inequality involving inner product and an othonormal set of vectors

$\newcommand{\ip}[2]{\left\langle #1,#2 \right\rangle}$

Here is the statement of the problem:

Suppose that $V$ is a real inner product space with an inner product $\langle\cdot,\cdot\rangle$, and let $\left\{ e_1,\dots,e_k\right\}$ be an orthonormal set of vectors in $V$. For any vector $v\in V$, show that

$$\sum_{i=1}^k \ip{v}{e_i}^2 \le ||v||^2$$

when does the equality hold?

My attempt: I'm bascially stuck from the beginning. All of my approaches indirectly had the (incorrect) assumption that $V$ can be expressed as a direct sum of subsapce of $V$ spanned by $\{e_1,\dots,e_k\}$ and its orthogonal complement, which I know is true for finite dimensional vector spaces. But I spotted that there were no assumption about $V$ being finite dimensional.

But then I couldn't think of any effective approach afterwards. Any help would be appreciated!

First we prove Pythagoras Theorem for inner product space $$\left\|\sum \limits_{i=1}^{k}a_i e_i\right\|^2=\sum \limits_{i=1}^{k}|a_i|^2 \|e_i\|^2 \hspace{5 mm}$$ $e_i$ is the orthonormal base.
So $\sum\limits_{i=1}^{k}|\langle v,e_i\rangle|^2\leqslant \|v\|^2$
By Pythagoras, $\|v\|^2 = \|\sum_i \langle v,e_i\rangle e_i\|^2 + \|v - \sum_i\langle v,e_i\rangle e_i\|^2$.
• So I guess you are assuming that $\sum_i \langle v,e_i\rangle e_i$ and $v-\sum_i \langle v,e_i\rangle e_i$ are orthogonal? But isn't that equivalent to saying, that for any $v\in V$ there is a component of $v$ lying in the space spanned by $\{e_1,\dots,e_k\}$ and a component that is orthogonal to that space, which means $V$ can be expressed as direct sum of these two spaces? But that is not true for infinite dimensional $V$ in general, is it? – user160738 May 29 '15 at 5:40
• We don't have to assume; just compute their inner product and you'll see. This is essentially a proof that $V$ can be so decomposed. The difficulty with infinite-dimensional $V$ would only be if there were infinitely many $e_i$, so that the sums there were actually limits, and maybe the space is not complete. – user21467 May 29 '15 at 5:52