Show that : $\|x\|^{2}\ge\sum\limits_{i=1}^{n}|\langle x,x_{i}\rangle |^2$

Question

Let $\mathbb{H}$ be a Hilbert space and $x_{1},x_{2},x_{3},...,x_{n}$ orthogonal vectors from $\mathbb{H}$ , $x\in\mathbb{H}$ then prove that :

$$\|x\|^{2}≥\displaystyle\sum_{i=1}^{n}|\langle x,x_{i}\rangle |^{2}$$

For all $x\in\mathbb{H}$

I don't know how I start in the proof, I don't have any hints in this type of questions. If someone know a book or PDF where I can find like this problem?

Answer 1

Hint: Write $$x=x_0 +\sum_{j=1}^n \langle x, x_j \rangle x_j$$ where $x_0$ is in the orthogonal complement of $\operatorname{span}\{ x_1, \dots, x_n \}$. Then use $\Vert x \Vert^2 =\langle x, x \rangle $ and use that $x_j\perp x_i$ for $i,j\in \{0, \dots , n\}$ and $i\neq j$.

Answer 2

We wish to show that $$\frac{\sum_{i=1}^n|\langle x,x_i\rangle|^2}{\|x\|^2}\leq 1$$ $$LHS=\frac{\sum_{i=1}^nx^Tx_ix_i^Tx}{\|x\|^2}$$ $$=\frac{x^T(\sum_{i=1}^nx_ix_i^T)x}{\|x\|^2}$$ $$=\frac{x^TAx}{x^Tx}$$ where $A$ is an idempotent matrix.

EDIT 1 We know that maximum value of $\frac{x^TAx}{x^Tx}$ is absolute value of the maximum eigen value of $A$, which is $1$ here, because $A$ is an idempotent matrix.

EDIT 2 I am assuming that the vectors $x_i,1\leq i\leq n$ are orthonormal, the proposition is not true otherwise, as noted in comments by Peter Melech and Severin Schraven. Also, the above argument is valid only for finite dimensional Hilbert Space.

Answer 3

Noting $$ 0\le \|x-\sum_{i=1}^n\left<x,x_i\right>x_i\|^2=\|x\|^2-2\left<x,\sum_{i=1}^n\left<x,x_i\right>x_i\right>+\|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2 $$ one has $$ 2\sum_{i=1}^n\left<x,x_i\right>^2\le\|x\|^2+\|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2. $$ Since $$ \|\sum_{i=1}^n\left<x,x_i\right>x_i\|^2=\sum_{i=1}^n\left<x,x_i\right>^2 $$ one has $$ \sum_{i=1}^n\left<x,x_i\right>^2\le\|x\|^2. $$