Showing that $\bigg\| x - \sum_{k=1}^n \lambda_ke_k\bigg\| \geq \bigg\|x-\sum_{k=1}^n\langle x,e_k\rangle e_k \bigg\|$

Question

Exercise :

Let $\{e_1,e_2,\dots, e_n\}$ be an orthonormal set over the Hilbert space $H$. Show that : $$\bigg\| x - \sum_{k=1}^n \lambda_ke_k\bigg\| \geq \bigg\|x-\sum_{k=1}^n\langle x,e_k\rangle e_k \bigg\|$$ for every $\lambda_1,\lambda_2,\dots,\lambda_n \in \mathbb R$.

Attempt :

Let $x \in H$ and $\varepsilon > 0$. Then, we can find $\lambda_1,\dots, \lambda_n \in \mathbb R$ such that :

$$\bigg\| x - \sum_{k=1}^n \lambda_ke_k\bigg\| < \varepsilon$$

But the element $w = x-\sum_{k=1}^n\langle x,e_k\rangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = \langle e_1,\dots, e_n \rangle$ and then it will be

$$\bigg\| x-\sum_{k=1}^n\langle x,e_k\rangle e_k \bigg\| \leq \bigg\| x - \sum_{k=1}^n \lambda_ke_k\bigg\| < \varepsilon$$

and thus the inequality which we were asked to show is proven.

Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !

Answer 1

Notice that $r = x - \sum_{i=1}^n \langle x, e_i \rangle e_i$ is orthogonal to each vector $e_i$. If $\lambda_1,\lambda_2,\ldots, \lambda_n \in \mathbb R$, then \begin{align} \| x - \sum_i \lambda_i e_i \|^2 &= \|x - \sum_i \langle x, e_i \rangle e_i + \sum_i \langle x, e_i \rangle e_i - \sum_i \lambda_i e_i \|^2 \\ &= \| r + \sum_i (\langle x, e_i \rangle - \lambda_i) e_i \|^2 \\ &= \|r\|^2 + \sum_i | \langle x, e_i \rangle - \lambda_i | \|e_i\|^2 \qquad \text{(by Pythagorean theorem)}\\ & \geq \|r\|^2. \end{align}