# Pythagoras theorem with infinitely many orthonormal vectors in a Hilbert space

My professor says that if $$\{e_k\}^\infty_{k=1}$$ is an orthonormal and $$f=\sum^ \infty_{k=1}a_ke_k$$ for some coefficients $$a_k$$, then $$\lVert f\rVert^2=\sum|a_k|^2$$ and this is a simple consequence of the Pythagoras theorem. I think this is incorrect.

• $||f||^2 = (f,f) = (\sum {a_i e_i}, \sum {a_i e_i})$ Dec 15, 2022 at 10:21
• Check out this MIT lecture, Pr. Zwiebach derives the result you ask properly with the Kronecker function ($\delta_{i,j}$) : youtube.com/… Dec 15, 2022 at 11:35
• @niobium This is an infinite series and we don't even have $a_k=\langle f,e_k\rangle$.
– user912011
Dec 16, 2022 at 0:12
• Yes we do have $a_k = (f,e_k)$ try to do the dot product, there is a lot of cancellation since the $e_i$ vectors are orthonormal... Dec 17, 2022 at 9:54

Define $$f_n = \sum_{k=1}^n \alpha_i e_i$$. As $$(e_i,e_j)=\delta_{i,j}$$, by Pythagoras we have that $$||\alpha_i e_i+ \alpha_j e_j||^2=||\alpha_i e_j||^2 + ||\alpha_j e_j||^2$$. Extending this $$n$$ times we find \begin{align*}||f_n||^2 & = ||\sum_{i=1}^n \alpha_i e_i||^2 \\ &= \sum_{i=1}^n ||\alpha_i e_i||^2 \\ &= \sum_{i=1}^n |\alpha_i|^2 ||e_i||^2 \\ &=\sum_{i=1}^n |\alpha_i|^2 \end{align*} Then, as $$f_n \to f$$ we have that $$||f_n|| \to ||f||$$ so $$||f||^2 = \lim_n ||f_n||^2 = \lim_n \sum_{i=1}^n |\alpha_i|^2 = \sum_{i=1}^{\infty} |\alpha_i|^2$$
The other method, which I think is being hinted at in the comments relies on the fact that if $$x_n \to x$$ and $$y_n \to y$$ then $$(x_n,y_n) \to (x,y)$$. From this we can derive the following \begin{align*} ||f||^2 &= (\sum_{i=1}^{\infty} \alpha_i e_i , \sum_{j=1}^{\infty} \alpha_j e_j) \\ &= (\lim_n \sum_{i=1}^{n} \alpha_i e_i , \lim_n \sum_{j=1}^{n} \alpha_j e_j) \\ &= \lim_n (\sum_{i=1}^{n} \alpha_i e_i , \sum_{j=1}^{n} \alpha_j e_j) \\ &= \lim_n \sum_{i=1}^n \alpha_i (e_i , \sum_{j=1}^{n} \alpha_j e_j) \\ &= \lim_n \sum_{i=1}^n \sum_{j=1}^n \alpha_i \bar{\alpha}_j (e_i , e_j) \\ &= \lim_n \sum_{i,j=1}^n \alpha_i \bar{\alpha}_j \delta_{i,j} \\ &= \lim_n \sum_{i=1}^n \alpha_i \bar{\alpha}_i \\ &= \sum_{i=1}^{\infty} |\alpha_i|^2 \end{align*}