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.

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

1 Answer 1


There are two ways to show this. I assume your issue with the statement has to do with the fact it is an infinite sequence.

Your professor was talking about doing something like this:

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*}


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