Bessel's Inequality
Let $(X, \langle\cdot,\cdot\rangle )$ be an inner product space and $(e_k)$ an orthonormal sequence in $X$. Then for every $x \in X$ : $$ \sum_{1}^{\infty} |\langle x,e_k\rangle |^2 \le ||x||^2$$
where $\| \cdot\|$ is of course the norm induced by the inner product.
Now suppose we have a sequence of scalars $a_k$ and that the series $$ \sum_{1}^{\infty} a_k e_k = x $$
converges to a $x \in X$.
Lemma 1
We can easily show that $a_k=\langle x,e_k\rangle $
(i'll do it fast)Proof. Denote $s_n$ the sequence of partial sums of the above series, which of course converges to $x$. Then for every $j<n$ , $ \langle s_n, e_j\rangle = a_j$ and by continuuity of the inner product $a_j=\langle x,e_j\rangle $
Lemma 2 We can also show that since $s_n$ converges to $x$, then $σ_n = |a_1|^2 + … + |a_2|^2 $ converges to $\|x\|^2 $ :
Proof. $\|s_n\|^2 = \| a_1 e_1 +…+a_2 e_2\|^2 = |a_1|^2 + … |a_n|^2 $ since $(e_k)$ are orthonormal (Pythagorean). But $||s_n||^2$ converges to $||x||^2 $ , which completes the proof.
So we showed the following $$\sum_1^{\infty} |a_k|^2= \sum_1^\infty |\langle x,e_k\rangle |^2 = ||x||^2$$
Confusion
So the equality holds for Bessel inequality, for $x$. We arbitrarily chose $a_k$, so does that mean the the equality holds for all $x \in X$ ? Obviously not, otherwise it would be Bessel's equality. What am I getting wrong?
Best Answer
What you are missing is that the statement says orthonormal sequence and not orthonormal basis. When the sequence is a basis, you get Parseval's equality. But the inequality holds for "partial sums".
If you already have $\sum_k a_ke_k=x$, then of course you get an equality.