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 $$\sum_1^{\infty} |a_k|^2= \sum_1^\infty |\langle x,e_k\rangle |^2 = ||x||^2$$


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?

  • 1
    $\begingroup$ Perhaps $x$ isn't of the form $\sum a_ke_k$? $\endgroup$ – Angina Seng May 9 '17 at 17:47
  • $\begingroup$ @LordSharktheUnknown Why wouldn't it be, set all $a_k = 1$ and you'll get $ \sum e_k$ which is a $x \in X$ $\endgroup$ – Dimitris May 9 '17 at 17:51

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.

  • $\begingroup$ I'm not familiar with Parseval's equality yet, sorry if I got it wrong: saying that $\sum a_k e_k = x $ means that $e_k$ must be an orthonormal basis? Or I can have that $\sum a_k e_k =x \in X$ even if $e_k$ isn't a basis, but that can't happen for all $x$? $\endgroup$ – Dimitris May 9 '17 at 17:56
  • $\begingroup$ If you have an arbitrary orthonormal sequence, why would you assume that you can write $x=\sum_ka_ke_k$? Of course, if $x$ is in the closure of the span of $\{e_k\}$, then the equality will hold, as you show. $\endgroup$ – Martin Argerami May 9 '17 at 17:58
  • $\begingroup$ I see, well I don't see why I shouldn't assume that. $\sum a_k e_k$ is always an element of X, since X is a vector space. $\endgroup$ – Dimitris May 9 '17 at 17:59
  • $\begingroup$ I don't assume that a random $x$ can be written that way, I say that the sum is some $x$ in $X$ $\endgroup$ – Dimitris May 9 '17 at 18:03
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    $\begingroup$ But then you are using $x$ for two diferent things; the theorem says "for every $x\in X$". Let's do this: why don't you take $X=\mathbb C^3$, with the usual inner product, and take the orthonormal sequence to be $\{e_1,e_2\}$, and $x=(1,2,3)$. $\endgroup$ – Martin Argerami May 9 '17 at 18:05

Bessel's inequality is equality for some $x$ iff $x$ is in the closed linear span $\overline{[\{ e_k \}]}$ of the orthonormal elements $\{ e_n \}$. You don't have to assume that $\sum_k \langle x,e_n\rangle e_n$ converges to $x$, but that is the final conclusion if $x \in \overline{[\{ e_k \}]}$. This conclusion follows because $$\|x-\sum_{k=1}^{n}\langle x,e_n\rangle e_n\| =\inf_{\alpha_1,\cdots,\alpha_n} \|x-\sum_{k=1}^{n}\alpha_k e_k\|$$


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