Weak convergence of orthonormal sequences in Hilbert space

[kisengue]

So, I've found the result that orthonormal sequences in Hilbert spaces always converge weakly to zero. I've only found wikipedia's "small proof" of this statement, though I have found the statement itself in many places, textbooks and such.

I've come to understand that this property follows from the Bessel inequality, and I've worked out many of the details, so I feel that I understand the Bessel inequality itself quite well. What I don't get is how the inequality gives us the weak convergence - the proof on wikipedia only states that "Therefore, [tex] |\langle e_n, x \rangle |^2 \rightarrow 0 [/tex]" after stating the Bessel inequality. It doesn't make sense to me - how is this information gleaned?

 

[Fredrik]

Can you at least link to the Wikipedia page? Are you asking how to see that ##|\langle e_n, x \rangle |^2 \rightarrow 0##, or how to see that this means that ##e_n\to 0## with respect to the weak topology?

 

[micromass]

You do know that if the series

[tex]\sum_{n=0}^{+\infty}{a_n}[/tex]

converges, that [itex]a_n\rightarrow 0[/itex]??

 

[kisengue]

The proof (sorry for not linking it immediately). Fredrik, I'm asking the first of those two - the second I understand.

Micromass: I didn't think of that... but of course. Of course. Damn it. Now I get it, I think.

 

[blue_raver22]

First of all, congratulations on your progress in understanding the Bessel inequality and its implications in Hilbert spaces. It is a fundamental result in functional analysis that has many important applications.

To answer your question, let's start by recalling the definition of weak convergence. In a Hilbert space, a sequence of vectors {x_n} converges weakly to a vector x if for any y in the Hilbert space, the inner product \langle x_n, y \rangle converges to \langle x, y \rangle. In other words, the sequence converges weakly if its inner product with any vector in the Hilbert space converges to the inner product with the limit vector.

Now, let's look at the Bessel inequality. It states that for any orthonormal sequence {e_n} in a Hilbert space, the sum of the squared inner products |\langle e_n, x \rangle |^2 is bounded above by the squared norm of x. This means that for any vector x, the inner products with the orthonormal sequence {e_n} cannot grow too large. In fact, the sum of these inner products is always smaller than the squared norm of x.

So, how does this help us understand weak convergence? Let's take a closer look at the statement on Wikipedia: "Therefore, |\langle e_n, x \rangle |^2 \rightarrow 0". This means that for any vector x, the inner product with the orthonormal sequence {e_n} tends to zero. In other words, the inner products get closer and closer to zero as n increases. And as we established earlier, this is exactly what we need for weak convergence.

Think of it this way - the Bessel inequality tells us that the inner products with the orthonormal sequence {e_n} are bounded above by the squared norm of x. And since this bound tends to zero, it means that the inner products themselves must also tend to zero. This is what we mean by weak convergence.

I hope this explanation helps you understand how the Bessel inequality leads to weak convergence in Hilbert spaces. Keep up the good work in your understanding of functional analysis!