Does the Sequence {sin(kx)} Converge Weakly to 0 in L^2(0,1)?

[lmedin02]

Homework Statement

Prove that the sequence [itex]\{sin(kx)\}[/itex] converges weakly to [itex]0[/itex] in [itex]L^2(0,1)[/itex].

Homework Equations

A sequence of elements [itex]\{f_k\}[/itex] in a Banach space [itex]X[/itex] is to converge weakly to an element [itex]x\in X[/itex] if [itex]L(f_k)→L(f)[/itex] as [itex]k→∞[/itex] for each [itex]L[/itex] in the dual of [itex]X[/itex].

The Attempt at a Solution

If the sequence was orthogonal on [itex](0,1)[/itex] then I can apply Bessel's inequality to show that the sequence does converge to [itex]0[/itex]. But this sequence in not orthogonal on [itex](0,1)[/itex]. So I don't know how to approach it anymore. Showing that a sequence is weakly convergent involves calculating an integral in [itex]L^p(a,b)[/itex]. What general idea can I use to calculate these integrals?

 

[mighty2000]

Thank you for your question. To prove that the sequence \{sin(kx)\} converges weakly to 0 in L^2(0,1), we can use the definition of weak convergence in a Banach space. This definition states that a sequence \{f_k\} in a Banach space X converges weakly to an element x\in X if L(f_k)→L(f) as k→∞ for each L in the dual of X. In this case, the dual of L^2(0,1) is L^2(0,1) itself.

To prove weak convergence, we need to show that for any L∈L^2(0,1), the sequence \{sin(kx)\} converges to 0 weakly. This means we need to show that the limit as k→∞ of the inner product <sin(kx),L> is equal to 0. Using the definition of the inner product in L^2(0,1), we have:

<sin(kx),L> = ∫_0^1 sin(kx)L(x) dx

Since L(x) is in L^2(0,1), we know that it is square integrable on (0,1). This means that the integral above is well-defined and finite. Now, as k→∞, the frequency of the sine function increases and the graph of sin(kx) becomes more and more oscillatory. This results in the integral above being "averaged out" over the interval (0,1), leading to a value close to 0. In fact, the integral will converge to 0 as k→∞, as the oscillations become more and more frequent.

Therefore, we can conclude that for any L∈L^2(0,1), the sequence \{sin(kx)\} converges weakly to 0, and thus the sequence converges weakly to 0 in L^2(0,1). This proves the statement that was given.

I hope this explanation helps. Please let me know if you have any further questions or need clarification. Keep up the good work in your studies!