Convergence of a cosine sequence in Banach space

[Jaggis]

Does the sequence \{f_n\}=\{\cos{(2nt)}\} converge or diverge in Banach space C(-1,1) endowed with the sup-norm ||f||_{\infty} = \text{sup}_{t\in (-1,1)}|f(t)|?

At first glance my intuition is that this sequence should diverge because cosine is a period function. But how to really prove that?

I have tried to approach the problem from the point of view that if the sequence converges it must be a Cauchy sequence, since C(-1,1) is a normed space. It would remain to show that \{f_n\} is not a Cauchy sequence.

For indexes n and n+1 it holds:


||f_n-f_{n+1}||_{\infty} = ||\cos{(2nt)} - \cos{(2(n+1)t)}||_{\infty} = \text{sup}_{t\in (-1,1)}|\cos{(2nt)} - \cos{(2(n+1)t)}|.

But how to see that this won't become smaller and smaller for n large enough, exactly? What I tried next was to argue that if \{\cos{(2nt)}\} converges, it must converge towards some constant value that cosine can have (i.e. a number from the interval [-1,1]) because there is no limit of the form \cos{(2kt)} for any k \in N. Since ||\cos{(2nt)}||_{\infty} = 1 for all indexes, the constant must be f = 1 or f = -1 because the limit function must have the same norm as all other members of the sequence. But then ||f_n - f||_{\infty} = ||\cos{(2nt)}\pm 1||_{\infty} = 2 for all indexes and thus \{f_n\} is not a Cauchy sequence. I'm not sure whether my thinking here is correct.

My other theory was to fix t \in (-1,1) and consider indexes n and m, for whom n< m, as degrees of freedom. Then it holds that

||f_n(t)-f_{m}(t)||_{\infty} = \text{sup}|\cos{(2nt)} - \cos{(2mt)}|

and there are arbitrarily many (especially large) n and m for whom \text{sup}|\cos{(2nt)} - \cos{(2mt)}| = 2 for any fixed t \in (-1,1) . Since n, the smaller of the indexes, can get larger and larger while ||f_n(t)-f_{m}(t)||_{\infty} = 2 still holds, the sequence \{f_n\} cannot be a Cauchy sequence.

What do you think?

 

[micromass]

Does the sequence converge pointswise?

 

[mathman]

Examine what happens at t=\frac{\pi}{4}. Value cycles 0,-1,0,1. Sequence can't converge.

 

[Jaggis]

Examine what happens at t=\frac{\pi}{4}. Value cycles 0,-1,0,1. Sequence can't converge.

Ah! It is as simple as that. Yes, convergence would have to hold for all t \in [-1,1] so it is enough to find an individual t that won't satisfy it.

 

[wrobel]

but it is weakly convergent :)

 

[mathman]

but it is weakly convergent :)

http://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf

I haven't thought it out completely, but it looks like it is weakly convergent to g(t)=0 for all t.

 

[Hawkeye18]

http://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf

I haven't thought it out completely, but it looks like it is weakly convergent to g(t)=0 for all t.

No, it does not converge weakly in $C(-1,1)$. It converge weakly to 0 in $L^p$, $1\le p <\infty$, and converges to 0 in weak-* topology in $L^\infty$, all thanks to Riemann--Lebesgue Lemma.

But there is no convergence in the weak topology of $C(-1,1)$: if you consider the functional $\delta=\delta_0$ on $C(-1,1)$, $\delta(f)=f(0)$, then $\delta(f_n)=1$ for all $n$. But for any functional $\Phi$ corresponding to an absolutely continuous measure, $$\Phi(f):= \int_{-1}^1 f(x)\phi(x) dx, \qquad \phi \in L^1(-1,01), $$ we have $\Phi(f_n)\to 0$, again thanks to Riemann--Lebesgue Lemma.

PS. And of course, if the sequence diverges in weak topology, it diverges in norm as well.

 

[wrobel]

But there is no convergence in the weak topology of C(−1,1)C(-1,1): if you consider the functional δ=δ0\delta=\delta_0 on C(−1,1)C(-1,1), δ(f)=f(0)\delta(f)=f(0),

O, indeed, even it has already been written:

Examine what happens at t=π4t=\frac{\pi}{4}. Value cycles 0,-1,0,1. Sequence can't converge.

 

[mathman]

No, it does not converge weakly in $C(-1,1)$. It converge weakly to 0 in $L^p$, $1\le p <\infty$, and converges to 0 in weak-* topology in $L^\infty$, all thanks to Riemann--Lebesgue Lemma.

But there is no convergence in the weak topology of $C(-1,1)$: if you consider the functional $\delta=\delta_0$ on $C(-1,1)$, $\delta(f)=f(0)$, then $\delta(f_n)=1$ for all $n$. But for any functional $\Phi$ corresponding to an absolutely continuous measure, $$\Phi(f):= \int_{-1}^1 f(x)\phi(x) dx, \qquad \phi \in L^1(-1,01), $$ we have $\Phi(f_n)\to 0$, again thanks to Riemann--Lebesgue Lemma.

PS. And of course, if the sequence diverges in weak topology, it diverges in norm as well.

Your statement is confusing. What has the delta function have to do with anything?

 

[Hawkeye18]

Your statement is confusing. What has the delta function have to do with anything?

Delta function is a bounded linear functional on $C(-1,1)$.

 

[mathman]

Delta function is a bounded linear functional on $C(-1,1)$.

Delta function is not bounded. Linear functionals consist of members of L1(-1,1). \lim_{n \to \infty}\int_{-1}^{1}g(t)cos(2nt)dt=0 for any L1 function g(t).

 

[Hawkeye18]

Delta function is a bounded linear functional on $C(-1,1)$: $$|\delta (f)| = | f(0)| \le \sup_{x\in(0,1)} |f(x)| =: \|f\|_{C(-1,1)}.$$

 

[micromass]

Delta function is not bounded. Linear functionals consist of members of L1(-1,1).

Definitely not true. The delta function is bounded. The dual of $C[-1,1]$ is given by the Riesz representation theorem and consist basically of Stieltjes integrals/measures (depending on how you look at it). https://en.wikipedia.org/wiki/Riesz–Markov–Kakutani_representation_theorem

 

[mathman]

Definitely not true. The delta function is bounded. The dual of $C[-1,1]$ is given by the Riesz representation theorem and consist basically of Stieltjes integrals/measures (depending on how you look at it). https://en.wikipedia.org/wiki/Riesz–Markov–Kakutani_representation_theorem

My misunderstanding. There is a concept of weak-* convergence, where L1 is used rather than the measures you described.