# A Convergence of a cosine sequence in Banach space

1. Jun 4, 2016

### 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?

2. Jun 4, 2016

### micromass

Staff Emeritus
Does the sequence converge pointswise?

3. Jun 4, 2016

### mathman

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

4. Jun 4, 2016

### Jaggis

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.

5. Jun 5, 2016

### wrobel

but it is weakly convergent :)

6. Jun 5, 2016

7. Jun 5, 2016

### Hawkeye18

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.

8. Jun 6, 2016

### wrobel

O, indeed, even it has already been written:

Last edited: Jun 6, 2016
9. Jun 6, 2016

### mathman

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

10. Jun 6, 2016

### Hawkeye18

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

11. Jun 7, 2016

### mathman

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).

12. Jun 7, 2016

### Hawkeye18

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

13. Jun 8, 2016

### micromass

Staff Emeritus
14. Jun 8, 2016

### mathman

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