Dense subset

**slevvio** · January 6th 2011, 12:53 PM

Let $X$ be the space of continuous functions $f:[0,\pi] \rightarrow \mathbb{C}$ with norm || $f|| = \int_0^\pi |f(x) |dx$ . I am trying to determine whether the linearly independent set $\text{span}\{\sin nx\}_{n \in \mathbb{N}}$ is dense in $X.$ Somebody gave me a hint that it is not dense and to look at the constant function 1. So that means I am trying to show that there is some open ball around the function 1, which contains no element of $\text{span}\{\sin nx\}_{n \in \mathbb{N}}$ .

This means that I need to show $\int_0^\pi |1 - \sum_{i=0}^r \alpha_i \sin (n_i x) | dx \ge \epsilon$ for some $\epsilon >$ 0.

Has anyone got any hints regarding how to go about this? Thanks very much

**Rebesques** · January 9th 2011, 03:23 AM

How about expressing cosx in terms of span{sinkx}?

**slevvio** · January 9th 2011, 06:05 AM

That is not possible since the set $\{f \mid f=\sum_{i=0}^r \alpha _i f_i, \text{ where }f_i = \sin n_i (\cdot) \text{ or } \cos n_i(\cdot) , r >0, \alpha_i \in \mathbb{C}, n_i\in\mathbb{N}\}$ is linearly independent.

This means of course that $|\cos x - \sum_{i=0}^r \alpha_i \sin n_i (x) |>0$ . But its not necessarily bigger than a constant $\epsilon$ which I choose at the start

**Rebesques** · January 9th 2011, 06:21 AM

Hmmm... What if x is near 0?

**slevvio** · January 9th 2011, 06:51 AM

at 0, the difference is 1, but how does the affect the integral between 0 and pi ? similarly at pi the difference is 1.

**Rebesques** · January 9th 2011, 07:02 AM

Well... If you suppose that f=cosx lies in span{sinkx}, then, for every a>0 there is a finite sum S(a), such that
||f-S(a)||<a. But since f(0)=1 and all finite sums are zero at x=0, in a sufficiently small right interval about 0, we can have f-S(a)>a/pi.
Integrating gives a>a.

**slevvio** · January 9th 2011, 07:13 AM

But I know in advance that cosx is not in that span, from linear independence.

I am trying to prove the statement " $\exists f \in C([0,\pi],\mathbb{C}) \text{ such that } \exists \epsilon > 0 \text{ such that }\forall g\in \text{span}\{\sin nx\}_{n \ge 0} , \int_0^{\pi} |f(x)-g(x)|\ge \epsilon$ "

**Rebesques** · January 9th 2011, 10:28 AM

If you already know that cosx is not in the span, then the span is not dense... right? Why bother proving more? :S

**slevvio** · January 9th 2011, 10:47 AM

But it might be in the closure of the span? Is this is a standard result you are using?

**Drexel28** · January 9th 2011, 07:26 PM

Originally Posted by slevvio

But it might be in the closure of the span? Is this is a standard result you are using?

I have been trying to think of a clever continuous linear functional which is constant on $\text{span}\{\sin(nx)\}$ but not constant on $\mathcal{C}\left([0,1],\mathbb{C}\right)$ . This of course would contradict density. Any ideas?

**Opalg** · January 10th 2011, 12:28 AM

Originally Posted by slevvio

Let $X$ be the space of continuous functions $f:[0,\pi] \rightarrow \mathbb{C}$ with norm $\|f\| = \int_0^\pi |f(x) |dx$ . I am trying to determine whether the linearly independent set $\text{span}\{\sin nx\}_{n \in \mathbb{N}}$ is dense in $X.$ Somebody gave me a hint that it is not dense and to look at the constant function 1. So that means I am trying to show that there is some open ball around the function 1, which contains no element of $\text{span}\{\sin nx\}_{n \in \mathbb{N}}$ .

This means that I need to show $\int_0^\pi |1 - \sum_{i=0}^r \alpha_i \sin (n_i x) | dx \ge \epsilon$ for some $\epsilon >$ 0.

Has anyone got any hints regarding how to go about this? Thanks very much

The ingredients for a solution are already here in the comments of Rebesques and Drexel28. Define a continuous linear functional $\phi$ on $X$ by $\phi(f) = \int_0^\pi f(x)\cos x\,dx$ . Then $\phi(\sin nx) = 0$ for all n, but $\phi(\cos x)\ne0$ . So the function $\cos x$ is not in the kernel of $\phi$ , which is a closed subspace of $X$ .

[I have moved this thread, because it's really a problem in functional analysis rather than linear algebra. The solution depends on the fact that $L^\infty[0,\pi]$ is the dual of $L^1[0,\pi]$ .]

**slevvio** · January 10th 2011, 09:42 AM

but $\phi(\sin nx )$ is not zero when n is even

the map $\theta(f) = \int_0^{\pi} f(x) sin(x) dx$ seems to work for $cos(2x).$ . Only sinx is not in the kernel here, but that could never be a sequence tending to cos(2x) anyway.

**slevvio** · January 10th 2011, 10:14 AM

yes that map works ! Thanks all I see how it works

**Opalg** · January 10th 2011, 10:55 AM

Originally Posted by slevvio

but $\phi(\sin nx )$ is not zero when n is even

Dammit, you're right!

And what's more, I'm changing my mind about whether this result is true.

I now think that every continuous function on $[0,\pi]$ can be approximated in the $L^1$ -norm by a linear combination of the functions $\sin kx$ . In fact, let $f$ be a continuous function on $[0,\pi]$ , and extend $f$ to an odd function $g$ on $[-\pi,\pi]$ , (so that $g(-x) = -f(x)$ ). Then the Fourier series of $g$ consists only of sine terms. Let $s_n(g)$ denote the sum of the first n terms of that series. Then $\|s_n(g) - g\|_2\to0$ as $n\to\infty$ , where $\|\,.\,\|_2$ is the $L^2$ -norm, given by $\|h\|_2 = \sqrt{\int_{-\pi}^\pi|h(x)|^2dx}$ . By the Cauchy–Schwarz inequality, if $\|h\|_1$ is the $L^1$ -norm of h, given by $\|h\|_1 = \int_{-\pi}^\pi|h(x)|\,dx$ , then

$\|h\|_1 = \int_{-\pi}^\pi|h(x)|\cdot1\,dx\leqslant\|h\|_2\|1\|_2 = \sqrt{2\pi}\|h\|_2$ .

It follows that $\|s_n(g)-g\|_1\leqslant\sqrt{2\pi}\|s_n(g)-g\|_2\to0$ , and if you restrict the interval of integration to $[0,\pi]$ , this tells you that $\int_0^\pi|s_n(f)-f|\,dx\to0$ . In other words, $f$ can be approximated in the $L^1[0,\pi]$ -norm by a linear combination of functions $\sin kx$ .

**Rebesques** · January 10th 2011, 02:16 PM

I'm not sure about this last argument Op, how can we restrict the interval to [0,pi]?

Now, about Slev's question.
How about using a little bit of Arzela-Ascoli to arrive at a contradiction?

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