Weak convergence of $f_n(x)=\cos (2 \pi n x)$

Question

Consider the space $H=C([0,1])$ with the norm $\|f\|_2=\left(\int_0^1|f(x)|^2 d x\right)^{1 / 2}$ and the sequence $f_n(x)=\cos (2 \pi n x)$ in $H$.

Show that for every $g \in H,\left\langle f_n, g\right\rangle \longrightarrow 0$.

I early see the following awnser:

"We use that the space of trigonometric polynomials is dense in $H$. For every $\epsilon>0$, there exists a trigonometric polynomial such that $\|g-p\|_2<\epsilon$. It follows from orthogonality that for sufficiently large $n,\left\langle f_n, p\right\rangle=0$. We obtain $$ \left|\left\langle f_n, g\right\rangle\right| \leq\left|\left\langle f_n, p\right\rangle\right|+\left|\left\langle f_n, g-p\right\rangle\right| \leq\left\|f_n\right\|_2\|g-p\|_2<\epsilon / \sqrt{2} . $$ This proves that $\left\langle f_n, g\right\rangle \rightarrow 0$."

But, I don't understand the point of "follows from orthogonality that for sufficiently large $n,\left\langle f_n, p\right\rangle=0$", I think that maybe this awnser be wrong. If is correct, what I missing?

Answer 1

The fact that $\int^1_0\cos(2\pi n t) f(t)\,dt\xrightarrow{n\rightarrow\infty}0$ for every $f\in C([0,1])$ can be deduced from the fact that $\{f_n(t)=\cos(2\pi n t):n\in\mathbb{Z}_+\}$ being an orthogonal family under the inner product $\langle f, g\rangle =\int^1_0 fg$ (see Bessel's inequality); however, there is a more general result that does not depend on orthogonality. In fact the same conclusion holds for any sequence if the form $g_n(t)=g(nt)$ where $g$ is $1$-periodic and $\int^1_0g=0$. This is a result by Fejer

Fejér's formula: Suppose $g$ is a bounded measurable $T$-periodic function on $\mathbb{R}$ ($T>0$). For any integrable function $f$ (denoted by $f\in\mathcal{L}_1(\mathbb{R})$) and numeric sequence $a_n\in\mathbb{R}$, $$ \lim_n\int_\mathbb{R} f(x)g(nx+a_n)\,dx=\Big(\frac{1}{T}\int^T_0 g\Big)\int_\mathbb{R} f \tag{1}\label{one} $$

In the case of the OP, take $g(x)=\cos(2\pi x)$ and for any $\phi\in C([0,1])$ define $f(x)=\phi(x)$ for $0\leq x\leq 1$ and $f(x)=0$ otherwise.

Answer 2

Take $n\in\Bbb N$. If $m\in\Bbb N\setminus\{n\}$, then$$\langle f_n,\cos(2\pi mx)\rangle=\langle f_n,\sin(2\pi mx)\rangle=0.$$Therefore, if $p(x)$ is a linear combination of functions of the form $\cos(2\pi mx)$ and of the form $\sin(2\pi mx)$ with $m<n$, you have $\langle f_n,p\rangle=0$. That is, $\langle f_n,p\rangle=0$ if $n>\deg p$. So, the argument is correct.