Why does the convergence theorem of Fourier series hold in $L^2$ but fail in $L^1$?

675 Views Asked by At

First consider a periodic continuous function $g \in C^0(\mathbb{T})$. We can find a sequence of trigonometric polynomials that converges uniformly to $g$ (e.g. convolution with Fejer kernels). Thus, the same sequence converges to $g$ in both $L^1$ and $L^2$ norm.

Now, $C^0(\mathbb{T})$ is dense in both $L^1(\mathbb{T})$ (wrt $L^1$ norm) and $L^2(\mathbb{T})$ (wrt $L^2$ norm). This together with the previous paragraph imply the linear span of $\{e^{in\theta}\}_{n\in \mathbb{Z}}$ is dense in both $L^1(\mathbb{T})$ and $L^2(\mathbb{T})$, so functions in both spaces can be written as trigonometric series.

However, it is well known that the Fourier series of every function $f \in L^2(\mathbb{T})$ converges to $f$ in $L^2$ norm, whereas there exist functions in $L^1(\mathbb{T})$ whose Fourier series diverge in $L^1$ norm (and in fact also diverge everywhere).

Why is that? Why doesn't the completeness of $\{e^{in\theta}\}_{n\in \mathbb{Z}}$ in $L^1$ imply the convergence of Fourier series?

1

There are 1 best solutions below

2
On BEST ANSWER

There is a differnce between the statements

a) linear span of $(x_n)$ is dense in the normed linear space $X$

b) every element of $X$ is of the form $\sum a_nx_n$.

It is true that the functions $e^{2\pi i n\theta}$ span dense subspace of $L^{1}$ but it is not true that the Fourier series of any $L^{1}$ function converges in $L^{1}$.

However a) and b) become equivalent if $X$ is a Hilbert space and $(x_n)$ is an orthonormal set.