Given the Dirichlet function defined as:
$$f(x) = \begin{cases}0 & x \in \mathbb{Q} \\ 1 & x \in \mathbb{R} \setminus \mathbb{Q}\end{cases}$$
Find the corresponding Fourier series.
Before starting, I believe that $f(x)$ is periodic for any real period greater than zero. Therefore I feel free to choose $2\pi$ as period. Hence the Fourier series will have the following form:
$$F(x) = a_0 + \sum_{n=1}^\infty a_n \cos nx + b_n \sin nx$$
With the following coefficients:
$$a_0 = \frac1{2\pi} \int_{-\pi}^\pi f(x) dx \\
a_n = \frac1\pi \int_{-\pi}^\pi f(x) \cos nx dx \\
b_n = \frac1\pi \int_{-\pi}^\pi f(x) \sin nx dx$$
My function is not Riemann-integrable, however is Lebesgue-integrable. Knowing the following theorem…
$$I = I_1 \cup I_2 \\
g(x) = \begin{cases}a & x \in I_1 \\ b & x \in I_2\end{cases} \\
\int_I g(x) dx = \int_{I_1} g(x) dx + \int_{I_2} g(x) dx$$
… I can easily calculate the coefficients $a_0 = 1, a_n = b_n = 0$. Therefore the Fourier series corresponding to the Dirichlet function is:
$$F(x) = 1$$
Which does not converge to the function.
This is what I discovered. Is my result exact? Is the procedure right?
Best Answer
Since you define the Fourier series via Lebesgue integral, you should be able to observe the following general fact:
Indeed, modifying a function on a null set does not change any of the integrals that determine the Fourier coefficients.
In your example $f$ is identically $1$, $g$ is the Dirichlet function, and $T$ can be any rational number.
Yes, the Fourier series of a discontinuous function need not converge to that function pointwise. (Even for some continuous functions the pointwise convergence fails, though examples are harder to come by.)