Proof of behavior of Fourier Series at point of discontinuity

Question

The following fact about Fourier series seems to be well known, but I'm having trouble finding a proof of it online.

Let $f:[-\pi,\pi] \to \Bbb R$ be piecewise differentiable (i.e. differentiable except at finitely many points, at which the function might fail to be continuous). Let $S(f)$ denote the full Fourier series of $f$, $$ S(f)(x) = \sum_{n = - \infty}^\infty \hat f(n) e^{inx}. \tag{1} $$ The limit $S(f)(x)$ exists at all points $x \in [-\pi,\pi]$. Moreover, if $f$ has a jump-discontinuity at $x_0$, then $$ S(f)(x_0) = \frac{f(x_0^+) + f(x_0^-)}{2}, \tag{2} $$ where $f(x_0^+)$ and $f(x_0^-)$ denote one sided limits of $f$ at $x_0$.

My main concern is not the convergence of the series; I am chiefly interested in a justification of Equation (2) above. I have seen the "demonstration" of this phenomenon as it applies to the unit step function, but I want an argument that this should hold more generally. I'd be thrilled if someone provided a proof, but a suitable online reference would also be sufficient.

Answer 1

Here's a partial answer. To begin, I'll prove the following result.

Lemma 1: Let $\sigma$ denote the signum function, $$ \sigma(x) = \begin{cases} -1 & x < 0\\ 0 & x = 0\\ 1 & x > 0. \end{cases} $$ We find that $S(\sigma)$ exists, with $S(\sigma) = \sigma$.

Proof outline: First, a quick note: it is easy to see that if $S(\sigma)$ exists, then it must hold that $S(\sigma)(0) = 0$. Indeed, the fact that $\sigma$ is an odd function ensures that each truncated Fourier series $$ S_N(\sigma)(x) := \sum_{n=-N}^N \hat f(n)e^{inx} $$ is itself odd. Thus, $S_N(\sigma)(0) = 0$ for all $N$, so that $\lim_{N \to \infty} S_N(\sigma)(0) = 0$.

As for proof of existence, it suffices to compute the Fourier series directly. One finds that $$ S_N(f) = \frac 4{\pi} \sum_{n\text{ odd}, n \leq N} \frac 1n \sin\left( \frac{n x}{2}\right). $$ At all $x$, this the above is the partial sum for a convergent alternating series. It remains to argue/show that this function converges to $\sigma(x)$ at the non-zero values of $x \in [-\pi,\pi]$.

I'll use the following facts without proof.

Lemma 2: Given functions $f,g$ with convergent Fourier series, $S(f + g) = S(f) + S(g)$

Lemma 3: If $f$ is continuous over $[-\pi,\pi]$ and differentiable except at finitely many points, then $S(f) = f$.

From there, the general result may be argued as follows. Given a suitable $f$ satisfying the hypothesis of the statement, we can build a piecewise constant function $g(x) = \sum_{j=1}^n a_j \sigma (x - x_j)$, where $x_1,\dots,x_n$ are the points at which $f$ is discontinuous and $a_j = \frac 12 (f(x_j^+) - f(x_j^-))$, such that $f - g$ is continuous. By combining the corollary of Lemma 1 and the other lemmas, one can deduce the more general result.

Answer 2

If you know the Riemman-Lebesgue Lemma for integrable functions, this is easier.

Riemman-Lebesgue: If $f$ is integrable in $[a,b]$, we have $\lim\int_a^b f(\theta)e^{\frac{-2\pi i n\theta}{b-a}}d\theta=0$.

For an elementary proof of the Riemman Lebesgue Lemma I would recommend the third chapter of Stein's Fourier Analysis book.

With this out of the way, let us consider a piecewise $C^1$ function $f$. Let $D_n(x):=\sum_{n=-N}^N e^{inx}$ $=\sin((N+1/2)x)/\sin(x/2)$ be the Dirichlet Kernel. In this case, the truncated Fourier series is given by a convolution with $D_n$, namely, $S_N(f)(x)=D_n * f(x)$. This will be more useful for our purposes, because:

$$\left| \int_{-\pi}^{\pi}f(x_o-x)D_n(x)dx-\frac{f(x_o^+)+f(x_o^{-})}{2}\right|\leq \left|\int_{-\pi}^0 (f(x_o-x)-f(x_o^{+}))D_n(x)dx\right|+\left|\int_{0}^\pi (f(x_o-x)-f(x_o^{-}))D_n(x)dx\right|$$

This inequality follows from the triangle inequality and also from an explicit computation that shows that $\int_{-\pi}^0 D_n(x)dx=\int_0^\pi D_n(x) dx=\frac{1}{2}.$ I will prove the first term can be taken to be as small as one wants:

$$ \left|\int_{-\pi}^0 (f(x_o-x)-f(x_o^{+}))D_n(x)dx\right|=\left|\int_{-\pi}^0 2\frac{(f(x_o-x)-f(x_o^{+}))}{x}\frac{\frac{x}{2}}{\sin(x/2)}\sin((N+1/2)x)dx\right|$$

If we define $g(x)=2\frac{(f(x_o-x)-f(x_o^{+}))}{x}\frac{\frac{x}{2}}{\sin(x/2)}$ to be equal to $2 f'(x_o^{+})$ at $x_o$, we gain continuity at this point. Therefore, our function is integrable. By the Riemman Lebesgue's Lemma, for any integrable function $h$:

$$\lim_N \int_{-\pi}^0 h(x)\sin(2N x)dx=\lim_N \int_{-\pi}^0 h(x)\cos(2N x)dx=0$$

Expanding $\sin((2N+1)x)$ and using the identities above, it can also be proven that:

$$\lim_N \int_{-\pi}^0 h(x)\sin((2N+1) x)dx=\lim_N \int_{-\pi}^0 h(x)\cos((2N+1) x)dx=0$$

These four identities enable us to write:

$$ \lim_N \int_{-\pi}^0 h(x)\sin(Nx)dx=\lim_N \int_{-\pi}^0 h(x)\cos(Nx)dx=0 $$

Because $g(x)\sin((N+1/2)x)=g(x) \sin(Nx)\cos(x/2)+g(x)\sin(x/2)\cos(Nx)$, we have that our first paranthesis can be made as small as we want as a corollary of the Riemman-Lebesgue Lemma:

$$\left|\int_{-\pi}^0 (f(x_o-x)-f(x_o^{+}))D_n(x)dx\right|\leq \left| \int_{-\pi}^0 g(x)\cos(x/2)\sin(Nx)dx\right| +\left| \int_{-\pi}^0 g(x)\sin(x/2)\cos(Nx)dx\right|\rightarrow 0$$