Conditions for convergence for a Fourier series

Question

Let $f$ and $f'$ be piecewise continuous on the interval (-p,p); that is, let $f$ and $f'$ be continuous except at a finite number of points in the interval and have only finite discontinuities at these points. Then the Fourier series of $f$ on the interval converges to $f(x)$ at a point of continuity. At a point of discontinuity the Fourierseries converges to the average $$\frac{f(x^{+})+f(x^{-})}{2}$$ How do we prove this statement?

Answer 1

Let $S_n(f)(x)$ denote the sequence of partial sums of the Fourier series associated to $f$. That's, $f(x) \sim S_n(f)(x)$.

We first introduce the so-called Dirichlet kernel:

$$D_n(t):= \frac12 + \sum_{k=1}^n \cos(kt)$$

The Dirichlet kernel has a nice closed form and satisfies an equality:

$$ D_n(t) = \begin{cases} \large{\frac{\sin \left( \frac{2n+1}2 t\right)}{2\sin(t/2)}}, &\text{if $t \notin 2 \pi \mathbb Z$} \\ \large \frac{2n+1}2, &\text{if $\exists k \in \mathbb Z, \ t = 2k\pi$} \end{cases}$$

Second, we recall the Riemann-Lebesgue lemma:

If $f$ is piecewise smooth on an interval $[\alpha,\beta]$, then $\forall \mu \in \mathbb R$:

$$\lim_{\lambda \to \infty} \int_{\alpha}^{\beta} f(v)\cos(\lambda v + \mu)dv = \lim_{\lambda \to \infty} \int_{\alpha}^{\beta} f(v)\sin(\lambda v + \mu)dv = 0$$

We have for $x$ not a multiple of $2 \pi$:

$$\small {S_n(f)(x)} = \frac1{2\pi} \int_{-\pi}^{\pi}f(v)dv + \frac1{\pi}\sum_{k=1}^n \left(\int_{-\pi}^{\pi}f(v)\cos(kv)dv\right)\cos(kx) +\left( \int_{-\pi}^{\pi}f(v)\sin(kv)dv\right)\sin(kx) \\ =\frac1{2 \pi}\int_{-\pi}^{\pi}f(v)dv + \frac1{\pi} \sum_{k=1}^n\int_{-\pi}^{\pi}f(v)\{\cos(kv)\cos(kx) + \sin(kv)\sin(kx)\}dv \\ = \frac1{\pi}\int_{-\pi}^{\pi}f(v)\{\frac12 + \sum_{k=1}^n\cos(k(v - x))\}dv = \frac1{\pi}\int_{-\pi}^{\pi}f(v)D_n(v-x)dv$$

Using periodicity and the change of variable $t = v-x$:

$$S_n(f)(x) = \frac1{\pi}\int_{0}^{\pi}\{f(x-t) + f(x+t)\}D_n(t)dt$$

Now, using:

$$1 = \frac2{\pi}\int_{0}^{\pi}D_n(t)dt$$

We have:

$$\left| S_n(f)(x) - \frac12(f(x^+) + f(x^-)) \right| = \left| \frac1{\pi} \int_{-\pi}^{\pi} D_n(t)\{f(x+t) - f(x^+) + f(x-t) - f(x^-)\}dt \right|$$

Now, as $f$ is piecewise smooth on $(-\pi,\pi)$, we may choose $\delta > 0$, such that $f'$ is continuous on the intervals $(x-\delta,x)$ and $(x,x+\delta)$.

Put:

$$I_1 = \frac1{\pi} \int_{0}^{\delta} \{f(x+t)-f(x+) + f(x-t) - f(x^-)\}D_n(t)dt $$

And:

$$I_2 = \frac1{\pi} \int_{\delta}^{\pi}\{f(x+t)-f(x^+) +f(x-t) - f(x^-)\}D_n(t)dt$$

Note that $f'$ is bounded due to piecewise continuity, then $\exists$ $M>0$ such that: $|f'(u)| \le M$, $\forall \ u$.

As $0 \le t \le \delta$, we may use MVT on $(x,x+t)$ and $(x - t, x)$ to $f$ to find $c_1, c_2 \in (0,1)$ such that:

$$f(x+t) - f(x^+) = t \times f'(x + c_1 t)$$

$$f(x-t) - f(x^-) = -t \times f'(x - c_2 t)$$

We also make use of the above expression of $D_n(t)$ to find:

$$I_1 = \frac1{\pi} \int_{0}^{\delta} \{tf'(x+c_1t) - tf'(x - c_2t) \}\frac{\sin((2n+1)/2)}{2\sin(t/2)}dt$$

Then:

$$|I_1| \le \frac1{\pi} \int_{0}^{\delta}2Mt \frac1{2\sin(t/2)}dt$$

Now, using Jordan's inequality, we obtain:

$$|I_1| \le M \delta$$

Now, let $\epsilon > 0$ be given. We may modify our above $\delta$ so that it happens that both: $f'$ is continuous on $(x-\delta, x)$ and $(x,x+\delta)$, and $\delta < \epsilon/2M$.

Thus, $|I_1| < \epsilon/2$.

Now, note that:

$$t \longrightarrow \frac{f(x+t) -f(x^+) + f(x-t) - f(x^-)}{2\sin(t/2)}$$

is piecewise smooth on $(\delta,\pi)$. Then, by Riemann-Lebesgue lemma, $|I_2| \to 0$.

Hence, $\exists$ $n_0 \in \mathbb N$, such that $|I_2| < \epsilon/2$.

Therefore, $\forall$ $n \ge n_0$, we have:

$$\left| S_n(f)(x) - \frac12(f(x^+) + f(x^-)) \right| \le |I_1| + |I_2| < \epsilon/2 + \epsilon/2 = \epsilon$$

This shows that:

$$\lim_{n \to \infty} S_n(f)(x) = \frac12(f(x^+) + f(x^-))$$

Which was required to prove.