Pointwise Convergence of Fourier Series

Partial sums of Fourier series

Given a function $f\in L^1[-\pi,\pi]$, we can construct its Fourier series in terms of complex exponentials, \[ (\ast) \quad f(x)\sim \sum_{n=-\infty}^\infty \alpha_ne^{inx},\ \alpha_n=\frac{1}{2\pi} \int_{-\pi}^\pi f(t)e^{-int}dt. \] What we are interested in is finding conditions on $f$ sufficient for the series to converge pointwise to $f$. Assume that $f$ is a 2π-periodic, piecewise continuous function. For the Fourier series in $(\ast)$, the partial sum $S_N(x)$ is given by \[ \begin{align} (\ast \ast) \quad S_N(x) &= \sum_{n=-N}^N \alpha_n e^{inx} \\ &= \int_{-\pi}^\pi \big(\tfrac{1}{2\pi}\sum_{n=-N}^N e^{in(x-t)}\big)f(t)dt \\ &= \int_{-\pi}^\pi D_N(x-t)f(t)dt, \ \text{where}\ D_N(u):=\frac{1}{2\pi} \sum_{n=-N}^N e^{inu}. \end{align} \] The function $D_N$ is called the Dirichlet kernel. $D_N$ has important properties. The first three listed below are easy to derive. The fourth requires some algebra and we will establish it later.

Properties of DN

  1. $D_N(u) = \frac{1}{2\pi} + \frac{1}{\pi}\sum_{n=1}^N \cos(nu)$.
  2. $D_N(u)$ is even and $2\pi$ periodic.
  3. $\int_{-\pi}^\pi D_N(u)du = 1$ and $\int_0^\pi D_N(u)du = \frac12$.
  4. $D_N(u) = \frac{\sin((N+\frac12)u)}{2\pi \sin(\frac{u}{2})}$.

The properties given for DN help put each partial sum SN in a form that will make it easier to work with. First of all, in the form of SN above, we change variables to u = t-x and use the fact that DN is even to get $S_N(x) = \int_{-\pi-x}^{\pi-x} f(u+x)D_N(u)du$. We will need the lemma below to go any further.

Lemma.   Let $g$ be a $2\pi$ periodic function (a.e.) that is integrable on each bounded interval in $\mathbb R$. Then, $\int_{-\pi+c}^{\pi+c} g(u)du$ is independent of $c$. In particular, $\int_{-\pi+c}^{\pi+c} g(u)du=\int_{-\pi}^\pi g(u)du$.

Proof: Exercise.

Fix $x$. Both $D_N(u)$ and $f(x+u)$ are $2\pi$-periodic in $u$, and so by the Lemma with $c=-x$, \[ S_N(x) = \int_{-\pi-x}^{\pi-x}f(u+x)D_N(u)du= \int_{-\pi}^{\pi}f(u+x)D_N(u)du. \] We can change variables in the integral going from $u$ to $-u$. The result is that \[ S_N(x) = -\int_{\pi}^{-\pi} f(x-u)D_N(-u)du = \int_{-\pi}^\pi f(x-u)D_N(u)du. \] (Recall that $D_N$ is even.) Add these two expressions together and divide by 2. The integrand in the result, $(f(x+u)+f(x-u))D_N(u)/2$, is even. Using the fact that an integral of an even function over a symmetric interval is twice the integral of the function over half the interval, we get, after canceling factors of 2, that \[ (\dagger) \quad S_N(x) = \int_0^\pi (f(x+u)+f(x-u))D_N(u)du. \] The next ingredient in our proof is the Riemann-Lebesgue Lemma, which we now state:

Riemann-Lebesgue Lemma.  If $f$ is in $L^1[a,b]$, then $ \lim_{\lambda\to \infty} \int_a^bf(x)\cos(\lambda x)dx = \lim_{\lambda\to \infty} \int_a^bf(x)\sin(\lambda x)dx = 0$.

Proof: We will prove the cosine version. Let $f$ be a continuous, piecewise smooth function on $[a,b]$. Then, after integrating by parts, we have \[ \int_a^b f(x)\cos(\lambda x)dx = \frac{1}{\lambda}(f(b)\sin(\lambda b) -f(a)\sin(\lambda a)) - \frac{1}{\lambda}\int_a^b f'(x)\sin(\lambda x)dx. \] From this, it easily follows that \[ (\ddagger)\quad \bigg|\int_a^b f(x)\cos(\lambda x)dx\bigg| \le \frac{1}{|\lambda|}\bigg(|f(b)|+|f(a)|+\int_a^b |f'(x)|dx\bigg). \] Letting $\lambda\to \infty$ above then yields $\lim_{\lambda \to \infty}\int_a^b f(x)\cos(\lambda x)dx=0$. Since linear splines are continuous piecewise smooth functions, the result holds for them. Now, let $f$ be in $L^1[a,b]$. We are given that $C[a,b]$ is dense in $L^1[a,b]$. Thus, given $\epsilon>0$, we can find $g\in C[a,b]$ such that $\|f-g\|_{L^1[a,b]} < \epsilon/3$. In addition, if $s$ is the linear spline interpolant to $g$, with knots equally spaced a distance $\delta_n=\frac{b-a}{n}$ apart, then $\|g-s\|_{C[a,b]}\le \omega(g;\delta_n)$ and, consequently, $\|g-s\|_{L^1[a,b]} \le (b-a)\omega(g;\delta_n)$. We may then choose $n$ so large that $\|g-s\|_{L^1[a,b]}<\epsilon/3$. By our estimate $(\ddagger)$ to $s$, we also have $\big|\int_a^b s(x)\cos(\lambda x)dx\big| \le \frac{C_s}{|\lambda|}$. If we choose $|\lambda| > 3C_s/\epsilon$, then $\big|\int_a^b s(x)\cos(\lambda x)dx\big| < \epsilon/3$. Finally, we can write $f=(f-g)+(g-s) +s$, from which it follows that \[ \bigg|\int_a^b f(x)\cos(\lambda x)dx\bigg| < \|f-g\|_{L^1[a,b]}+ \|g-s\|_{L^1[a,b]}+\bigg|\int_a^b s(x)\cos(\lambda x)dx\bigg| < \epsilon/3+ \epsilon/3+\epsilon/3 = \epsilon, \] which proves the lemma. $\square$

Pointwise Convergence Theorem.  If f is a 2π-periodic piecewise continuous function that has a right-hand derivative $f'(x+)$ and a left hand derivative $f'(x-)$ at $x$, then \[ \lim_{N\to \infty}S_N(x) = \left\{ \begin{array}{cl} f(x) & \text{if }f \ \text{is continuous at }x.\\ \tfrac12 (f(x+)+f(x-)) & \text{if }f\ \text{has a jump discontinuity at } x. \end{array} \right. \] Proof: The trick here is to put the the error SN(x) - f(x) or SN(x) - (f(x+)+f(x-))/2 in the right form. Since in the continuous case one just has f(x+)=f(x-)=f(x), we can do both cases using $E_N(x) = S_N(x) - \frac12 (f(x+)+f(x-))$. From the third property of DN listed above and the final form of $S_N$ given in $(\dagger$), we obtain $E_N(x) = \int_0^\pi (f(x+u) + f(x-u) - f(x+) - f(x-))D_N(u)du$. Use property 4 of $D_N(u)$ to put the error in the form \[ \begin{aligned} E_N(x) =& \int_0^\pi (f(x+u) + f(x-u) - f(x+) - f(x-)/(2\pi \sin(u/2)) \sin((N+\tfrac12)u)du \\ =& \int_0^\pi F(u) \sin((N+\tfrac12)u)du, \end{aligned} \] where $F(u) := (f(x+u) + f(x-u) - f(x+) - f(x-)/(2\pi \sin(u/2))$. Since x is fixed, we can ignore the dependence of the expression above on $x$. L'Hospital's rule and the fact that $f$ has right and left derivatives at x imply that $ \lim_{u \downarrow 0} F(u) = (f'(x+) - f'(x-))/\pi$, so F has a right hand limit at u = 0. This and the piecewise continuity of f and 1/sin(u/2) on the open interval (0,π] mean that F(u) is piecewise continuous on [0,π]. From the Riemann Lebesgue Lemma we have that $\lim_{N\to \infty}E_N(x) = 0.$ $\ \square$

An example. In class we showed that if $f(x) = \left\{ \begin{array}{cl} 1 & 0 \le x \le \pi, \\ -1 & -\pi <\le x< 0 \end{array} \right.$, then, in the Fourier series for $f$, all of the $a_n$'s are $0$ and also, when $n$ is even, the $b_n$'s are $0$. For odd $n$, $b_n=\frac{4}{n \pi}$. Thus the Fourier series for $f$ satisfies \[ \sum_{n\ \text{odd}} \frac{4}{n \pi}\sin(nx) = \sum_{k=1}^\infty \frac{4}{(2k-1) \pi}\sin\big((2k-1)x\big) = \left\{ \begin{array}{cl} f(x) = 1, & \text{for } \ 0< x < \pi \\ \frac12 \big(f(0+)+ f(0-)\big) = 0,& \text{for } x=0 \\ f(x) = -1, & \text{for } -\pi < x< 0. \end{array} \right. \] As an illustration of what this means, let's set $x = \pi/2 $. For this value of $ x $, we have $1 = \sum_{k=1}^\infty \frac{4}{(2k-1) \pi}\sin\big((2k-1)\pi/2\big)$. Since $\sin\big(\frac{(2k-1)\pi}{2}\big)= (-1)^{k+1}$, we have, after multiplying both sides by $\pi/4$, \[ \frac{\pi}{4} = \sum_{k=1}^\infty \frac{4(-1)^{k+1}}{2k-1} = 1 - \frac13 + \frac17 - \frac19 + \cdots, \] which is a famous series that was used to compute $\pi$. Since the series has terms that alternate and decrease in absolute value, the error made truncating the series after $K$ terms is $\frac{1}{2K+1}$. Thus to compute $\pi/4$ to within $0.01$, we would need to take $K\ge 50$.

The Formula for the Dirichlet Kernel

We want to establish property 4 of DN(u). Let z = eiu, so zn = einu for n = -N, ..., N. We can write the Dirichlet kernel this way:
2π DN(u) = (1 + z + z2 + ... + zN) + (1 + z-1 + ... + z-N) - 1.
From this, we get that
2π(z-1)DN = zN+1 -1 - z-N + z - z + 1 = zN+1 - z-N,
2π(eiu-1)DN = ei(N+1)u - e-iNu,
2i eiu/2 sin(u/2)(2π) DN = 2i eiu/2 sin((N+1/2)u) .
Dividing by the appropriate factors yields the formula that we want.

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