Fourier Series

So here's the natural next question. Take a nice function, a continuous one, and pick a point $x$. Does $S_N f(x)$ converge to $f(x)$? It feels like it should. The answer is no, and the reason is one of the prettiest uses of the Banach space machinery: we never write down the bad function. We show that the partial sums, viewed as linear functionals, have norms that run off to infinity, and the Banach–Steinhaus theorem hands us a continuous function that breaks them.

Then we see two ways out. Averaging the partial sums (Fejér's idea) fixes everything for continuous functions. And for integrable functions, the coefficients themselves always fade to $0$ (Riemann–Lebesgue), though not every sequence that fades to $0$ is a list of coefficients. Throughout, $T = [-\pi, \pi]$ with its ends glued, functions on $T$ are $2\pi$-periodic, and every integral over $T$ carries the same $\frac{1}{2\pi}$ as the inner product: $\hat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) e^{-int} \, dt$ and $\|f\|_1 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(t)| \, dt$.

Partial Sums as an Integral

First, a name for the thing we're asking about. The coefficients make sense for any integrable $f$, since $|f(t) e^{-int}| = |f(t)|$, so the partial sums do too.

The trick that makes everything on this page work: write each $\hat{f}(n)$ as its integral and move the finite sum inside. What's left inside the integral is one fixed function, the same for every $f$, and it gets a name.

Pairing $e^{ikt}$ with $e^{-ikt}$ gives $2 \cos kt$, so $D_N(t)$ is $1 + 2 \cos t + \cdots + 2 \cos Nt$. It's real and even. Before the general facts, one small kernel by hand.

That pattern (a tall spike of height $2N + 1$ at $0$, then wiggles that dip below zero) is what the closed form below makes exact. The lemma collects the three facts we'll use over and over.

Part (c) is the picture to keep: $S_N f(x)$ is a weighted average of $f$ around $x$, with weights $D_N(x - t)$. Part (b) says the weights total $1$, which is what an average should do. But the weights aren't all positive, and that's where the trouble will come from. Watch the average get computed as the kernel slides along a plateau.

Away from the edges of the plateau the red and blue lobes mostly cancel, and $S_8 f$ sits near $f$. Near an edge, a blue lobe hangs over the side where $f = 0$ on one side and $1$ on the other, so the average gets pushed past $f$. At the edge itself, half the kernel's weight lands on each side. Here's that last effect computed exactly.

That's no contradiction with anything: $f$ isn't continuous at $0$, and changing $f(0)$ doesn't change a single coefficient. One more property of the kernel, the one the whole page turns on, is yours to prove.

Now compute a few values yourself. The closed form and the cosine sum must agree, so use whichever is quicker, and check with the other.

Before the theory gets heavy, go poking at single points. Pick a function, drag $x$, and watch the sequence $S_0 f(x), S_1 f(x), \ldots$ underneath. Try the jumps, and try just beside them.

Everything there behaves at continuous points: the dots settle on $f(x)$. The overshoot beside a jump is real, though (it's called the Gibbs phenomenon), and it never shrinks as $N$ grows. It only gets squeezed closer to the jump. Keep that in mind; Fejér will fix it.

How Big $D_N$ Gets

Here's why the size of the kernel matters. Take $f$ continuous with $|f| \le 1$. By part (c) of the lemma at $x = 0$ (and $D_N$ even), $$\begin{aligned} |S_N f(0)| &= \left|\frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) D_N(t) \, dt\right| \\ &\le \frac{1}{2\pi} \int_{-\pi}^{\pi} |D_N(t)| \, dt = \|D_N\|_1. \end{aligned}$$ So if $\|D_N\|_1$ stayed bounded, every bounded $f$ would have bounded partial sums at $0$. Part (b) says the signed integral is $1$. The absolute integral is another story, even for $N = 1$.

For larger $N$ the lobes multiply, and each one contributes. Comparing $\sin(t/2)$ with $t/2$ turns that into a clean lower bound.

So the absolute area grows without bound, but only logarithmically. The spike gets taller like $N$; the norm creeps up like $\log N$. Here's the race.

The bound is honest but not sharp: the true $\|D_N\|_1$ runs about $1.27$ above $\frac{4}{\pi^2}\log N$ for large $N$, so it's the growth rate that matters, not the constant. Feel how slow a logarithm is.

Now package "the value of the partial sum at $0$" as a functional on the continuous functions, and compute its norm. The estimate at the start of this section already gives $\|D_N\|_1$ as an upper bound. The point of the next proposition is that nothing smaller works.

The $f_k$ are smoothed-out copies of "the sign of $D_N$": they're as close as a continuous function of size $1$ can get to agreeing with $D_N$ in sign everywhere. Here they are sharpening.

Now you be the worst input. Steps aren't continuous, but rounding a step's corners changes $\Lambda_N$ by as little as you like, so the best step tells you $\|\Lambda_N\|$ too.

Notice what happened when $N$ went from $3$ to $6$: the best value went up. That's the lemma in action, and it's worth saying carefully what it does and doesn't give you.

A Continuous Function Whose Fourier Series Diverges

We have a family of bounded linear functionals $\Lambda_N$ on $C(T)$ whose norms go to infinity. Two facts turn that into a bad function. First, $C(T)$ with $\|\cdot\|_\infty$ is a Banach space: a uniformly Cauchy sequence of continuous $2\pi$-periodic functions converges uniformly, and a uniform limit of continuous $2\pi$-periodic functions is continuous and $2\pi$-periodic. Second, the Banach–Steinhaus theorem: if $X$ is a Banach space and $\{\Lambda_\alpha\}$ is a family of bounded linear maps from $X$ to a normed space, then either $\sup_\alpha \|\Lambda_\alpha\| < \infty$, or the set of $x \in X$ with $\sup_\alpha \|\Lambda_\alpha x\| = \infty$ is a dense $G_\delta$ in $X$.

A dense $G_\delta$ set is a countable intersection of open sets that is dense. By Baire's theorem it's "most" of the space, in a precise sense, and in particular it's not empty. Put the two facts together.

Notice what the proof didn't do: it never built $f$. Paul du Bois-Reymond wrote down an explicit continuous function with a divergent Fourier series in 1873, and it's a page of careful estimates. Banach–Steinhaus gets existence in a paragraph, and gets more: the bad functions are dense. Every continuous function has a bad one as close as you like in the sup norm. Nothing is special about $0$, either. Translating $f$ moves the bad point anywhere.

One bad point per function, anywhere you like. The Spicy Bonus pushes this to a single continuous $f$ whose Fourier series diverges on a dense, uncountable set of points. How much worse can it get?

Fejér's Rescue

The partial sums average $f$ against $D_N$, and the negative lobes of $D_N$ are the problem. Fejér's fix is to average the partial sums themselves. A sequence that wobbles can still have averages that settle down, and here the averaging turns the bad kernel into a good one.

Count how often each coefficient shows up. The term $\hat{f}(k) e_k$ is in $S_n f$ exactly when $n \ge |k|$, which is $N + 1 - |k|$ of the $N + 1$ partial sums. So $$\sigma_N f = \sum_{|k| \le N} \left(1 - \frac{|k|}{N + 1}\right) \hat{f}(k) e_k :$$ the same coefficients, tapered linearly to $0$ at frequency $N + 1$. That's easy to compute with.

Just as $S_N f$ was an integral against $D_N$, $\sigma_N f$ is an integral against the average of the kernels.

By the same count as above, $K_N(t) = \sum_{|k| \le N} \left(1 - \frac{|k|}{N + 1}\right) e^{ikt}$. The smallest case already shows the difference.

That's not luck: $K_N$ is a square for every $N$. The lemma records it, along with the two other things a good averaging kernel needs.

Positive, total mass $1$, and the mass piles up at $0$: (c) says the part of $K_N$ outside any window $|t| < \delta$ shrinks like $\frac{1}{N}$. An average against such a kernel is a genuine local average of $f$, and a continuous $f$ has no choice but to be close to its local averages.

Read the proof again with $D_N$ in place of $K_N$ and watch it break: the absolute value can't move inside without turning $D_N$ into $|D_N|$, whose integral is $\|D_N\|_1 \to \infty$. Positivity is the whole difference. You can see it in the overshoot.

The red partial sums peak near $1.18$ at every $N$; the green Fejér means never pass $1$. Positivity buys safety, but not speed. Compare the two on a few functions, smooth ones included.

The no-overshoot property is worth proving once, since it's the whole reason the green curves behave.

Fejér's theorem also settles a debt. The basis theorem on Orthonormal Bases borrowed Stone–Weierstrass for exactly one fact, and now we can pay it back with our own hands.

The second half is the one the rest of the page runs on. Before that, some Fejér means to compute. Remember the taper: frequency $k$ gets weight $1 - \frac{|k|}{N + 1}$, and anything past $N$ is gone.

The Riemann–Lebesgue Lemma

Now leave pointwise convergence and look at the coefficients themselves. For $f \in L^2(T)$, Bessel's inequality makes $\sum |\hat{f}(n)|^2$ finite, so $\hat{f}(n) \to 0$. An $L^1$ function can be far too spiky for that argument. Look at one with a jump first.

The integral $\int f(t) e^{-int} \, dt$ goes to $0$ because $e^{-int}$ wiggles faster and faster, and against a fixed $f$ the wiggles cancel. A step function is the easy case; the lemma says every integrable $f$ is.

That's the standard approximation argument: prove it for a dense class where it's trivial, then pass to the limit with a uniform bound. Watch the cancellation happen on an interval with lopsided ends.

The bars sit under $\frac{1}{\pi n}$, the same rate as the example. Try to fight the cancellation yourself. You can make one coefficient big, but you can't keep all of them big.

Here's the lemma in the form it usually shows up on a prelim, on a set instead of a function.

A warning before the next exercise. Riemann–Lebesgue is about $e^{-int}$, which averages to $0$ over a period. A fast oscillation that doesn't average to $0$, like $|\sin nt|$, doesn't integrate to $0$ against $f$; it integrates to its average times $\int f$. Same proof, one new idea.

So the recipe for $\lim \int_a^b f(t) \, g(nt) \, dt$ (with $[a, b] \subseteq [-\pi, \pi]$, applying the exercise to $f\chi_{[a, b]}$) is: the average of $g$ over a period, times $\int_a^b f$. For $g = \cos$ or $\sin$ the average is $0$, and that's Riemann–Lebesgue. Try it.

Coefficients Don't Fill $c_0(\mathbb{Z})$

Riemann–Lebesgue says $\hat{f}$ always lands in $c_0(\mathbb{Z})$, the two-sided version of $c_0$ from Bounded Operators: the sequences $(c_n)_{n \in \mathbb{Z}}$ with $c_n \to 0$ as $n \to \pm\infty$, normed by $\|c\|_\infty = \sup_n |c_n|$. It's a Banach space for the same reason $c_0$ is. Two questions: does $\hat{f}$ determine $f$, and does every sequence in $c_0(\mathbb{Z})$ come from some $f$? The first answer is yes, almost everywhere.

"Almost everywhere" is the best possible: changing $f$ on a null set changes no integral, so no coefficient. For continuous functions, though, almost everywhere is everywhere, and that gives a clean convergence test.

Now the second question. The map $\Lambda f = \hat{f}$ from $L^1(T)$ to $c_0(\mathbb{Z})$ is linear, bounded (we saw $|\hat{f}(n)| \le \|f\|_1$), and, by uniqueness, injective. If it were also onto, it would be a bounded linear bijection between Banach spaces. The bounded inverse theorem (another consequence of Baire, proved on Baire Category & Its Consequences) would then make the inverse bounded too. The Dirichlet kernels say it can't be.

Like the divergence theorem, this one is an existence proof: it says a sequence in $c_0(\mathbb{Z})$ misses the range without naming one. Here's the size of the contradiction in actual numbers.

Same kernels, side by side: the coefficients never get bigger, the function's $L^1$ norm never stops growing.

So deciding whether a given sequence is a coefficient sequence takes some care. Two tools decide most cases: if it doesn't tend to $0$ at both ends, Riemann–Lebesgue rules it out; if it's in $\ell^2(\mathbb{Z})$, Riesz–Fischer hands you an $f \in L^2(T) \subseteq L^1(T)$. Check both ends.

Mixed Practice

Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.

Next, the proofs. Each one below is a step from a proof on this page with a blank in it, and the wrong fills are the steps that look right and aren't.

Then the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types that need their answer choices will say so.

Last, the Spicy Bonus. The divergence theorem found one bad point for most continuous functions. Run it at countably many points at once, and Baire's theorem gives a single continuous function that misbehaves on a big set.