Fourier Series Convergence

Consider the one-sided Fourier partial sums

$\displaystyle S_N(x):=\sum_{n=1}^{N}\frac{e^{inx}}{n},\quad x\in \mathbb T =[0, 2\pi).$

This is one of the first Fourier series in which several different notions of convergence visibly separate from one another.

The coefficients satisfy $\displaystyle \sum_{n=1}^{\infty}\frac1{n^2}<\infty,$ but $\displaystyle \sum_{n=1}^{\infty}\frac1n=\infty.$ Thus the coefficient sequence belongs to $\ell^2$ , but not to $\ell^1$ . Square summability will give an $L^2$ limit. The failure of absolute summability means that uniform convergence is not automatic. Indeed, we will see that the series converges at every nonzero point of the circle but diverges at $x=0$ .

The example is useful because it reveals four different ideas which should not be confused. Orthogonality constructs the $L^2$ limit. Oscillation and summation by parts prove pointwise convergence away from the singular point. A subsequence argument identifies the pointwise sum with the $L^2$ function almost everywhere. Finally, the comparison with $1/n^2$ shows how one additional power of decay changes a singular $L^2$ function into a continuous function.

L_2 convergence

For $M>N$ , orthogonality of the exponentials gives

$\displaystyle \begin{aligned} ||S_M-S_N||^2 &=\int_0^{2\pi}\Big |\sum_{n=N+1}^{M}\frac{e^{inx}}{n}\Big |^2dx \\ &=2\pi\sum_{n=N+1}^{M}\frac1{n^2}. \end{aligned}$

The tail on the right tends to $0$ as $M,N\to\infty$ . Hence $(S_N)$ is Cauchy in $L^2(\mathbb T)$ , and completeness gives a function $f\in L^2(\mathbb T)$ such that $S_N\longrightarrow f$ in $L^2(\mathbb T)$ . This is the construction of the function with Fourier coefficients

$\displaystyle \widehat f(n)=\begin{cases}\frac1n,&n\ge1,\\ 0,&n\le0.\end{cases}$

The statement $S_N\to f$ in $L^2$ means $\int_0^{2\pi}|S_N(x)-f(x)|^2 dx\longrightarrow0.$

It says that the average squared error tends to zero. It does not yet say that $S_N(x)$ converges for each individual $x$ . This distinction is essential: convergence in $L^2$ is an averaged statement, while pointwise convergence is a statement at each separate point.

Pointwise Convergence

Fix $x\not\equiv0\pmod{2\pi}$ . The exponential terms now rotate around the unit circle rather than all pointing in the same direction. Their partial sums are geometric sums:

$\displaystyle A_k(x):=\sum_{j=1}^{k}e^{ijx}=\frac{e^{ix}(1-e^{ikx})}{1-e^{ix}}.$

Since $|1-e^{ikx}|\le2$ , we have

$\displaystyle |A_k(x)|\le\frac{2}{|1-e^{ix}|}.$

For a fixed nonzero $x$ , this is a bound independent of $k$ . The partial sums of the phases stay bounded because of cancellation. This is exactly what fails at $x=0$ . Apply summation by parts to the tail of the series. For $M>N$ ,

$\displaystyle \sum_{n=N+1}^{M}\frac{e^{inx}}{n} =\frac{A_M(x)}{M}-\frac{A_N(x)}{N+1}+\sum_{n=N+1}^{M-1}A_n(x)\Big (\frac1n-\frac1{n+1}\Big ).$

Using the bound for $A_n(x)$ , we obtain

$\displaystyle \begin{aligned} \Big |\sum_{n=N+1}^{M}\frac{e^{inx}}{n}\Big | &\le\frac{2}{M|1-e^{ix}|} +\frac{2}{(N+1)|1-e^{ix}|} \\ &\quad+\frac{2}{|1-e^{ix}|} \sum_{n=N+1}^{M-1}\Big (\frac1n-\frac1{n+1}\Big ). \end{aligned}$

The final sum telescopes, and therefore

$\displaystyle \Big |\sum_{n=N+1}^{M}\frac{e^{inx}}{n}\Big | \le\frac{4}{(N+1)|1-e^{ix}|}.$

For every fixed $x\ne0$ , the right-hand side tends to $0$ as $N\to\infty$ . Thus $\displaystyle \sum_{n=1}^{\infty}\frac{e^{inx}}{n}$ converges for every $x\not\equiv0\pmod{2\pi}$ . The estimate is also uniform on every closed arc which avoids $0$ . On such an arc, $|1-e^{ix}|$ is bounded below by a positive constant. Hence the series converges uniformly on $[\epsilon, 2\pi -\epsilon)$

Its sum is therefore continuous away from $0$ . At the exceptional point, there is no oscillation: $S_N(0)=\sum_{n=1}^{N}\frac1n.$ This is the harmonic series, so it diverges. The obstruction is completely concrete: at $x=0$ , every phase $e^{inx}$ equals $1$ , and the cancellation disappears.

Let the pointwise limit function be

$\displaystyle g(x):=\sum_{n=1}^{\infty}\frac{e^{inx}}{n},\quad x\ne0.$

The preceding argument shows that $S_N(x)\to g(x)$ at every nonzero point. On the other hand, we already know that $S_N\to f$ in $L^2(\mathbb T)$ . We now show that the two limits agree almost everywhere.

Choose a subsequence $S_{N_j}$ such that $||S_{N_j}-f||_{2}^2<\frac{1}{2^{3j}}.$ Define $E_j:=\{x\in\mathbb T:|S_{N_j}(x)-f(x)|>2^{-j}\}.$ Chebyshev’s inequality gives

$\displaystyle |E_j| \le2^{2j}||S_{N_j}-f||_2^2 <\frac{1}{2^{j}}.$

Thus $\displaystyle \sum_{j=1}^{\infty}|E_j|<\infty.$ For every $J$ , define $F_J:=\bigcup_{j\ge J}E_j.$ Then

$\displaystyle |F_J|\le\sum_{j\ge J}|E_j|,$

and the right-hand side tends to $0$ as $J\to\infty$ . Therefore the set of points belonging to infinitely many of the $E_j$ has measure zero. For almost every $x$ ,

$\displaystyle |S_{N_j}(x)-f(x)|\le \frac{1}{2^{j}}$

for all sufficiently large $j$ (depending on x). Hence $S_{N_j}(x)\longrightarrow f(x)$ for almost every $x$ .

But whenever $x\ne0$ , the full sequence $S_N(x)$ converges to $g(x)$ . Its subsequence must converge to the same value. Therefore $f(x)=g(x)$ for almost every $x\in\mathbb T$ . Thus $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{e^{inx}}{n}$ for almost every $x$ .

There are three genuinely separate ingredients here. Square summability constructs $f$ in $L^2$ . Summation by parts constructs the pointwise sum $g$ away from $0$ . The subsequence argument proves that these constructions describe the same function almost everywhere.

Computing the function

For $0<r<1$ , the damped series converges $\displaystyle \sum_{n=1}^{\infty}\frac{r^ne^{inx}}{n}$ absolutely to $-\log(1-re^{ix}).$ This is the power-series identity $\displaystyle -\log(1-w)=\sum_{n=1}^{\infty}\frac{w^n}{n},\quad |w|<1,$ with $w=re^{ix}$ . For $0<x<2\pi$ , the original undamped series already converges. Letting $r\uparrow1$ therefore gives

$\displaystyle \sum_{n=1}^{\infty}\frac{e^{inx}}{n} =-\log(1-e^{ix}).$

Now $\displaystyle 1-e^{ix} =2\sin\left(\frac{x}{2}\right)e^{i(x/2-\pi/2)}, \quad 0<x<2\pi.$ Since $\sin(x/2)>0$ on this interval,

$\displaystyle \log(1-e^{ix}) =\log(2\sin(\frac{x}{2})) +i(\frac{x}{2}-\frac{\pi}{2} ).$

Hence $\displaystyle \sum_{n=1}^{\infty}\frac{e^{inx}}{n} =-\log (2\sin(\frac{x}{2} ) +i\frac{\pi-x}{2}, \quad 0<x<2\pi.$ Taking real and imaginary parts gives $\displaystyle \sum_{n=1}^{\infty}\frac{\cos(nx)}{n} =-\log(2\sin(\frac{x}{2})), \quad 0<x<2\pi,$ and $\displaystyle \sum_{n=1}^{\infty}\frac{\sin(nx)}{n} =\frac{\pi-x}{2}, \quad 0<x<2\pi.$ The imaginary part is a sawtooth function. The real part has a logarithmic singularity at $x=0$ . Since $2\sin\left(\frac{x}{2}\right)=x+O(x^3)$ as $x\to0$ , we have $-\log(2\sin(\frac{x}{2}) =-\log|x|+O(1).$ The function is unbounded near $0$ , but it still belongs to $L^2$ , because

$\displaystyle \int_0^\varepsilon|\log x|^2,dx<\infty.$

This is a concrete example of an $L^2$ function with a true point singularity.

Another Example:

Now consider $\displaystyle T_N(x):=\sum_{n=1}^{N}\frac{e^{inx}}{n^2}.$ The coefficients are absolutely summable: $\displaystyle \sum_{n=1}^{\infty}\frac1{n^2}<\infty.$ For $M>N$ , we have

$\displaystyle \sup_{x\in\mathbb T} |T_M(x)-T_N(x)| \le\sum_{n=N+1}^{M}\frac1{n^2}.$

The right-hand side tends to $0$ independently of $x$ . Hence $T_N$ converges uniformly on the entire circle. Its limit

$\displaystyle T(x):=\sum_{n=1}^{\infty}\frac{e^{inx}}{n^2}$

is continuous and $2\pi$ -periodic. In particular, $\displaystyle T(0)=\sum_{n=1}^{\infty}\frac1{n^2} =\frac{\pi^2}{6}.$

The difference from the $1/n$ example is now sharp. The coefficient sequence $1/n$ is in $\ell^2$ but not in $\ell^1$ , so one obtains only an $L^2$ function and pointwise convergence away from the singularity. The coefficient sequence $1/n^2$ is in $\ell^1$ , so uniform convergence gives a continuous function immediately. However, continuity is not the same as differentiability. On every closed arc avoiding $0$ , the differentiated series converges locally uniformly: $\displaystyle T'(x)=i\sum_{n=1}^{\infty}\frac{e^{inx}}{n}.$ Therefore, for $0<x<2\pi$ ,

$\displaystyle T'(x) =-\frac{\pi-x}{2} -i\log\left(2\sin\left(\frac{x}{2}\right)\right).$

Thus $T$ is continuous everywhere, but its derivative has a logarithmic singularity at $0$ . The extra power of decay is enough for continuity, but differentiation removes that power and returns us to the borderline $1/n$ situation.

Coefficient decay and smoothness

For a general Fourier series

$\displaystyle \sum_{n\in\mathbb Z}a_ne^{inx},$

there is a useful hierarchy. If $\sum_{n\in\mathbb Z}|a_n|^2<\infty,$ then the coefficients define an $L^2(\mathbb T)$ function, and the Fourier partial sums converge in $L^2$ . If $\sum_{n\in\mathbb Z}|a_n|<\infty,$ then the series converges uniformly, and its sum is continuous. More generally, if $\sum_{n\in\mathbb Z}|n|^k|a_n|<\infty,$ then the series can be differentiated $k$ times term by term. Indeed,

$\displaystyle \frac{d^k}{dx^k}\sum_{n\in\mathbb Z}a_ne^{inx} =\sum_{n\in\mathbb Z}(in)^ka_ne^{inx},$

and the final series converges uniformly. The resulting function belongs to $C^k(\mathbb T)$ . There is also a converse direction. If $f$ has $k$ integrable derivatives, then integration by parts gives $\displaystyle \widehat f(n)=\frac{\widehat{f^{(k)}}(n)}{(in)^k}, \quad n\ne0.$ Hence $\displaystyle |\widehat f(n)| \le\frac{||f^{(k)}||_{1}} {2\pi|n|^k}.$ Smoothness forces decay of Fourier coefficients. But the reverse implication is weaker at the endpoint. The estimate $|a_n|\ll|n|^{-k}$ does not automatically make $\sum|n|^k|a_n|$ convergent. Borderline logarithms and the phases of the coefficients can matter.

Carleson Theorem

The direct proof above used the special structure of the coefficients $1/n$ . The bounded geometric sums and summation by parts gave a concrete proof of convergence at every nonzero point. For a general $L^2$ Fourier series there may be no analogous elementary cancellation estimate.

Carleson’s theorem gives the general result. If $f\in L^2(\mathbb T)$ has Fourier coefficients $\widehat f(n)$ , define the symmetric partial sums

$\displaystyle \mathcal S_Nf(x):=\sum_{|n|\le N}\widehat f(n)e^{inx}.$

Then $\displaystyle \mathcal S_Nf(x)\longrightarrow f(x)$ for almost every $x\in\mathbb T$ . A stronger formulation is the maximal inequality

$\displaystyle ||\sup_{N\ge1}|\mathcal S_Nf| ~||_2 \le C||f||_2.$

This is a deep theorem. Orthogonality alone gives $L^2$ convergence, but it does not control the pointwise maximal function. Carleson’s theorem supplies precisely that missing control and turns $L^2$ information into almost-everywhere convergence.

Carleson’s theorem therefore implies almost-everywhere convergence using the maximal estimate. The direct argument is stronger in this special case: it proves convergence at every point except $x=0$ . In general, almost-everywhere convergence cannot be replaced by everywhere convergence. There are continuous functions whose Fourier series diverge at some points. On the other hand, $L^1$ is too large: there are integrable functions whose Fourier series diverge almost everywhere. The $L^2$ theorem is therefore a delicate and important threshold.

The series $\displaystyle \sum_{n=1}^{\infty}\frac{e^{inx}}{n}$ lies exactly between the easy absolute-convergence world and the general $L^2$ world. It is not absolutely summable, so uniform convergence fails and divergence is possible. It is square-summable, so an $L^2$ limit exists. Its particular oscillation is strong enough to prove convergence at every nonzero point, while the loss of oscillation at $0$ produces the harmonic-series divergence. The resulting function has a logarithmic singularity, but that singularity remains square-integrable. The series $\displaystyle \sum_{n=1}^{\infty}\frac{e^{inx}}{n^2}$ has one extra power of decay. That is enough for absolute and uniform convergence, hence for continuity. But one derivative removes that extra power and reveals the same logarithmic singularity seen in the $1/n$ example.

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