Editing Convergence of Fourier series (section)

==Pointwise convergence==
{{see also|Dirichlet conditions|Dini test}}
[[File:Sawtooth Fourier Analysis.JPG|thumb|280px|Superposition of sinusoidal wave basis functions (bottom) to form a sawtooth wave (top); the basis functions have [[wavelength]]s λ/''k'' (''k''=integer) shorter than the wavelength λ of the sawtooth itself (except for ''k''=1). All basis functions have nodes at the nodes of the sawtooth, but all but the fundamental have additional nodes. The oscillation about the sawtooth is called the [[Gibbs phenomenon]]]]
There are many known sufficient conditions for the Fourier series of a function to converge at a given point ''x'', for example if the function is [[Differentiable function|differentiable]] at ''x''. Even a jump discontinuity does not pose a problem: if the function has left and right derivatives at ''x'', then the Fourier series converges to the average of the left and right limits (but see [[Gibbs phenomenon]]).

The '''Dirichlet–Dini Criterion''' states that:<ref>Teschl, Corollary 8.9</ref> if ''ƒ'' is 2{{pi}}–periodic, locally integrable and satisfies

:<math>\int_0^{\pi} \left| \frac{f(x_0 + t) + f(x_0 - t)}2 - \ell \right| \frac{\mathrm{d}t }{t} < \infty,</math>

then (S<sub>''n''</sub>''f'')(''x''<sub>0</sub>) converges to ℓ.  This implies that for any function ''f'' of any [[Hölder condition|Hölder class]] ''α''&nbsp;> 0,  the Fourier series converges everywhere to ''f''(''x'').

It is also known that for any periodic function of [[bounded variation]], the Fourier series converges. In general, the most common criteria for pointwise convergence of a periodic function ''f'' are as follows:

* If ''f'' satisfies a Holder condition, then its Fourier series converges uniformly.<ref>Teschl, Theorem 8.12</ref>
* If ''f'' is of bounded variation, then its Fourier series converges everywhere. If ''f'' is additionally continuous, the convergence is uniform.<ref>Teschl, Theorem 8.14</ref>
* If ''f'' is continuous and its Fourier coefficients are absolutely summable, then the Fourier series converges uniformly.<ref>Follows from the [[Weierstrass M-test]]</ref>

There exist continuous functions whose Fourier series converges pointwise but not uniformly.<ref> Zygmund, ''[[Trigonometric Series]]'', vol. 1, Chapter 8, Theorem 1.13, p.&nbsp;300</ref>

However, the Fourier series of a [[continuous function]] need not converge pointwise. Perhaps the easiest proof  uses the non-boundedness of Dirichlet's kernel in ''L''<sup>1</sup>('''T''') and the Banach–Steinhaus [[uniform boundedness principle]]. As typical for existence arguments invoking the [[Baire category theorem]], this proof is nonconstructive. It shows that the family of continuous functions whose Fourier series converges at a given ''x'' is of [[Baire space|first Baire category]], in the [[Banach space]] of continuous functions on the circle. 

So in some sense pointwise convergence is ''atypical'', and for most continuous functions the Fourier series does not converge at a given point. However [[Carleson's theorem]] shows that for a given continuous function the Fourier series converges almost everywhere.

It is also possible to give explicit examples of a continuous function whose Fourier series diverges at 0: for instance, the even and 2π-periodic function ''f'' defined for all ''x'' in [0,π] by<ref>Teschl Example 8.6 or {{cite book |last=Gourdon |first= Xavier|title= Les maths en tête. Analyse (2ème édition)|language= french| date=2009 |publisher= Ellipses|page=264 |isbn=978-2729837594}}</ref>

:<math>f(x) = \sum_{n=1}^{\infty} \frac{1}{n^2} \sin\left[ \left( 2^{n^3} +1 \right) \frac{x}{2}\right].</math>

In this example it is easy to show how the series behaves at zero. Because the function is even the Fourier series contains only cosines:

:<math>f(x) \sim \sum_{m=0}^\infty C_m \cos(mx).</math>

The coefficients are:

:<math>\begin{align}
C_m&=\frac 2\pi\sum_{n=1}^{\infty} \frac{1}{n^2}\int_0^\pi \sin\left[ \left( 2^{n^3} +1 \right) \frac{x}{2}\right]\cos{mx}\,\mathrm{d}x\\
& =\frac 1\pi\sum_{n=1}^{\infty} \frac{1}{n^2}\int_0^\pi \left\{\sin\left[ \left( 2^{n^3} +1-2m\right) \frac{x}{2}\right]+\sin\left[ \left( 2^{n^3} +1+2m\right) \frac{x}{2}\right]\right\}\,\mathrm{d}x\\
& =\frac 1\pi\sum_{n=1}^{\infty} \frac{1}{n^2} \left\{\frac 2{2^{n^3} +1-2m}+\frac 2{2^{n^3} +1+2m}\right\}\\
\end{align}</math>

As {{mvar|m}} increases, the coefficients will be positive and increasing until they reach a value of about <math>C_m\approx 2/(n^2\pi)</math> at <math>m=2^{n^3}/2</math> for some {{mvar|n}} and then become negative (starting with a value around <math>-2/(n^2\pi)</math>) and getting smaller, before starting a new such wave. At <math>x=0</math> the Fourier series is simply the running sum of <math>C_m,</math> and this builds up to around

:<math>\frac 1{n^2\pi}\sum_{k=0}^{2^{n^3}/2}\frac 2{2k+1}\sim\frac 1{n^2\pi}\ln 2^{n^3}=\frac n\pi\ln 2</math>

in the {{mvar|n}}th wave before returning to around zero, showing that the series does not converge at zero but reaches higher and higher peaks.