Editing Dirichlet–Jordan test (section)

==Dirichlet–Jordan test for Fourier series==
Let <math>f(x)</math> be complex-valued [[Lebesgue_integral#Complex-valued_functions|integrable]] function on the interval <math>[-\pi,\pi]</math> and the [[Series_(mathematics)#Partial_sum_of_a_series|partial sums]] of its Fourier series <math>S_nf(x)</math>, given by
<math display="block">S_nf(x) = \sum_{k=-n}^nc_k e^{ikx},</math>
with [[Fourier coefficients]] <math>c_k</math> defined as
<math display="block"> c_k = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-ikx}\, dx.</math>
The Dirichlet-Jordan test states that if <math>f</math> is of [[bounded variation]], then for each <math>x \in [-\pi,\pi]</math> the limit <math>S_nf(x)</math> exists and is equal to{{sfn|Zygmund|Fefferman|2003|p=57}}{{sfn|Lion|1986|pp=281–282}}
<math display="block">\lim_{n \to \infty} S_nf(x) =\lim_{\varepsilon\to 0}\frac{f(x+\varepsilon)+f(x-\varepsilon)}{2}.</math>
Alternatively, Jordan's test states that if <math>f\in L^1</math> is of bounded variation in a neighborhood of <math>x</math>, then the limit of <math>S_nf(x)</math> exists and converges in a similar manner.{{sfn|Edwards|1979|p=156}}  

If, in addition, <math>f</math> is continuous at <math>x</math>, then
<math display="block">\lim_{n \to \infty} S_nf(x) = f(x).</math>
Moreover, if <math>f</math> is continuous at every point in <math>[-\pi,\pi]</math>, then the convergence is [[uniform convergence|uniform]] rather than just [[pointwise convergence|pointwise]].

The analogous statement holds irrespective of the choice of period of <math>f</math>, or which [[Fourier_series#Synthesis|version of the Fourier series]] is chosen.