# Dirichlet Conditions and Fourier Analysis.

I read in my text book that the Dirichlet conditions are sufficient conditions for a real-valued, periodic function $f(x)$ to be equal to the sum of its Fourier series at each point where $f$ is continuous.

However, it further stated that although the conditions are sufficient but they are NOT necessary.

Why are the Dirichlet conditions "not necessary" ?

Example : One of the Dirichlet conditions state that the function can not have infinite discontinuities. Hence we can not express, a function like $tan x$ in terms of a Fourier series since (as it appears) violates one of the conditions. So, why is that they are 'not necessary'?

P.S.: The Wikipedia Link to the Dirichlet conditions.

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$\tan{x}$ is not continuous at the point at which it appears to violate the Dirichlet conditions. –  Ron Gordon Jan 3 '13 at 5:14
Yes, that is why we can not express it in terms of Fourier series. It has an infinite discontinuity at pi/2 and hence can not be expanded in the interval -pi to pi. The question is why are the Dirichlet conditions deemed 'NOT NECESSARY'. –  noir1993 Jan 3 '13 at 5:22
Something like $f(x)=\sin(1/x)$ –  Alex R. Jan 3 '13 at 19:22
I think the confusion stems from the usage of "not necessary" in mathematical writing, which is different from colloquial English. Saying that a condition is not necessary does not mean that it's something we can throw out. See necessity and sufficiency –  user53153 Jan 4 '13 at 3:57

The conditions are "not necessary" because no one proved a theorem that if the Fourier series of a function $f(x)$ converge pointwise then the function satisfies the Dirichlet conditions.
Some general information: the issues of the pointwise converges of Fourier series are very delicate matter. As far as I understand we still do not have the exact condition to fully identify the class of functions whose Fourier series converge. It was proved, as @chaohuang pointed in another answer, that if $f\in L^2$, then the Fourier series converge. On the other hand, A. Kolmogorov provided an example of a function $f\in L^1$ whose Fourier series do not converge at any point. I am sure you can find much more information on these issues in any serious text on Fourier analysis.
Another sufficient condition is $f(x) \in \mathbb{L}_2(-\frac{1}{2}L,\frac{1}{2}L)$.