Fourier series at discontinuities I was reading about Fourier series and have a doubt concerning it. The book I am reading from does not seem to help. As I understand, $\{e_0=\frac{1}{\sqrt{2}},e_1=sin(x),e_2=cos(x),e_3=sin(2x),e_4=cos(2x)\cdots\}$ is a basis for the inner product space of piecewise continuous functions in $[-\pi,\pi]$ with inner product $<f,g>=\frac{1}{\pi}\int_{-\pi}^{\pi}f\bar g$. Hence any function in this space may be represented by $f=\sum_{k=0}^{\infty}<f,e_k>e_k$. My question is what happens at points of discontinuity x. As f is identical with the series (which by the way is unclear to me as to why it coverges) shouldn't f(x) be identical with the series at x, i.e. $f(x)=\sum_{k=0}^{\infty}<f,e_k>e_k(x)$. But Dirichlet's theorem (stated without proof in my book) says that at points of discontinuity, the series $\sum_{k=0}^{\infty}<f,e_k>e_k(x)$ converges to $\frac{f(x-)+f(x+)}{2}$ and not to f(x). Why is this so? Thanks.
 A: The suggestion by Hans to look in Chernoff's paper is a very good one.  I haven't looked at it in a long time but I do believe it is the source of the following argument.  (Note added later:  it does indeed appear to be the source of this approach.)
The following theorem is well-known.
Theorem. Let $f$ be Lebesgue integrable, $2\pi$-periodic, and suppose that
$$
\frac{f(x)-f(x_0^{\pm})}{|x-x_0|}
$$
exists as $x\rightarrow x_0$, where $f(x_0^{\pm})$ denotes the right and left hand limits respectively.  Then the Fourier series at $x_0$, $\hat{f}(x_0)$ converges to $f(x_0)$.
Proof hint.  Consider the Fourier series of the auxilliary function $g(x) = f(x)/(e^{ix}-1)$.
Using this theorem we can prove the following, which is the main concern of your question.
Corollary.  Suppose $f$ has a jump discontinuity at zero, and suppose the average
$$
\frac{f(x)+f(-x)}{x}
$$
exists as $x\rightarrow0$.  Then the Fourier series at zero converges to this limit.
Note: By only talking about the average instead of the left and right hand limits, we don't formally need them to individually exist.
Proof.  By subtracting a constant from $f$, we may assume that $f(0^+) = -f(0^-)$.  Denote by $S_n(x)$ the $n$-th partial sum of the (symmetric) Fourier series of $f$.  Then
$$
S_n(0) = \int_{-\pi}^{\pi} f(x) \Big[\frac{1}{2\pi}\sum_{k=-n}^{n}e^{ikx}\Big] dx = \int_{-\pi}^{\pi} f(x) D_n(x) dx
$$
where $D_n$ is the Dirichlet kernel.  Note that it is even, so in fact
$$
S_n(0) = \frac{1}{2}\int_{-\pi}^{\pi} \big[f(x) + f(-x)\big]D_n(x) dx.
$$
Applying the theorem above to the average $\frac{1}{2}\big[f(x) + f(-x)\big]$ allows us to conclude the result.
For more details, I would strongly recommend to look in the paper of Chernoff.
A: Because Fourier Series are approximating a discontinuous function with a set of continuous basis functions.  No matter how many are used the estimate must remain continuous at $f(x)$.  Also note that due to Gibbs Phenomenon, the point at $x+$ converges to about 9% above $f(x+)$ (relative to the jump $|f(x+) - f(x-)|$.
A: Remember, the basic mode of convergence given by Fourier series is mean-square convergence.  This is true, provided that the function you are computing a Fourier expansion for is square-integrable.  Convergence at points is a separate issue.  
