Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Just a little point that might help you to think it: suppose $f_1(x)$ has a discontinuity at $x=a$, and that $f_1(a)=f_1(a-)$. Let $f_2(x)$ be identical to $f_1(x)$ except that $f_2(a)=f_2(a+)$. Further, let $f_3(x)$ be identical except that $f_3(a)$ is another (whatever) value. All these functions have the same Fourier series. – leonbloy Jul 1 '11 at 3:53
Find a book which does contain the proof, and read it! ;-) – Hans Lundmark Jul 1 '11 at 6:29
There's also Chernoff's famous short paper:… – Hans Lundmark Jul 1 '11 at 6:30
@Hans I started writing my answer below before looking at the paper, but did have a vague recollection of it. It turns out that in the end, the argument is identical to that found in Chernoff's paper. Do you know if this is the original source of the argument? – Glen Wheeler Jul 1 '11 at 10:54
@Glen: I guess so, but I'm not an expert at all. – Hans Lundmark Jul 1 '11 at 11:09
up vote 4 down vote accepted

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.

share|cite|improve this answer
the "average" should be $\frac{f(x)+f(-x)}{2}$ – foaly Jun 17 '15 at 12:20

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 $|f(x+) - f(x-)|$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.