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My question relates to the properties of the Fourier series of a function, $f: \mathbb{R} \to \mathbb{R}$. I know from an elementary course in differential equations (for engineers) that, for all practical purposes, if $f$ satisfies the Dirichlet conditions, then the Fourier series of f, $\mathcal{S}(f)$, is equal to $f$ everywhere, except at jump discontinuities, where it equals the "average value of $f$." That's fine.

This leads me to wonder: supposing $f$ has finitely many jump discontinuities, is $\mathcal{S}(f)$ a continuous function? (Can we even talk about the function $\mathcal{S}(f)$ independently of $f$? It may be that my question stems from my own naivete.)

I asked my instructor and he said that, no, $\mathcal{S}(f)$ can't possibly be continuous since we know that, being equal to $f$ except at the jumps, that it has jumps discontinuities itself. I thought that sufficed.

However, $\mathcal{S}(f)$ is the sum of continuous functions (in general, sines, cosines, and a constant). Shouldn't it therefore be itself a continuous function? Granted, it is an infinite sum of continuous functions so it may be that I am unfamiliar with theory concerning the continuity properties of infinite sums of continuous functions.

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You've answered it. An infinite sum of continuous functions, even if it converges everywhere, may not be continuous. – Umberto P. Jun 7 '13 at 16:10
With a few epsilons and deltas one can prove that the sum of TWO continuous functions is continuous. To go from there to the conclusion that the sum of FINITELY many continuous functions is continuous is one of the simplest exercises in mathematical induction that there are. This example shows that going from there to infinitely many is problematic. – Michael Hardy Jun 7 '13 at 17:09
Thanks a lot. That's cleared up... for now. – fourierwho Jun 11 '13 at 2:33

The mathematics behind this, it ends up, are subtle, and as such this was the subject of heated debate at the turn of the 20th century. In fact, much of the foundation we now have for analysis came directly out of the question of Fourier series.

However, to answer your question, the answer is no: the infinite sum of continuous functions does not always give you a continuous function. In fact, you don't even need to consider an $f$ with jump discontinuities; just consider the Fourier series of $f(x)=x$, which gives you the sawtooth curve.

If you take a second semester of real analysis (usually when they go over this stuff), you'll learn that there are conditions under which you can guarantee the continuity of the limit of a continuous sequence of functions. For example, if a sequence of continuous functions "converges uniformly", then the limit of that sequence is itself a continuous function. The finite cases, as it ends up, fall under the umbrella of uniformly convergent sequences; but Fourier series tend not to behave so nicely.

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At this point another iteration of what I said in comments may not be needed.

Fourier published his first results on these series in the early 19th century. It was not until 1966, that the following was proved:

If $\displaystyle\int_0^{2\pi} |f(x)|^2\,dx < \infty$ then the Fourier series of $f$, evaluated at $x$ converges to $f(x)$ for almost every value of $x$. "Almost every" means that the measure of the set of exceptions---points at which convergence fails---is zero.

(One writes $|f(x)|^2$, not $f(x)^2$, since $f(x)$ is allowed to be complex, not necessarily real.)


One moral of the story: In some cases, questions about convergence of Fourier series take a lot of work to answer!

Here's a much easier result, but still a fair amount of work: Let $\mathcal{S}_n(f)$ be the $n$th partial sum in the Fourier series.

Then if $\displaystyle\int_0^{2\pi} |f(x)|^2\,dx<\infty$ then $\displaystyle\lim_{n\to\infty}\int_0^{2\pi} \left|\mathcal{S}_n(f)(x) - f(x)\right|^2\,dx=0$.

Convergence of Fourier series is not a trivial matter.

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It is, however, relatively easy to prove that $\mathcal S_n(f)$ has a subsequence that converges p.w. a.e. to $f$. But as N. L. Carothers writes in his book Real Analysis: The study of pointwise convergence of Fourier series has a long and checkered history - to paraphrase Halmos, its history includes "almost 200 years of barking up the wrong tree." – kahen Jun 7 '13 at 17:33
Did the part you're quoting from Carothers end with the word "history" or did he include the quote from Halmos about "barking up the wrong tree"? – Michael Hardy Jun 7 '13 at 17:38
Everything after the colon – kahen Jun 7 '13 at 17:39

Suppose we completely get rid of the assumption that $f$ has finitely many discontinuities and instead ask if the Fourier series of a continuous function is necessarily continuous. One can prove (using the techniques of functional analysis) that for pretty much all continuous functions $f$ on $[0,2\pi]$, the Fourier series of $f$ diverges at pretty much all of the points in $[0,2\pi]$ (both in the Baire category sense). A proof can be found in Rudin Real and Complex Analysis Thm 5.12.

A sufficient condition for pointwise convergence of the Fourier series to the function is that the function be Holder continuous. This is the reason why classical solutions to differential equations have the nice property that you can work with the Fourier series: typically such a solution will have a continuous first derivative, which implies Holder continuity of order 1. If you ever get to the point where you are working with non-classical solutions (which come up a lot--think about solving for a Dirac delta for example) then you typically cannot just work with the Fourier series pointwise.

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An infinite sum of continuous functions need not be continuous. This is because the limit of continuous functions need not be continuous. Consider the function $x^n$ on the interval $[0,1]$. It is continuous for every positive integer $n$. However, the limit as $n \to \infty$ is 0 for all $x<1$, but 1 at $x=1$, so the limit function has a jump at $x=1$.

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