# Equality of fouriers expression

When we are resolving a periodic piecewise continuous function in a fouriers series of harmonic terms, as follows: f(x)= a0/2 +Σancos nΠx/L + Σ bnsin nΠx/L, period 2L

we are essentially expressing a sectionally discontinuous function (with finite no of such finite discontinuities, if at all) by a continuous function, arent we? As in the series is assuming the mean value of the the function at evry point, to define itself. Correct? So how does the equality hold, if you consider the definition rigorously? Im at a loss here :’(

By Fejer's theorem, the series converges to (1/2)(f(x+) + f(x-)) at a discontinuity ( & not f(x)).

exactly. thats what i mean by 'assuming mean value at every point'. so the series IS defined at discontinuous points of the function, or points of no definition. so i return to my question: how does the equality hold?

Your question "how does an equation hold?" is not very clear.

It holds so that the left and right side are equal? Anyway, if you were interested in some rigor assumptions that guarantee the convergence, here's something:

http://mathworld.wolfram.com/DirichletFourierSeriesConditions.html

When I read about convergence from one book, it used an assumption about bounded variation property, but it could be it is the same thing as this finite number of extrema property. You can formulate the assumptions in different ways.

Mute
Homework Helper
A sum of infinitely many analytic functions may indeed by a non-analytic function. This is the crux of how a partition function in statistical mechanics can represent more than one phase.

Of course, this doesn't necessarily resolve your question, because, for example, it is well known that at the discontinuous points the fourier series overshoots the value of one by some amount. This is known as the Gibbs phenomenon.

In any event, it may be best to consider the fourier series not as being equal to the original function, but as being a representation of it, which turns out to be perfect for continuous functions but has some imperfections for discontinuous ones.

jbunniii
Homework Helper
Gold Member
we are essentially expressing a sectionally discontinuous function (with finite no of such finite discontinuities, if at all) by a continuous function, arent we? As in the series is assuming the mean value of the the function at evry point, to define itself. Correct? So how does the equality hold, if you consider the definition rigorously? Im at a loss here :’(

Very simple, equality does NOT hold at points of discontinuity. That is, if

$$a_n = \int f(x) e^{-inx} dx$$

it isn't necessarily true that

$$f(x) = \sum_n a_n e^{inx}$$

for all $x$.

In fact, it isn't necessarily true if $f$ is continuous. The best that can be said in general is that if $f$ is continuous (or more generally, $f \in L^p$ for some $1 < p < \infty)$, then its Fourier series converges to $f(x)$ for almost every $x$. That result was not known until the 1960s (Carleson's theorem).

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In fact, it isn't necessarily true if $f$ is continuous. The best that can be said in general is that if $f$ is continuous (or more generally, $f \in L^p$ for some $1 < p < \infty)$, then its Fourier series converges to $f(x)$ for almost every $x$.

You shouldn't state claims like this. You never know if there is still something that you don't know about yet.

There exist assumptions which will guarantee point wise convergence too. If $f$ is continuous and has the bounded variation property at some environment of $x_0$, then the Fourier inverse transform will converge towards the value $f(x_0)$.

jbunniii
There exist assumptions which will guarantee point wise convergence too. If $f$ is continuous and has the bounded variation property at some environment of $x_0$, then the Fourier inverse transform will converge towards the value $f(x_0)$.