# Rudin lecture — if f(x) is not integrable on some interval, does it not have a Fourier Series expansion on that interval?

I found an old lecture on YouTube given by Walter Rudin (1990, in Wisconsin), and towards the beginning he mentions that if $$f(x)$$ were not integrable, on some interval, it would be obvious that it doesn't have a Fourier Series on that interval, since if you looked at the Fourier coefficient given by

$$\frac{1}{\pi}\int_0^\pi f(x) \cos(nx) dx$$

it wouldn't be meaningful, given that $$f$$ is not integrable.

I was a bit confused because I'm wondering whether the product $$f(x)\cos(nx)$$ could be integrable, and so one could indeed compute the coefficients.

So my question is: if $$f(x)$$ is not integrable, how do we know that the product $$f(x)\cos(nx)$$ is not integrable?

If $$f$$ is not integrable, then the first fourier coefficient obtained by considering $$n=0$$ (i.e., the mean) is not defined.

Other Fourier coefficients are likely to be undefined as well (but some could very well exist, as Kirill already pointed out).

In any case, the existence of the Fourier Series for such a function is doomed.

The point is to guarantee that the Fourier coefficients exist, not to make really fine distinctions about what might happen if the guarantee is violated. Furthermore, you would be hard pressed to find a reasonable function for which this is an issue, if it exists at all. The class of integrable functions is already so wide as to include everything you might want to compute the Fourier coefficients of anyway.

The statement as you quoted it doesn’t seem right because $$1/\cos x$$ isn’t integrable but has some finite Fourier coefficients.

Applying Cauchy-Schwartz inequaility, we have: \begin{aligned} \left(\int_{0}^{\pi}f(x)\cos{(x)}\,dx \right)^2&\leq \left(\int_{0}^{\pi}|f(x)\cos{(x)}|\,dx \right)^2\\ &\leq \int_{0}^{\pi}f(x)^2\,dx \int_{0}^{\pi}\cos{(x)}^2\,dx=\frac{\pi}{2}\int_{0}^{\pi}f(x)^2\,dx \end{aligned}

And if $$f(x)\notin {L}^2(x)$$ then the Fourier coefficients are not bounded.

• This isn’t right, if the rhs is unbounded, the lhs might still exists and be bounded, did you mean to derive a $\geq$ inequality? Eg $f=1/\cos x$. – Kirill Dec 24 '18 at 21:28
• Also, usually the condition on f is that it should be absolutely integrable, but in your answer it has to be square integrable instead, eg take $f=1/\sqrt x$. – Kirill Dec 24 '18 at 21:39