If $f$ vanishes at all its points of continuity, then $\|f\| = 0.$

Question

Denote by $\mathcal R$ the set of all Riemann integrable functions on $[0,2\pi]$ with norm given by $$ \|f\| = \bigg( \frac{1}{2\pi}\int_0^{2\pi}|f(x)|^2dx\bigg)^{1/2}$$

If $f\in \mathcal R$ vanishes at all its points of continuity, then $\|f\| = 0$.

This exercise appeard on a book on Fourier analysis. The unique idea that I had was to use parseval identity: $\|f\|^2 = \sum |c_n|^2$ and then using the identity $f(x)= \sum c_n e^{inx}$ to prove that all Fourier coeffcients must be zero. But this is not possible with $f$ only being continuous at $x$. We must know more about its derivative.

Any help?

Answer 1

According to Lebesgue's classic theorem: A bounded function is Riemann integrable on $[a,b]$ iff it is continuous at almost every point of $[a,b]$. In fact, this theorem of Lebesgue is where he came up with the idea of measuring more general sets, and integrating more general functions.

In your case, if $f$ is Riemann integrable on $[a,b]$ and it vanishes at all points of continuity, then $f$ is zero almost everywhere. So you can choose Riemann sums that are all $0$ that converge to the integral, which gives you $0$ for $\int_a^b f(t)\,dt$. The same is also true of $\int_a^b f(t)^2dt$.

Answer 2

What if $f(x) = \left. \begin{cases} \sin x & \text{if $x$ is irrational,} \\ 0 & \text{otherwise} \end{cases} \right\} \text{?}$

Then $f$ is continuous only at $x\in\{0,\pi,2\pi\},$ and it vanishes there. And $\|f\|^2 = 1/2.$

PS: The question initially said "integrable" and later got expanded to "Riemann-integrable". If it's Riemann integrable, then the function above will not serve, since that is not Riemann-integrable.