If a function is compactly supported, then its Fourier series converge?

Question

Suppose that $f$ is continuous on $\mathbb{R}$. Show that $f$ and the Fourier transform $\hat{f}$ cannot both be compactly supported unless $f=0$.

Hint: Assume $f$ is supported in $[0,1/2]$. Expand $f$ in a Fourier series in the interval $[0,1]$, and note that as a result, $f$ is a trigonometric polynomial.

I tried to solve this following the hint. Assume both $f$ and $\hat{f}$ are compactly supported, then if we follow the hint we get $$f(x)=\sum_{n=-\infty}^\infty \hat{f}(n)e^{2\pi inx}.$$ But since $\hat{f}$ is compactly supported, for large $n$, $\hat{f}(n)=0$, and we get that $f$ is a trigonometric polynomial, which is not compactly supported. However, my question is, for this to work out, we need the identity in the above, which is not guaranteed unless, say $f$ is $C^1$, but we only have that $f$ is continuous, so how can I get the result? I would greatly appreciate any help.

I've seen other solutions from this website based on the fact that $\hat{f}$ is not holomorphic, but in this book, holomorphic functions are not yet introduced. So I'm wondering if perhaps compactly supported guarantees the convergence of the Fourier series?

Answer 1

If you have learned about $L^2$ convergence of Fourier series then you have that $f(x)=\sum_n c_n e^{2\pi i n x}$ in $L^2([0,1]$ (and only finitely many terms in the sum). But both sides are continuous and agree on a set of full measure. They are therefore identical. Since $f$ is identical zero on $[\frac12,1]$ both must be zero. Without $L^2$ (or holomorphic), I am not sure how to argue.

Answer 2

Even with H.H. Rugh's observation, there's a problem with your proof. We have

$$f(t) = \sum_{-\infty}^{\infty}\hat f (n)e^{2\pi i n t}$$

in $L^2[0,1],$ not in $L^2(\mathbb R).$ We're still OK, but the point is that $f$ equals a trig. poly $T$ a.e. in $[0,1].$ This implies $T=0$ on $[1/2,1],$ which is impossible for the trig. poly. $T$ unless $T \equiv 0.$