Let \(f \in L^1(S^1)\) such that \(\widehat{f} \in \ell^1(\mathbb{Z})\). Prove that \(f \in C(S^1)\) (that is, \(f\) is equal almost everywhere to a continuous function on the circle \(S^1\)).
Solution
Write \(\widehat f(n)=\frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,dx\) and let \(g(x)\da\sum_{n\in\ZZ}\widehat f(n)e^{inx}\).
1. \(g\) is continuous on \(S^1\).
\(\abs{\widehat f(n)e^{inx}}=\abs{\widehat f(n)}\) and \(\sum_n\abs{\widehat f(n)}<\infty\), so by the Weierstrass \(M\)-test the series converges uniformly on \(\RR\). Its terms are continuous and \(2\pi\)-periodic, so \(g\) is too.
2. \(\widehat g(k)=\widehat f(k)\) for all \(k\in\ZZ\).
Uniform convergence allows integrating term by term, and \(\frac1{2\pi}\int_{-\pi}^\pi e^{i(n-k)x}\,dx\) is \(1\) for \(n=k\) and \(0\) otherwise.
3. Q.E.D.
By step 2, \(f-g\in L^1(S^1)\) has all Fourier coefficients \(0\), so \(f-g=0\) almost everywhere by the uniqueness theorem for Fourier coefficients of \(L^1(S^1)\) functions, a consequence of Fejér’s theorem [@SS03a]. By step 1, \(f\) equals the continuous function \(g\) almost everywhere.