Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 1 / 11F / Solution

Let $S_N(x)={\sum }_{n=1}^N[a_n\cos (nx)+b_n\sin (nx)]$. Under the first coefficient bounds,This proves the uniform Cauchy condition on the whole real line and hence uniform convergence to $f$. It is the Weierstrass M-test applied with the summable majorant $2c{n}^{-1-\epsilon }$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequalityproves continuity. Also $S_N(x+2\pi )=S_N(x)$ for every $N$, so taking limits gives $f(x+2\pi )=f(x)$.

Under the stronger coefficient bounds, the derivative series has summable uniform majorant $2c{n}^{-1-\epsilon }$. Thus $S_N^{\prime }$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the fundamental theorem of calculus gives $S_N(x)-S_N(0)={\int }_0^x{S}_N^{\prime }(t)dt$. Uniform convergence permits passage to the limit inside this finite integral, since the error is at most $|x|\Vert S_N^{\prime }-g{\Vert }_{\infty }$. Therefore $f(x)-f(0)={\int }_0^{x}g(t)dt$, and continuity of $g$ gives $f^{\prime }=g$. This justifies Termwise differentiation of a Fourier series rather than assuming it: