Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 16E / Solution

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. First compute the Fourier coefficients of the periodized sum. Absolute integrability givesThus Fubini's theorem justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ giveReindexing a convergent periodization gives periodicity; the integral argument always gives its periodic representative almost everywhere. This proves the Fourier-coefficient content of the Poisson summation formula.

There is a genuine qualification to the printed pointwise claim: the periodization of an integrable function need not agree pointwise with its Fourier series under the stated hypotheses alone. Set $F(0)=1$ and $F(x)=0$ for $x\ne 0$. This is absolutely integrable, and on one closed period the periodization has only two possibly nonzero summands, so the series is uniformly convergent. Yet $f(0)=1$, while $\hat{F}(m)=0$ for every $m$. The claimed pointwise equality at zero is therefore false. This counterexample also works for Riemann integrability, not only for an almost-everywhere convention.

The coefficient identity gives a rigorous general replacement: the weighted Fejér sumsconverge to the periodization in the integral norm, and at every continuity point. Indeed they are its convolution with the normalized Fejér kernel; that kernel is nonnegative, has integral one, and has integral tending to zero outside every fixed neighbourhood of zero. For convergence in the L1 norm, split the convolution error into small translations, controlled by continuity of translation in the integral norm, and the remaining tail. The same split gives pointwise convergence at a continuity point. If the periodization is piecewise continuously differentiable, the ordinary Fourier series has the usual midpoint limit; if it is continuous and its coefficients are absolutely summable, the displayed unweighted formula holds pointwise. These are sufficient interpretations of the intended identity.

For the specified exponential example all the needed Fourier series convergence properties do hold. Split its Fourier transform at zero to obtainIts periodization is continuous, with uniformly geometrically bounded tails, and the sampled coefficients are absolutely summable. The absolutely summable coefficients define a uniformly convergent Fourier series with the computed coefficients. Its continuous sum equals the continuous periodization: their difference has every Fourier coefficient zero, hence is zero in the L2 norm by Fourier completeness, and continuity upgrades that to equality everywhere. Therefore its ordinary Poisson summation formula is valid. Evaluate it at zero:Thus the fully justified requested sum is