Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 7 / 3

Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

Use $\hat{f}(\lambda )=\int f(t)e^{-i\lambda t}dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb{R})$ with the continuous representative furnished by Fourier inversion. Write $F=\hat{f}$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the Cauchy-Schwarz inequality on its finite support, sois continuous and defined at every $t$. The integer samples are precisely the Fourier coefficients of $F$ with the sign convention appropriate to the basis $e^{-in\lambda }$:The completeness and Parseval identity for the exponential orthonormal basis on $[-\pi ,\pi ]$ give $F={\sum }_{n\in \mathbb{Z}}f(n)e^{-in\lambda }$ in $L^2$ andThe last equality is the Plancherel theorem. Apply the inverse integral to the finite symmetric Fourier partial sums. Sincewe obtain the Nyquist–Shannon sampling theorem in this normalization:where the quotient is one at $t=n$. This is a sampling expansion by periodic Fourier projection, not merely an almost-everywhere inversion. Indeed, the inverse integral of an error $G\in L^2[-\pi ,\pi ]$ is bounded, uniformly in $t$, by $\Vert G{\Vert }_2/\sqrt{2\pi }$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb{R}$. Moreover, the Parseval identity applied to $e^{it\lambda }$ givesThe Cauchy-Schwarz inequality therefore makes the sampling series absolutely convergent at every $t$, with its absolute tail bounded uniformly by $({\sum }_{|n|>N}|f(n){|}^2{)}^{1/2}$. Thus the integer samples recover the function everywhere.

For the nonuniqueness beyond this bandwidth, fix $0<{ϵ}_0<\min (ϵ,\pi )$ and a nonzero smooth function $\Phi$ compactly supported in $(-{ϵ}_0,{ϵ}_0)$. Let $h$ be its inverse Fourier transform and putThe function $h$, and hence $g$, is a Schwartz function. The two shifted spectral supports are disjoint, so $\hat{g}\ne 0$, but its support is contained strictly inside $(-\pi -ϵ,\pi +ϵ)$. Every integer sample of $g$ is zero. Therefore $g$ and the zero function have identical samples but are distinct, providing a strictly supercritical sampling alias counterexample for every positive bandwidth enlargement.