Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 22F (Analysis of Functions)

Let $w(\xi )=(1+|\xi {|}^2{)}^{s/2}$. The map $U:f\mapsto w\hat{f}$ is linear, and $\Vert U(f+g){\Vert }_2\le \Vert Uf{\Vert }_2+\Vert Ug{\Vert }_2$ proves the vector-subspace claim. The stated inner product is $\int w^2\hat{f}\overline{\hat{g}}$. By the Plancherel theorem it is positive definite.

More strongly, $U:H^s\to L^2$ is onto: for $h\in L^2$, $w^{-1}h\in L^2$, since $w\ge 1$, and $f={\mathcal{F}}^{-1}(w^{-1}h)$ satisfies $Uf=h$. Thus $U$ is an isometric linear bijection, and completeness of $L^2$ givesThis argument includes $s=0$ if zero is allowed in the printed ${\mathbb{R}}_{+}$ convention.

Take $f=1_{[0,1]}$. Its Fourier transform iswith value one at zero. It is bounded near zero and is $O(|\xi {|}^{-1})$ at infinity, so $(1+{\xi }^2{)}^s|\hat{f}{|}^2=O(|\xi {|}^{2s-2})$ is integrable for $s<1/2$. ThereforeIf a continuous function agreed with this indicator almost everywhere, continuity would force it to equal zero on $(-\infty ,0)$ and one on $(0,1)$: any violation has a nonempty open neighbourhood of positive measure. The one-sided limits at zero would then differ, a contradiction. Hence no continuous representative exists.

The defining integral is well-defined and bounded for every dimension: $|\mathcal{F}f|\le \Vert f{\Vert }_1$ and the multiplier is supported on a finite-volume ball. However the asserted general-dimensional $L^1$ conclusion is false for the Euclidean radial cutoff printed in the original PDF. The nonintegrability of a radial triangular Fourier cutoff in three dimensions is a source issue, not an omitted conjugate or OCR substitution.

For a direct counterexample in $n=3$, take the Schwartz function $f(x)=e^{-\pi |x{|}^2}$, whose Fourier transform is the same Gaussian. With $r=|x|$ and $a=2\pi r$, radial integration givesThe amplitude $b(\rho )=\rho (1-\rho /R)e^{-\pi {\rho }^2}$ vanishes at both endpoints, and $b^{\prime }(R)=-e^{-\pi R^2}$. Two integrations by parts, with a further one to bound the remainder, giveThe integral ${\int }_1^{\infty }{r}^2|F_R(r)|dr$ diverges logarithmically for every fixed $R>0$. For example, restrict to the periodic subintervals on which $|\sin (2\pi Rr)|\ge 1/2$; the error is smaller than half the leading term at sufficiently large $r$. Thus $F_R\notin L^1({\mathbb{R}}^3)$, even for this smooth integrable $f$.

The familiar intended argument is valid in one dimension. There the inverse Fourier kernel isand $F_R=K_R\ast f$ by Fubini's theorem. Its mass outside any fixed neighbourhood of zero tends to zero by scaling, soby continuity of translations in $L^1$. For arbitrary dimension, replacing the radial cutoff by the product ${\prod }_{j=1}^n(1-|{\xi }_j|/R{)}_{+}$ gives a tensor product of these integrable kernels and the same approximate identity proof. These are explicit valid repairs, not claims that the PDF specified a product cutoff.

The implication is true in every dimension despite the failure of part (iii). The hypothesis gives $\mathcal{F}f\in L^2$, since its weight is at least one. Let $g$ be its inverse Fourier-Plancherel transform; then $g\in H^s$ by construction.

Use the Gaussian approximate identity $G_{\epsilon }(x)={\epsilon }^{-n}{e}^{-\pi |x{|}^2/{\epsilon }^2}$. Its Fourier transform is $e^{-\pi {\epsilon }^2|\xi {|}^2}$. Fourier inversion of this integrable multiplier times the bounded transform of $f$, justified by Fubini, gives the same function both as $G_{\epsilon }\ast f$ and as $G_{\epsilon }\ast g$. The first converges to $f$ in $L^1$, while the second converges to $g$ in $L^2$ by Plancherel and dominated convergence. Along a common sequence, select subsequences converging almost everywhere for both norms. Their limits agree, giving $f=g$ almost everywhere. ConsequentlyThis proves the requested conclusion without relying on the incorrect general-dimensional kernel assertion.