Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 1

Use normalized Fourier coefficients and the inner productThe functions $e_n(t)=e^{int}$ are orthonormal. Hence the Fourier partial sum $S_{N}f={\sum }_{|n|\le N}\hat{f}(n)e_n$ is the orthogonal projection onto the trigonometric polynomials of degree at most $N$. For any such polynomial $P$, orthogonality givesso $\Vert f-S_{N}f{\Vert }_2\le \Vert f-P{\Vert }_2$.

Given $\epsilon >0$, the permitted density result supplies a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. Once $N$ includes its degree,Therefore the Fourier partial sums converge to $f$ in the normalized $L^2$ norm, giving exactly the stated mean-square limit.

Apply part (i) to the continuous difference $h=f-g$. Every Fourier coefficient of $h$ vanishes, so every Fourier partial sum is zero. The orthogonal projection convergence from part (i) therefore gives $\Vert h{\Vert }_2=0$.

If $h(t_0)\ne 0$, continuity gives an interval on which $|h|$ is bounded below by a positive number. That interval would contribute positively to $\Vert h{\Vert }_2^2$, a contradiction. HenceThe continuity hypothesis upgrades equality almost everywhere to pointwise equality.

By the Weierstrass M-test, absolute summability of the Fourier coefficients makes ${\sum }_n\hat{f}(n)e^{int}$ uniformly convergent to a continuous periodic function $H$. Termwise integration is justified by that uniform convergence, and orthogonality givesPart (ii) then implies $H=f$. Thus the Fourier series converges uniformly to the original function, not merely to some continuous limit. In particular,

The absolute Fourier convergence from a square-integrable derivative uses more than the pointwise estimate $|\hat{f}(n)|=O(1/|n|)$. Periodic integration by parts givesBy Bessel's inequality,Now apply the Cauchy-Schwarz inequality:Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has $f^{\prime }\in L^2$, all hypotheses of part (iii) hold and its Fourier series converges uniformly.

Part (i) and the Cauchy-Schwarz inequality showDirect integration of this finite Fourier partial sum givesMoreover Bessel's inequality puts both coefficient sequences in ${\ell }^2$, so their product series is absolutely convergent by the Cauchy-Schwarz inequality. Consequently the cross form of Parseval's identity isTaking $g=f$ also gives equality of the squared function norm and the squared coefficient norm.

For the Hurwitz proof of the planar isoperimetric inequality, take a positively oriented regular simple closed curve of length $L$ and enclosed area $A$. Write its complex position as $z(t)=x(t)+iy(t)$, with $0\le t\le 2\pi$ proportional to arc length. Then $|z^{\prime }(t)|=L/(2\pi )$. Translate the curve to make its mean position zero, and write its Fourier coefficients as $c_n$, with $c_0=0$.

Green's theorem gives the signed area, and Parseval's identity computes it:Periodic integration by parts and Parseval's identity applied to $z^{\prime }$ giveSince $n\le n^2$ for every integer $n$,The sums converge absolutely: $\sum |n||c_n{|}^2\le (\sum |c_n{|}^2{)}^{1/2}(\sum n^2|c_n{|}^2{)}^{1/2}$. Equality forces $c_n=0$ unless $n=0$ or $1$; after the mean translation, $z(t)=c_1{e}^{it}$ is a circle. Conversely a circle attains equality. Thus circles uniquely attain equality, up to translation and orientation.

The same proof applies to a rectifiable simple closed curve using its Lipschitz arc length parametrization. Its derivative exists almost everywhere and belongs to $L^2$; periodic mollification converges to the curve in the function and derivative $L^2$ norms. This justifies the derivative coefficient identity, Parseval's identity and area integral by approximation. Reversing orientation, if necessary, makes the enclosed area positive. The printed name “Hurewitz” is read as Hurwitz.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.