Past exam of the analysis course of the University of Oxford / 2009 / B

A complete orthonormal sequence $(e_n)$ has $⟨e_n,e_m⟩={\delta }_{nm}$ and dense linear span; equivalently, the only vector orthogonal to every $e_n$ is zero. The Parseval identity says $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$, with expansion $x={\sum }_n⟨x,e_n⟩e_n$ in the Hilbert norm.

Put $G(t,z)=(1-2zt+z^2{)}^{-1/2}$ and $D=1-2zt+z^2$. Direct differentiation givesSubstitution yields $(1-t^2)G_{tt}-2t{G}_t+z^2{G}_{zz}+2z{G}_z=0$. Extracting the coefficient of $z^n$ in the generating function givesthe Legendre differential equation.

The generating function gives $P_n(-t)=(-1{)}^n{P}_n(t)$, so the odd function $f$ is orthogonal to every even-degree Legendre polynomial. For $n\ge 1$, integrate the self-adjoint form of the Legendre differential equation from zero to one:Thus ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. Differentiating the generating function in $t$ and then setting $t=0$ givesCombining these identities with the normalization of $p_n$ yieldsFor the Legendre expansion of the sign function, $\Vert f{\Vert }_2^2=2$, so the Parseval identity givesTherefore the requested sum equals

Expand both generating functions and use orthogonality of the Legendre polynomials, with ${\int }_{-1}^1{P}_n{P}_m=2{\delta }_{nm}/(2n+1)$. The allowed exchanges giveHence it depends only on $r=wz$. Explicitly,The series form alone already proves the required dependence.

With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.

The Hilbert projection theorem states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed linear subspace $M$, with $m\in M$ and $n\in M^{\perp }$. The vector $m$ is the unique nearest point in $M$.

The Riesz representation theorem states that every bounded complex-linear functional $F$ on a Hilbert space has a unique representation $F(x)=⟨x,y⟩$, with $\Vert F\Vert =\Vert y\Vert$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^{\perp }$ with $F(z)\ne 0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, soThis gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The Cauchy-Schwarz inequality gives $\Vert F\Vert \le \Vert y\Vert$, and evaluation at $y/\Vert y\Vert$ gives equality when $y\ne 0$.

For the adjoint, fix $v$. The functional $u\mapsto ⟨Tu,v⟩$ is bounded with norm at most $\Vert T\Vert \Vert v\Vert$. By the Riesz representation theorem, there is a unique vector $T^{\ast }v$ satisfyingFor scalars $a,b$, conjugate-linearity in the second argument shows that $T^{\ast }(av+bw)=a{T}^{\ast }v+b{T}^{\ast }w$, by uniqueness of the representing vector. The same norm estimate gives $\Vert T^{\ast }v\Vert \le \Vert T\Vert \Vert v\Vert$. Thus $T^{\ast }$ is a unique bounded linear operator, the adjoint operator.

For the adjoint of an integration operator anchored at zero, use the inner product $⟨f,g⟩=\int f\bar{g}$. On the positive half-interval, reversing the triangular integration order givesOn the negative half-interval the corresponding calculation givesThese exchanges are justified by the Fubini theorem, since on a finite interval $L^2$ functions belong to $L^1$ and the absolute double integral is at most $\Vert f{\Vert }_1\Vert g{\Vert }_1$. ThereforeAdding $Tg$ gives ${\int }_0^{\pi }g$ on the positive half and $-{\int }_{-\pi }^{0}g$ on the negative half, exactly the prescribed $Ug$. Values at zero do not matter in $L^2$. Consequently

Since $f(0)=0$, the fundamental theorem of calculus gives $f=T{f}^{\prime }$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued functionChoose any value at zero. This function is bounded and hence belongs to $L^2(-\pi ,\pi ]$. Since it is real, the adjoint operator identity yieldsOne can also check this by integration by parts separately on each half-interval: $g^{\prime }=-t^2$, $g(\pm \pi )=0$, and the two possible jump terms at zero vanish because $f(0)=0$.