Past exam of the analysis course of the University of Oxford / 2008 / B

Use the convention that the inner product is linear in its first argument. If $v\ne 0$, nonnegativity of $\Vert u-⟨u,v⟩v/\Vert v{\Vert }^2{\Vert }^2$ givesThe case $v=0$ is immediate. This Cauchy-Schwarz inequality shows that $f_v$ is bounded linear with norm at most $\Vert v\Vert$. For $v\ne 0$, evaluation at $v/\Vert v\Vert$ gives equality; for $v=0$ both norms vanish. The Banach dual is the space of all bounded linear functionals with the operator norm.

The Riesz representation theorem asserts that every $f\in V^{\prime }$ is uniquely $f(u)=⟨u,v⟩$ for some $v\in V$, with $\Vert f\Vert =\Vert v\Vert$. If $f=0$, take $v=0$. Otherwise put $W=\ker f$, a closed subspace, and use the projection theorem to obtain $w\in W^{\perp }$ with $f(w)=1$. Since $u-f(u)w\in W$, $⟨u,w⟩=f(u)\Vert w{\Vert }^2$. Thus $v=w/\Vert w{\Vert }^2$ represents $f$. If two vectors represent it, their difference has inner product zero with every vector, in particular itself, giving uniqueness. The norm statement was already proved.

The space $P_n$ is finite-dimensional and hence complete; evaluation at zero is bounded there by equivalence of finite-dimensional norms. Riesz therefore supplies exactly one $q_n\in P_n$. In any orthonormal polynomial basis,For the real Legendre functions the bars may be omitted.

The printed even-degree normalization has an error. Orthogonality forces a degree-two polynomial to be a scalar multiple of $3t^2-1$, whose squared integral on $[-1,1]$ is $8/5$. Hence any orthonormal choice has $|p_2(0){|}^2=5/8$, whereas the printed formula gives $3/8$. The factor under its square root must be $4m+1$ instead of $2m+1$. Signs of the individual polynomials can be chosen so that the values at zero are positive; these signs do not affect $q_n$ or its norm.

Put $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$. With the consistent orthonormal normalization the norm isThe finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1{)}^2{c}_m^2-(2m-1{)}^2{c}_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac{1}{2}{\sum }_{m=0}^M(2m+1)c_m^2$; that is not the norm of the actual orthonormal Legendre evaluation representer.

For a polynomial $r$ of degree $d$, the defining identity gives $⟨r,q_n⟩=r(0)$ for every $n\ge d$, so the requested limit is eventually constant. It does not extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\Vert q_n\Vert$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\ge 1/(2m+1)$, and the displayed sum dominates a divergent constant multiple of the harmonic series. The Uniform boundedness principle then provides an $L^2$ function for which ${\sup }_n|⟨r,q_n⟩|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.

Write $L_n(t)={\sum }_{k=0}^n{a}_k{t}^k$. Matching coefficients in the differential equation givesTaking $a_0=1$ yieldsThe top coefficient is $(-1{)}^n/n!\ne 0$, so this is a real polynomial of degree exactly $n$.

Put $D_n=L_n^{\prime }-L_n$. Integration by parts, with vanishing endpoint terms because of the factor $t$ at zero and exponential decay at infinity, givesThe last step uses $D_n+t{D}_n^{\prime }=t{L}_n^{\prime \prime }+(1-t)L_n^{\prime }-L_n=-(n+1)L_n$. The left side is symmetric in $m,n$. Interchanging them therefore gives $(n-m)\int L_n{L}_m{e}^{-t}=0$, proving orthogonality for distinct indices. These integrals are exactly $⟨{\phi }_n,{\phi }_m⟩$, and each ${\phi }_n$ is square-integrable and nonzero. Thus they form an orthogonal sequence in $L^2(0,\infty )$.

Multiplying the equations by $z^n$ and summing givesConversely equality of power-series coefficients recovers every polynomial equation. For the proposed expression,Substitution makes the PDE left side zero. Its coefficient of $z^n$ has value one at $t=0$, so it corresponds to the normalization $L_n(0)=1$ chosen above.

With the classical real coefficientsthe partial sum is $s_n^f=a_0/2+{\sum }_{j=1}^n(a_j\cos jx+b_j\sin jx)$. It is the orthogonal projection onto the first trigonometric modes. Completeness of the trigonometric orthonormal system therefore gives $\Vert s_n^f-f{\Vert }_2\to 0$.

For $f\in C^1$ periodic, integration by parts has zero endpoint terms and gives $a_j(f^{\prime })=j{b}_j(f)$, $b_j(f^{\prime })=-j{a}_j(f)$, and $a_0(f^{\prime })=0$. ThusFor periodic $C^2$ functions, another integration by parts gives $a_j(f)=-a_j(f^{\prime \prime })/j^2$ and $b_j(f)=-b_j(f^{\prime \prime })/j^2$. Riemann-Lebesgue makes the numerator sequences tend to zero, in particular bounded. The Weierstrass M-test and $\sum j^{-2}<\infty$ prove absolute and uniform convergence. The resulting continuous sum agrees with $f$ in $L^2$ by the first result. Two continuous functions equal almost everywhere are equal everywhere, so the pointwise sum is $f$ for every real $x$.

For the polynomial example the periodic extension is $C^2$: the function, its first derivative and its second derivative match at the endpoints. It is even, so all sine coefficients vanish. Direct integration givesFor example, integration by parts gives ${\int }_0^{\pi }{x}^2\cos (jx)dx=2\pi (-1{)}^j/j^2$ and ${\int }_0^{\pi }{x}^4\cos (jx)dx=4{\pi }^3(-1{)}^j/j^2-24\pi (-1{)}^j/j^4$, from which the cancellation of the $j^{-2}$ terms follows. ThereforeAt $x=0$, the left side vanishes, yielding

Write $\lambda =a+ib$ with $b\ne 0$. Since $aI-T$ is self-adjoint, $⟨(aI-T)v,v⟩$ is real. The cross term in the squared norm therefore vanishes:This proves the equality and lower bound, and makes both $S=\lambda I-T$ and $S^{\ast }=\overline{\lambda }I-T$ injective.

If $S{v}_n$ converges, the lower bound shows $v_n$ is Cauchy. Completeness gives $v_n\to v$ and boundedness gives $S{v}_n\to Sv$, so the range is closed. Also $(\mathrm{ran}S{)}^{\perp }=\ker S^{\ast }=0$, so the range is dense. Being both closed and dense, it equals $V$. Hence $S$ is bijective, and its lower bound gives

Define $f^{♯}(s)=\overline{f(-s)}$. Interchanging the integrals and substituting variables givesMoreover Cauchy-Schwarz in the inner integral givesTaking the supremum over unit $g_2$ proves $\Vert T_f{g}_1{\Vert }_2\le \Vert f{\Vert }_1\Vert g_1{\Vert }_2$, so $T_f$ is bounded with norm at most $\Vert f{\Vert }_1$, and the adjoint identity is justified.

It remains to ensure necessity is a condition on the kernel itself. If $T_h=0$ for some $h\in L^1$, convolve $h$ with a smooth compactly supported approximate identity ${\eta }_{\epsilon }\in L^1\cap L^2$. Since $T_h{\eta }_{\epsilon }=0$, all these convolutions vanish. But $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, by continuity of translations in $L^1$, so $h=0$ almost everywhere. Thus the map from kernels to convolution operators is injective. Applying this to $h=f-f^{♯}$ yields the exact criterionFor real-valued kernels this reduces to evenness; real-valuedness by itself is insufficient.