Past exam of the analysis course of the University of Oxford / 2008 (B4, DMAB 2643, DMCN 2643, DMPB 2643)

If $(a^{(n)})$ is Cauchy in the maximum norm on ${\mathbb{C}}^m$, every coordinate is Cauchy in $\mathbb{C}$ and has a limit $a_j$. Given $\epsilon >0$, for sufficiently large $n,k$ every coordinate difference is at most $\epsilon$. Letting $k\to \infty$ gives ${\max }_j|a_j^{(n)}-a_j|\le \epsilon$. Thus ${\mathbb{C}}^m$ with its maximum norm is complete.

Let $F(a)=\sum a_j{z}_j$. The triangle inequality givesTake $\delta =\sum \Vert z_j\Vert$. This estimate also makes $a\mapsto \Vert F(a)\Vert$ continuous on the maximum-norm unit sphere, which is compact by coordinatewise Bolzano-Weierstrass. Linear independence makes this function strictly positive there, so its minimum $\gamma$ is positive. Homogeneity then givesA Cauchy sequence in $X$ consequently has Cauchy coordinate vectors in ${\mathbb{C}}^m$. Their coordinate limit maps by the upper estimate to a limit in $X$. Therefore every finite-dimensional normed complex space is a Banach space.

The dual space $X^{\prime }$ consists of bounded linear maps $g:X\to \mathbb{C}$, with operator norm $\Vert g\Vert ={\sup }_{\Vert x\Vert \le 1}|g(x)|$. It is Banach even if $X$ is not: an operator-norm Cauchy sequence has pointwise limits, these define a bounded linear functional, and the uniform Cauchy estimate on the unit ball gives operator-norm convergence.

Restriction $R(g)=g{|}_Z$ is linear and contractive. If $Z$ is dense, for every $x\in X$ choose $z_n\in Z$ with $z_n\to x$. The bound $|g(z_n)|\le \Vert Rg\Vert \Vert z_n\Vert$ passes to the limit, proving $|g(x)|\le \Vert Rg\Vert \Vert x\Vert$. Hence $\Vert g\Vert =\Vert Rg\Vert$ and restriction is injective. Conversely, for $h\in Z^{\prime }$ define $g(x)=\lim h(z_n)$. The estimate $|h(z_n)-h(z_k)|\le \Vert h\Vert \Vert z_n-z_k\Vert$ gives existence and independence of the approximating sequence. It also gives linearity and $\Vert g\Vert \le \Vert h\Vert$. Thus restriction is an onto linear isometry of dual spaces.

For the sequence subspace in this part, the norm is the unweighted ${\ell }^1$ norm; only its membership condition uses the weight $j$. It contains all finitely supported sequences, hence is dense in ${\ell }^1$. Applying restriction to $X={\ell }^1$ and using its known dual yields

The complex Hahn-Banach theorem says that every bounded complex-linear functional on a linear subspace of a normed complex space extends to the whole space with the same norm.

For the separation assertion choose $x_0\notin M$ and $d=\mathrm{dist}(x_0,M)>0$. On $M+\mathbb{C}{x}_0$ define $g_0(m+\lambda x_0)=\lambda d$. The inequality $\Vert m+\lambda x_0\Vert \ge |\lambda |d$ makes its norm at most one; approximating the distance $d$ shows the norm is exactly one. Hahn-Banach gives an extension $g$ of norm one annihilating $M$.

Each geometric sequence satisfies $\Vert x_k{\Vert }_1=(1-|{\beta }_k|{)}^{-1}\le (1-\delta {)}^{-1}$. Let $M$ be their closed linear span. If $M\ne {\ell }^1$, the separation just proved gives a nonzero bounded functional annihilating all $x_k$. Represent it by $b\in {\ell }^{\infty }$ and formThe series converges locally uniformly and defines a holomorphic function. It vanishes at every ${\beta }_k$. These distinct points lie in the compact disk $|z|\le \delta <1$, so have an accumulation point inside the domain. The identity theorem gives $F=0$, hence every Taylor coefficient $b_j=0$, contradicting the functional's nonzero norm. Therefore

The spectrum of a bounded operator is $\sigma (T)=\{\lambda :\lambda I-T\text{ has no bounded two-sided inverse}\}$. Its point spectrum is the set of eigenvalues, where $\lambda I-T$ is not injective.

Here $T$ is linear and takes continuous functions to continuously differentiable functions. Also $\Vert Tx{\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$; the constant function one attains this bound at $t=1$, so $\Vert T\Vert =2$. Its range is contained in $C^1([0,1])$, a proper subset of $C([0,1])$, so zero is in its spectrum. It is not an eigenvalue: $Tx=0$ gives $x(0)=0$ at zero and $x=0$ upon differentiation.

For a nonzero eigenvalue, $\lambda x=Tx$ implies $x^{\prime }=x/\lambda$ and $(\lambda -1)x(0)=0$. A nonzero solution is therefore possible exactly at $\lambda =1$, with eigenfunctions $x(t)=c{e}^t$. Hence $p\sigma (T)=\{1\}$.

For the proposed resolvent putThe proposed $x$ is $y/\lambda +y(0)e^{t/\lambda }/[\lambda (\lambda -1)]+F/{\lambda }^2$. Consequently $x(0)=y(0)/(\lambda -1)$ andAdding $x(0)$ and subtracting from $\lambda x(t)$ gives exactly $y(t)$. Thus $\lambda I-T$ is onto for $\lambda \ne 0,1$. It is also injective by the eigenvalue calculation. The displayed formula defines a bounded map in the supremum norm, since its exponential kernels are bounded on the compact interval. It is therefore the bounded inverse, and

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.