Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 6 (Introduction to Functional Analysis)

The Baire category theorem states that in a complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of nowhere dense sets, where a set is nowhere dense if its closure has empty interior.

Here is the proof. Let $G_1,G_2,\dots$ be open dense sets and let $U$ be any nonempty open set. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the closed ball $B_1=\overline{B}(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty open set $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such thatThus the closed balls are nested and their radii tend to zero. For $j,k\ge m$, both centers lie in $B_m$, so $d(a_j,a_k)\le 2r_m$. They form a Cauchy sequence. Completeness gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap {\bigcap }_m{G}_m$. As $U$ was arbitrary, the intersection is dense.

To obtain the equivalent formulation, take complements of the closures of the nowhere dense sets; those complements are open dense. Conversely, complements of open dense sets are closed with empty interior. In particular, a countable closed cover of a nonempty complete metric space must contain a set with nonempty interior.

For each nonzero integer tuple $k=(k_1,\dots ,k_{n+1})$, putIf its first $n$ coordinates are all zero, then $k_{n+1}\ne 0$ and $H_k$ is empty. Otherwise it is a closed proper affine hyperplane. It has empty interior: from any of its points, an arbitrarily small displacement in the nonzero normal direction $(k_1,\dots ,k_n)$ leaves it. Thus every $H_k$ is nowhere dense.

There are countably many integer tuples, and ${\mathbb{R}}^n$ is complete. The Baire category theorem therefore makes the complement of their union dense and nonempty. Choosing any point there givesEquivalently, $1,x_1,\dots ,x_n$ are linearly independent over $\mathbb{Q}$: multiplying a rational relation by a common denominator produces just such an integer relation. This is an instance of Baire avoidance of countably many affine hyperplanes.

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized measure $dt/(2\pi )$. The Fourier partial sum at zero is the real linear functionalwhere the Dirichlet kernel isThe quotient follows by summing the finite geometric progression, with its removable value $D_N(0)=2N+1$. It is a real continuous periodic function. Ifthen $|L_{N}f|\le {\Lambda }_N\Vert f{\Vert }_{\infty }$. This bound is its exact operator norm even on the real continuous functions. Indeed, for $\delta >0$ setIt is real, continuous and periodic with norm at most one, andby dominated convergence. At zeros of the kernel the integrand is zero, and elsewhere its limit is $|D_N|$.

For $0<t\le \pi$, $\sin (t/2)\le t/2$. Using evenness and putting $v=(N+1/2)t$ givesFor the last inequality, split into the first $N$ complete intervals $[(j-1)\pi ,j\pi ]$: their sine integrals are two and $1/v\ge 1/(j\pi )$. The harmonic sum tends to infinity. This proves the Dirichlet kernel harmonic lower bound.

Choose $N$ with ${\Lambda }_N>K$, then choose $\delta$ sufficiently small that $L_N{f}_{N,\delta }>K$. ThusThis constructs continuous near-maximizers, rather than using the discontinuous sign of the kernel as the desired function.

The real space $C(\mathbb{T})$ is a Banach space in the uniform norm: a uniformly Cauchy sequence has a uniform pointwise limit by completeness of $\mathbb{R}$, and that uniform limit is continuous. Each functional $L_N$ from the preceding part is bounded, but their norms are unbounded.

For integers $m\ge 1$, defineEach $E_m$ is closed, being an intersection of inverse images of closed intervals under continuous functionals. It has empty interior. Otherwise some ball $B(f_0,r)$ would be contained in it. For every $h$ of norm at most one, both $f_0$ and $f_0+(r/2)h$ would belong to that ball, soTaking the supremum over $h$ would give $\Vert L_N\Vert \le 4m/r$ for all $N$, contradicting the preceding part.

Thus the $E_m$ are nowhere dense. The Baire category theorem says their union cannot cover $C(\mathbb{T})$. A function outside that union satisfiesso its Fourier series diverges at zero. In fact these functions form a dense $G_{\delta }$ subset of $C(\mathbb{T})$. This proves the relevant Uniform boundedness principle argument directly and establishes generic unbounded Fourier sums at a fixed point.

The hypotheses make $p$ a sublinear functional and $q$ a superlinear functional. In particular $p(0)=q(0)=0$. Apply the domination property to $y+y^{\prime }\in Y$ and the vector $x+x^{\prime }$:Subadditivity of $p$, with the inserted vectors $z$ and $-z$, givesSuperadditivity of $q$ gives $q(x+x^{\prime })\ge q(x)+q(x^{\prime })$. Combining these inequalities and moving terms yields exactlyEvery vector appearing in the comparison is permitted by the assumed domination, so no membership of $x,x^{\prime },z$ in $Y$ is needed.

We prove the extension by Zorn's lemma, retaining the full domination condition. Consider all pairs $(Y,S)$ extending $(Y_0,T_0)$, with $Y$ a subspace and $S$ linear, satisfying $S(y)\le p(x+y)-q(x)$ for every $x\in X,y\in Y$. Order these pairs by extension. A chain has an upper bound: take the union of its subspaces and maps. The union is a subspace, and its map is well defined and linear because any finitely many vectors lie together in one member of the chain. Its domination is inherited there. Thus a maximal extension exists.

Suppose its domain $Y$ is proper, and choose $z\notin Y$. A linear extension to $Y+\mathbb{R}z$ has the form $S_a(y+tz)=S(y)+ta$. Define the admissible values for a sandwich extension bywith $x,x^{\prime }\in X$ and $y,y^{\prime }\in Y$. Part (i) shows every quantity in the supremum is at most every quantity in the infimum, hence $A\le B$. Both bounds are real and finite: choosing zero vectors provides $A\ge -p(-z)$ and $B\le p(z)$, and the pairwise comparison then bounds them from the other sides. Choose $a\in [A,B]$.

For $t>0$, positive homogeneity and the upper bound on $a$, used with $x/t,y/t$, giveFor $t=-s<0$, the lower bound on $a$ applied to $x/s,y/s$ giveswhich rearranges to $S(y)-sa\le p(x+y-sz)-q(x)$. For $t=0$ this is the existing domination. Thus $S_a$ is a legitimate larger extension, contradicting maximality. The maximal domain is therefore all of $X$; call its map $T$.

Setting the first vector in the domination to zero yields $T(v)\le p(v)$. Setting it to $v$ and the domain vector to $-v$ yields $-T(v)\le -q(v)$. ConsequentlyThe one-dimensional extension has checked both signs of its coefficient; positive homogeneity alone does not permit treating negative coefficients as positive ones.

Take $Y_0=\{0\}$ and $T_0(0)=0$. The hypothesis $p(x)\ge q(x)$ is exactly the required initial domination $0\le p(x+0)-q(x)$. Applying the extension proved in part (ii) giveswith $U$ linear. There is no reason that $U$ must be nonzero; for instance $p=q=0$ forces $U=0$.

This is the sandwich form of the Hahn-Banach theorem. Conversely, any linear map between these two functionals satisfies the stronger mixed domination, since $U(y)=U(x+y)-U(x)\le p(x+y)-q(x)$. Thus the mixed inequality used during extension is the appropriate invariant, not an extra final restriction.

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

We construct the finite evaluation representation explicitly by Lagrange interpolation. This supplies all the polynomial facts needed for the bound.

Define the Chebyshev polynomial by $T_0=1,T_1=x$ andInduction shows that $T_m$ has degree $m$; the cosine addition identity also proves $T_m(\cos \theta )=\cos (m\theta )$. Therefore $\Vert T_n{\Vert }_{[-1,1]}=1$, and $T_n(-x)=(-1{)}^n{T}_n(x)$ follows from the recurrence. The case $n=0$ is immediate, since both $P$ and $T_0$ are constant.

For $n\ge 1$, choose the descending extremal nodesso $T_n(x_j)=(-1{)}^j$. Define the fundamental interpolation polynomialsThey satisfy ${\ell }_j(x_k)={\delta }_{jk}$. Consequentlyfor every polynomial of degree at most $n$: the difference has degree at most $n$ and vanishes at the $n+1$ distinct nodes, so it is zero. The root count used here follows by successively factoring $(t-x_j)$ from a polynomial having such a root; a nonzero degree-$n$ polynomial cannot have more than $n$ distinct roots. Thus the interpolation identity has been proved.

If $u>1$, every numerator factor in ${\ell }_j(u)$ is positive, while exactly $j$ denominator factors are negative. Hence $\mathrm{sgn}{\ell }_j(u)=(-1{)}^j$. Applying interpolation to $T_n$ itself givesPutting $M={\sup }_{[-1,1]}|P|$ in the interpolation formula yieldsFor $u<-1$, apply the proved positive-side inequality to $Q(t)=P(-t)$ at $-u>1$, and use the parity identity. The result isThis is sharp, with equality for $P=T_n$ or its scalar multiples.

In functional terms, the evaluation map $E_u:P\mapsto P(u)$ on the normed polynomial space has $\Vert E_u\Vert =|T_n(u)|$: the inequality gives the upper bound and $T_n$ attains it. For $u>1$, its normalized version has the exact representationThus Chebyshev interpolation represents external evaluation using $n+1$ nodes. The necessary finite representation and its norm are derived directly, so no convex-hull theorem is being invoked without proof.