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

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