Past exam of the functional analysis I course of the University of Oxford / 2023 / 3

Writing $\mathcal{B}(X)$ for the Banach space of bounded linear operators on $X$, the spectrum isHere $I$ is the identity operator. Since $X$ is a Banach space, the bounded inverse theorem makes this equivalent to failure of bijectivity of $\lambda I-T$.

The spectral radius isFor a nonzero complex Banach space, the spectrum of a bounded linear operator is nonempty and compact, so the supremum is a maximum. If the zero space is allowed, its empty spectrum is assigned spectral radius zero.

The Neumann series criterion states that $\Vert A\Vert <1$ for a bounded linear operator on a Banach space implieswith convergence in the operator norm. In particular, $\Vert T\Vert <1$ suffices for invertibility of $I-T$.

If $|\lambda |>\Vert T\Vert$, apply this criterion to $A=T/\lambda$. It givesThus no such $\lambda$ belongs to the spectrum, and

For each integer $j\ge 1$, if $|\lambda {|}^j>\Vert T^j\Vert$, then ${\lambda }^{j}I-T^j$ is invertible by the same argument. The commuting factorizationthen gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $({\lambda }^{j}I-T^j{)}^{-1}$. Hence every $\lambda \in \sigma (T)$ satisfies $|\lambda {|}^j\le \Vert T^j\Vert$. Taking the supremum over the spectrum and then the infimum over $j$ proves

For a continuous function $g$ on $[0,1]$, the L-infinity norm of its restriction to $(0,1)$ equals its supremum norm on $[0,1]$. Indeed, values arbitrarily close to a maximum occur on an interval of positive Lebesgue measure, even if the maximum lies at an endpoint. Also, continuous functions equal almost everywhere on $(0,1)$ are equal everywhere on $[0,1]$. Thus restriction embeds $C([0,1])$ isometrically as a linear subspace of $X$.

The point evaluation functional $E_{x_0}(g)=g(x_0)$ is complex-linear and satisfies $|E_{x_0}(g)|\le \Vert g{\Vert }_{\infty }$. The constant function one attains equality, so $\Vert E_{x_0}\Vert =1$. The complex Hahn-Banach theorem extends this bounded linear functional from that linear subspace to $X$, preserving its operator norm. Therefore

Write $\psi ={\psi }_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because Lp spaces identify functions equal almost everywhere. Its L-infinity norm is one.

The supplied estimate shows that $\psi$ cannot agree almost everywhere with any continuous function $q$: choosing $g=-q$ would give $0\ge |q(x_0)|+1$. Henceis a direct sum, so every element has a unique expression $g+c\psi$.

For any complex $a$ with $|a|\le 1$, define the linear functionalIf $c\ne 0$, the supplied estimate, applied to $g/c$, and absolute homogeneity of a norm giveFor $c=0$, the same bound follows from point evaluation. Thus $\Vert L_a\Vert \le 1$, and $L_a(1)=1$ makes its operator norm exactly one. The complex Hahn-Banach theorem extends $L_a$ to a norm-one linear functional on all of $X$, with value $a$ at $\psi$.

Conversely every norm-one extension satisfies $|{\ell }_{x_0}(\psi )|\le \Vert \psi {\Vert }_{\infty }=1$. We have therefore determined all possible values:In particular, taking $a=0$ and $a=1$ proves nonuniqueness of norm-preserving point evaluation extensions.

Because the linear functionals ${\ell }_0,{\ell }_1$ agree with endpoint point evaluations on continuous functions,The endpoint values also make $e_0,e_1$ linearly independent. By its definition, the image of a linear map $T$ lies in $E=⟨e_0,e_1⟩$. Conversely $T{e}_j=e_j$, so $T$ acts as the identity operator on $E$. ThusFor every $f$, $Tf\in E$, whence $T^{2}f=Tf$. It is a linear projection, and an annihilating polynomial is

Let $M=\Vert |e_0|+|e_1|{\Vert }_{\infty }$. Since both linear functionals have operator norm one, for every $f\in X$ the triangle inequality givesTaking the essential supremum proves $\Vert T\Vert \le M$.

For the reverse bound, continuity and the extreme value theorem give $x_{\ast }\in [0,1]$ with $|e_0(x_{\ast })|+|e_1(x_{\ast })|=M$. Choose complex $a_0,a_1$ of modulus one so that $a_j{e}_j(x_{\ast })=|e_j(x_{\ast })|$, taking $a_j=1$ if $e_j(x_{\ast })=0$. The continuous functionhas supremum norm one by the triangle inequality and its endpoint values. Since ${\ell }_j(f_{\ast })=a_j$,The output is continuous, so its L-infinity norm equals its supremum norm, including when $x_{\ast }$ is an endpoint. Therefore $\Vert T\Vert \ge M$, and the exact operator norm isThis is the norm of a two-endpoint interpolation projection.

From $T{e}_0=e_0\ne 0$, one is an eigenvalue. Zero is also an eigenvalue: the nonzero continuous function $q(x)=x(1-x)$ has both endpoint values zero, so $Tq=0$.

For $\lambda \notin \{0,1\}$, the linear projection identity $T^2=T$ gives an explicit bounded inverse:Multiplying on either side by $\lambda I-T$ yields $(I-T)+T=I$. There are no other points in the spectrum, and thus the spectrum of a nontrivial bounded projection here is

Let $M=\Vert T\Vert$, as calculated in part (ii). Its endpoint values ensure $1\le M<\infty$. Since $T^j=T$ for all $j\ge 1$,This verifies the spectral radius formula explicitly: