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

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: