 Solution

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: