 Solution

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.