I did the following exercise (given in my notes) can you tell me if my answer is correct? Thanks.
Exercise: Compute the operator norm of the continuous map $f \mapsto f$ when viewed:
(a) as a map $T: C^1([0, 1]) \to C([0, 1])$
(b) as a map $S: C([0, 1]) \to L^1_\mu([0, 1])$, where $\mu$ is Lebesgue measure on $[0, 1]$
(c) Compute the operator norm of the composition of the maps from (a) and from (b).
(d) Now restrict the maps in (a), (b) and (c) to the space of functions $f$ with $f(0) = 0$, and compute the operator norms again.
My answer:
(a) $\|T\| = \sup_{\|f\|_{C^1} \leq 1} \|f\|_\infty \leq \sup_{\|f\|_\infty \leq 1} \|f\|_\infty = 1$ since for example $f(x) = x \in C^1[0,1]$ and $\|f\|_\infty = 1$. OTOH, $\|T\| \geq 1$ since $\|Tf\| = \|x\| = 1$.
(b) $\|S\| = \sup_{\|f\|_\infty \leq 1} \|f\|_1 = 1$ since $\|f\|_1 \leq \int_{[0,1]} 1 d \mu$ so $\sup_{\|f\|_\infty} \|f\|_1 \leq 1$. But $f(x) = 1 \in C[0,1]$ and $\|1\|_\infty \leq 1$ and $\|1\|_1 = 1$.
(c) $\|ST\| \leq \|S\| \|T\| = 1$. We have $\|ST\| \geq 1$ since for $f(x) = 1$, $\| STf\| = \|Sf\| = 1$.
(d)
(d.a) $\|T\| = 1$ by the same argument as in (a).
(d.b) $\|S\| = 1$ because the sequence $f_n$ where $f_n$ is the function that is $nx$ on $[0, \frac1n]$ and $1$ on $[\frac1n, 1]$ is in $C[0,1]$ and $\|f_n\|_\infty \leq 1$ and $\|f_n\|_1 \to 1$ (by monotone convergence theorem)
(d.c) $\|ST\| \leq 1$ (same argument as (c)). Unfortunately, $f_n$ from (d.b) are not in $C^1[0,1]$. So I'm not sure what to do. Is $\|ST\| = 1$ in this case too? Or perhaps not?