This is Problem 2.7.10 from Erwin Kreyszig's Introductory Functional Analysis with Applications.
Let $C[0,1]$ denote the normed space of all (real- or complex-valued) functions defined and continuous on the closed unit interval $[0,1]$ of the real line $\mathbb{R}$, with the norm defined by
$$\Vert x \Vert \colon= \max_{t\in[0,1]} \vert x(t) \vert \, \, \forall x \in C[0,1]. $$
Let $S \colon C[0,1] \to C[0,1]$ and $T \colon C[0,1] \to C[0,1]$ be defined as follows: $$Sx(s) \colon= s \int_0^1 x(t) \, dt,$$ and $$ Tx(s) \colon= s \cdot x(s) $$ for all $x \in C[0,1]$.
Do $S$ and $T$ commute?
Find $\Vert S \Vert$, $\Vert T \Vert$, $\Vert ST \Vert$, and $\Vert TS \Vert$.
My effort: For any $x\in C[0,1]$, we have $$STx(s) = S(sx(s)) = s \int_0^1 tx(t) \, dt,$$ while $$TSx(s) = T\left(s \int_0^1 x(t) \, dt\right) = s^2 \int_0^1 x(t) \, dt.$$ Thus, $TS \neq ST$ because for the point $x\in C[0,1]$ defined as $x(t) \colon= 1$ for all $t \in [0,1]$, we see that, for all $s \in [0,1]$, $$TSx(s) = s^2,$$ but $$STx(s) = \frac{s}{2}.$$ Am I right?
To compute the norms, we see that, for any $x \in C[0,1]$, $$\Vert Sx \Vert_{C[0,1]} = \max_{s\in [0,1]} \left\vert s \int_0^1 x(t) \, dt \right\vert = \max_{s \in [0,1]} \left( \vert s \vert \cdot \left\vert \int_0^1 x(t) \, dt \right\vert \right) = \max_{s\in[0,1]} \vert s \vert \cdot \left\vert \int_0^1 x(t) \, dt \right\vert \\ = \left\vert \int_0^1 x(t) \, dt \right\vert \leq \int_0^1 \vert x(t) \vert \, dt \leq \int_0^1 \max_{\tau\in[0,1]} \vert x(\tau) \vert \, dt = \int_0^1 \Vert x \Vert_{C[0,1]} \, dt \\ = \Vert x \Vert_{C[0,1]} \int_0^1 dt = \Vert x \Vert_{C[0,1]}. $$ So, for all non-zero $x \in C[0,1]$, we have $$\frac{\Vert Sx \Vert_{C[0,1]}}{\Vert x \Vert_{C[0,1]}} \leq 1;$$ so $$\Vert S \Vert \leq 1.$$ But for $x \in C[0,1]$ such that $x(t) \colon= 1$ for all $t \in [0,1]$, we see that $$\frac{\Vert Sx \Vert_{C[0,1]}}{\Vert x \Vert_{C[0,1]}} = 1;$$ so $$\Vert S \Vert = 1.$$ Am I right?
Now, for any $x \in C[0,1]$, we have $$\Vert Tx \Vert_{C[0,1]} = \max_{s\in[0,1]} \vert s x(s) \vert = \max_{s\in[0,1]} \left( \vert s \vert \cdot \vert x(s) \vert \right) \leq \max_{s\in[0,1]} \vert x(s) \vert = \Vert x \Vert.$$ So $$\Vert T \Vert \leq 1.$$ But for $x = 1$, we see that $$\Vert Tx \Vert_{C[0,1]} = \max_{s\in [0,1]} \vert s \vert = 1,$$ and $$\Vert x \Vert_{C[0,1]} = 1.$$ So it follows that $$\Vert T \Vert = 1.$$ Am I right?
Am I right so far?
And if so, then what about $\Vert ST \Vert$ and $\Vert TS \Vert$?