let $T \in B(C([0,1])) $ defined by
$$ \forall f \in C([0,1]), (Tf)(x)=xf(x)$$
and consider that $B(C([0,1]))$ is equipped with the max-norm. Find operator norm $|T|$
Let $T:M_1 \to M_2$ a bounded linear operator then
$$|| T|| \sup_{x \in M_1 ,x \neq 0 } \frac{||Tx||_{M_1}}{||x||_{M_1}}
=\sup_{x \in M_1 , || x||_{M_1}=1} ||Tx ||_{M_2} $$
$$ \begin{aligned}
||T||&= \sup_{f \in C([0,1])} \frac{||T f ||_\infty}{||f||_\infty}
= \sup_{f \in C([0,1])} \frac{\max_{x \in [0,1]}|x f(x) ||}{\max_{x \in [0,1]}|f(x)|}
\\ &=\sup |x|=1
\end{aligned}$$
not sure that canceling out $f(x)$ is allowed. OR if the propostion of the operator norm was used corectly
Was suppused to show that it is bounded operator. $x, f(x)$ are continous their preimage is closede so its image is closed and bounded and we have that
$$\begin{aligned}
||Tx ||_\infty &= \max_{x \in [0,1]} |Tf(x)|=\max_{x \in [0,1]}|xf(x)|
\\ &\leq \max_{x \in [0,1]}|x| \max_{x \in [0,1]} |f(x)|
=1\max |f(x)|=||f||_\infty
\end{aligned} $$
So $||Tf||_\infty \leq ||f||_\infty$
Best Answer
Since $x\in[0,1]$, you have $$ |xf(x)|\leq|f(x)| $$ for all $x$, and so $\|Tf\|\leq\|f\|$. This shows that $T$ is bounded and that $\|T\|\leq1$. In your argument, you can directly say that $\max|xf(x)|\leq\max|f(x)$.
Now let $f\in C[0,1]$ be $f(x)=1$. Then $$\|Tf\|=\max\{|x|:\ x\in[0,1]\}=1$$ and $\|f\|=1$, so $\|Tf\|=\|f\|$, which implies that $\|T\|\geq1$. It follows that $\|T\|=1$.