Functional analysis, operator norm, continuous We observe $C[0,1]$ (the $\mathbb{C}$-vectorspace of all continous functions in $\mathbb{C}$ on $[0,1]$) as subspace of $L^p([0,1],\lambda)$ where $\lambda$ notes the Lebesgue-measure on $[0,1]$.

For $1<p\leq\infty$ is $T:(C[0,1],\|\cdot\|_p)\to (C[0,1],\|\cdot\|_1)$, $T(f)=f$ continuous, but $T^{-1}$ is not continuous.
  Determine the operatornorm of $T$.

To show that $T$ is continuous I can show, that $T$ is bounded. Therefore it exists a $c>0$, such that for every $f\in C[0,1]$ holds:
$\|T(f)\|_p\leq c\|f\|_1$
So, do I have to show, that $\|T(f)\|_p=(\int_0^1 |f|^p d\lambda)^{1/p}\leq c\int_0^1 |f|$ for some $c>0$, to get that it is bounded?
How can I find $T^{-1}$. It has to be $T^{-1}(T(f))=f$, since $T(f)=f$, it has to be $T^{-1}(f)=f$. Is that correct?
To find the operator norm of $T$, I have to give
$\sup_{\|f\|_p\leq 1} \|T(f)\|_1$.
Can you give me hints on the parts, to solve this question?
Thanks in advance.
 A: To show that $T$ is bounded you have to show that there exists $c > 0$ such that $$\|f\|_1 = \|Tf\|_1 \le c \|f\|_p, \quad \forall f \in C[0,1]$$
Indeed, using Hölder's inequality for $p$ and its conjugate exponent $q$ we obtain:
$$\|f_1\| = \int_0^1 |f| = \int_0^1 |f|\cdot 1 \stackrel{\text{Hölder}}{\le} \sqrt[p]{\int_0^1 |f|}\underbrace{\sqrt[q]{\int_0^1 1^q}}_{=1} = \|f\|_p$$
Thus, $T$ is bounded and $\|T\| \le 1$.
For $f \equiv 1$ we have:
$$\|T\| \ge \frac{\|Tf\|_1}{\|f\|_p} = \frac{\|f\|_1}{\|f\|_p} = \frac11 = 1$$
Thus, $\|T\| = 1$.
Since $T$ is in fact the identity operator, we of course have $T^{-1}(f) = f$. Let's show that $T^{-1}$ is not bounded, meaning there does not exist $c > 0$ such that $\|f\|_p \le c \|f\|_1$ for all $f \in C[0,1]$.
Assume such $c > 0$ exists. For $n \in \mathbb{N}$ define $f_n \in C[0,1]$ as $f_n(t) = t^n$, $\forall t \in [0,1]$.
We have:
$$\|f_n\|_1 = \int_0^1 t^n\,dt = \frac{1}{n+1}$$
$$\|f_n\|_p = \sqrt[p]{\int_0^1 t^{pn}\,dt} = \frac{1}{\sqrt[p]{pn+1}}$$
Thus:
$$c^p \ge \lim_{n\to\infty} \frac{\|f_n\|^p_p}{\|f_n\|^p_1} =\lim_{n\to\infty} \frac{(n+1)^p}{pn+1} = +\infty$$
Therefore, such a constant $c$ cannot exist so $T^{-1}$ is not bounded.
