identity operator isn't bounded Suppose we consider the identity operator between the spaces
$(C([0,1]),\| . \|_{\infty}) \rightarrow (C([0,1]),\| . \|_{1})$. Then the identity operator is bounded but its inverse isn't bounded. 
I am a little bit confused about this. So suppose we call the identity operator as $T : (C([0,1]),\| . \|_{\infty}) \rightarrow (C([0,1]),\| . \|_{1})$. If we calculate the norm in for this operator we find that 
$$\|T\| = \sup_{||f||_{\infty} = 1} \int_{t = 0}^{t = 1} |f(t)| dt$$
I am not sure how can we argue that this is bounded.
Also, I have troubles for the reverse side as well. That $T^{-1}$ isn't bounded. Can someone explain this?

Note that
$$
\|T\|=\sup_{\|f\|_\infty\le 1}\|Tf\|_1=\sup_{\|f\|_\infty\le 1}\int_0^1|f(t)|\,dt\le\sup_{\|f\|_\infty\le1}\int_0^1\|f\|_\infty\,dt=1,
$$
so $T$ is bounded.
To see that $T^{-1}$ is not bounded, it suffices to find a sequence of functions $(f_n)$ in $C[0,1]$ whose $L_1$ norms are less than $1$ but whose supremum norms grow without bound.
To do this, take $f_n$ to be the piecewise linear function on $[0,1]$ whose graph connects the points $(0,4n)$ to $(1/2n,0)$ to $(1,0)$, that is,

Then
$$
\|f\|_1 = \frac{(4n)(1/2n)}{2}=1
$$
but $\|f\|_\infty=4n$.
Consequently
$$
\|T^{-1}\|=\sup_{\|f\|_1\le 1}\|Tf\|_\infty\ge\sup_{n\in\mathbb{N}}\|f_n\|_\infty=\infty,
$$
so $T^{-1}$ is not bounded.
