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?
 A: 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.
A: In arbitrary normed spaces, a linear mapping is not automatically continuous, as it happens when there is a finite basis. As long as vector spaces have a homogeneous structure, their topology can be determined by a system of neighborhoods near 0. The same holds true for morphisms between normed spaces. It is enough to examine for continuity at 0. Say now, $X, Y$ are normed spaces and $f : X \longmapsto Y$ is a linear map. $f$ is continuous at 0 if and only if given $\epsilon > 0$ we can find $\delta > 0$ such that the following implication holds
$$ \|x\| < \delta \Longrightarrow \| f(x) \| < \epsilon $$
Due to linearity, this is completely equivalent 
$$ \|x\| < 1 \Longrightarrow \| f(x) \| < \epsilon/\delta $$
which is completely equivalent to saying
$$ \| f(x) \| < \epsilon/\delta \| x\| \; , \quad x \in X$$
Hence checking for continuity reduces to seeking for a constant $M \geq 0$ such that $ \| f(x) \| \leq M \| x\|$ (*) holds. The term bounded comes for the supremum
$$ \sup \left\{ \| f(x)\| ; \|x \| \leq 1 \right\}$$
being finite. This is equivalent to saying that $f$ is continous at 0, and continuous everywhere. 
Suppose now that $f$ is 1-1 and surjective. There is a function 
$g : Y \longrightarrow X$ such that $f \circ g = I$ and $g \circ f = I$.
But as long we are working in spaces having both a linear and a topological structure, we need to prove that $g$ is also a topological morphism (that is, continuous) to say that $g$ is indeed the inverse of $f$.
Suppose also that $X = Y$.
Again in contrast to the finite dimensional case that we had the nice dimension theorem, the assumption that $f$ is 1-1 or surjective wont drive us to find an inverse even set theoretically. There are numerous examples of linear mappings being surjective but not injective and vice versa. Even worse, the same linear space as set may be endowed with many different topologies making even the identity being not-continuous.
The standard method to prove continuity in a normed space context is to prove ($*$). The standard method to prove not-continuity is to prove that ($*$) doesnt hold, equivalently, to find some bounded sequence $\left(x_n\right)_{n\in\mathbb N}$ such that $\|f(x_n)\| \longrightarrow +\infty$
