Relations between normed spaces Is the application 
$$
Id:( C([0,1]), \|\cdot\|_{\infty})\to ( C([0,1]), \|\cdot\|_{1})
$$
open?
where $Id(f)=f$, $\|f\|_{\infty}=\sup\|f(x)\|$ and $\|f\|_1=\int |f(x)|dx$
 A: I found the problem with two different norms interesting, so I'll try to give a solution.
Let's put $\|\cdot\|_1$ in one of the spaces and call:
$$
X_1 = ( C([0,1]), \|\cdot\|_{1} )
$$
$$
X_\infty = ( C([0,1]), \|\cdot\|_{\infty} )
$$
Two identity linear operators get defined:
$$
Id_1:X_\infty\to X_1
$$
$$
Id_\infty:X_1\to X_\infty
$$
Asking if one of the operators is an open map is equivalent to ask if its inverse is continuous (as the pre-image of an open set is open for continuous operators).
Note that the $Id$ operators defined are the inverse of each other.
To check continuity we can check boundedness over the unit ball. Take:
$$
B_1 = \{f\in X_1: \|f\|_1<1\} = \{f\in C([0,1]): \int |f(x)|dx<1\}\subset X_1
$$
$$
B_\infty = \{f\in X_\infty: \|f\|_\infty<1\} = \{f\in C([0,1]): \sup\|f(x)\|<1\}\subset X_\infty
$$
We have that:

$$Id_1 \mbox{ is bounded.} $$

Take $f\in Id_1(B_\infty) \subset X_1$. We have:
$$\|f\|_1=\int |f(x)|dx <= \int \sup\|f(x)\| = \int \|f\|_{\infty} = \|f\|_{\infty}<1$$
So $Id_1$ is bounded (and hence continuous).

$$Id_\infty \mbox{ is not bounded.}$$

Consider  $\{f_n\}\subset X_1$ where 
$$
f_n = \begin{cases} n - n^2x, &x<\frac{1}{n} \\
 0, &x >= \frac{1}{n} \end{cases}
$$
It's easy to show that $f_n$ is continuous and $\|f_n\|_1 = 1/2 < 1$ so $\{f_n\}\subset B_1$. But $\|f_n\|_{\infty} >= |f_n(0)| = n$ so, for $n\to \infty$ we have $\|f_n\|_\infty\to \infty$. This means that $Id_\infty$ is not bounded.
It follows that $Id_1$ (the inverse of $Id_\infty$, which is not continuous) is not an open map, while $Id_\infty$ is an open map.
Note: being $X_\infty$ a Banach space, this also shows that $X_1$ is not a Banach space, otherwise being $Id_1$ continuous between two Banach spaces it would be also open for the Open Map Theorem, but it's not.
