$I_{id}:(C[0,1],\|\|_{\infty})\to (C[0,1],\|\|_{1}) $ is continuous and the open map? $I_{id}:(C[0,1],\|\|_{\infty})\to (C[0,1],\|\|_{1}) $ is continuous and  the open map?
$\|f\|_{\infty}=\sup_{x\in [0,1]}\{|f(x)|\}\le\int_{0}^{1} |f (x) |$ so it is continuous and homeomorphism so open map?
 A: Hints: $\max_{[0,1]} |x^n| = 1$, $\int_0^1 |x^n| dx = \frac{1}{n+1}$ and $\int_0^1 |f(x)| dx \leqslant \max_{[0,1]} |f(x)|$.
A: You have the direction of the inequality the wrong way around. For all $f\in C([0,1], \| \cdot \|_{\infty})$ we have $ \| I(f) \|_1 \leq \| f \|_{\infty}$ so $I$ is continuous. However your claim that it is a homeomorphism is incorrect. The supremum norm and the $1$ norm do not induce the same topology on $C([0,1]).$ For example, of functions $f_n$ which are the linear interpolations between the points $(0,n), (1/n^2, 0)$ and $(1,0)$ converges to $0$ in the $1$ norm but diverge in the supremum norm. Since $I$ is continuous, the preimage of open sets is open. If $I$ was an open map, $U$ would be open if and only if $I(U)$ was open i.e. the two topologies are equivalent. We just saw that was not the case, so it is not an open map.
A: No, you do not prove anything and your inequality does not work.
First of all, this map is linear so you only have to check that $\exists C>0: \forall f\in C([0,1]), \|f\|_1\leq C\|f\|_\infty$ in order to prove it is continuous, ok ?
Then, unfortunately, this map is not a homeomorphism. You can check that using the following sequence $(u_n)_{n\in \mathbb N}\subset C([0,1]), u_n(x)=x^n$: $$ u_n \underset{\|.\|_1}{\longrightarrow} 0 \text{ but } u_n\underset{\|.\|_\infty}{\not \longrightarrow} 0$$
Since $I_{id}$ is a bijection, if it was an open map then it would be a homeomorphism, which is not the case.
