Relations between normed spaces

Question

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 $\|f\|_1=\int |f(x)|dx$

Answer 1

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.