the identity mapping is not open Let $X=C[0,1]$ and define $I:(X,\|\cdot\|_{\infty})\longrightarrow (X,\|\cdot\|_1 )$. Need to show that $I$ is not open.
I was thinking to find an open set $A$ in $(X,\|\cdot\|_{\infty})$ whose image $I(A)=A$ is not open in $(X,\|\cdot\|_1)$.
But does that make sense to find a set that is open with respect to a norm but not open with respect to another norm, using the same Banach space $X$. Or $I$ may look for another method to prove this.
 A: Yes, your approach makes sense. There will indeed be sets that are open in one norm, but which are not open in the other.
In fact, $A = B_\infty(0; 1)$ should be a counterexample, if a counterexample exists. If $B_\infty(0; 1)$ is open with respect to $\|\cdot\|_1$, then the same is true for every $\infty$-ball. Since $\infty$-open sets are unions of $\infty$-balls, then this would make every $\infty$-open set also $1$-open, and $I$ would be an open map after all.
You can argue even further that $0 \in B_\infty(0; 1)$ cannot be contained in the $1$-interior. So, what would help is to show that $0$ is the $1$-limit of a sequence of functions in $C[0, 1] \setminus B_\infty(0; 1)$.
How do we construct such functions? It's not hard. The $1$-norm doesn't care about how large the function values get, so long as the size of the integral is managed. Functions with a sharp spike in the graph can have small norms, provided the spike is very thin. We need to construct functions with thin enough spikes so that the integral approaches $0$, while the peaks of the spikes give us function values larger than $1$.
With this in mind, let us define, for $n \ge 1$,
$$f_n(x) = \begin{cases} nx & \text{if } 0 \le x \le \frac{1}{n} \\ 2 - nx & \text{if } \frac{1}{n} < x \le \frac{2}{n} \\ 0 & \text{otherwise.}\end{cases}$$
Verify that $f_n$ is continuous. Note that $f_n(1/n) = 1$, which means that $\|f_n\|_\infty \ge 1$. But,
\begin{align*}
\|f_n\|_1 &= \int_0^1 |f_n(x)| \, \mathrm{d}x \\
&= \int_0^{\frac{1}{n}} nx \, \mathrm{d}x + \int_{\frac{1}{n}}^{\frac{2}{n}} 2 - nx \, \mathrm{d}x + \int_{\frac{2}{n}}^1 0 \, \mathrm{d}x \\
&= \left[\frac{n}{2}x^2\right]_0^{\frac{1}{n}} + \left[2x - \frac{n}{2}x^2\right]_{\frac{1}{n}}^{\frac{2}{n}} \\
&= \frac{1}{2n} + \frac{2}{n} - \frac{3}{2n} = \frac{1}{n},
\end{align*}
which tends to $0$. That is, $f_n \to 0$ in the $1$-norm, completing the proof.
The other way you could view this is by looking at the inverse map. Since $I$ is bijective, $I^{-1}$ will be continuous if and only if $I$ is open. Since $I^{-1}$ is linear, it suffices to show it's not bounded. So, we should be able to find a sequence of points in $B_1(0; 1)$ whose $\infty$-norm is unbounded. You can give that one some thought yourself.
A: Consider the functions $2x^n \in X, n=1,2,\dots$ As $n\to \infty,$ $2x^n\to 0$ in $(X,\|\cdot\|_1).$ Since $2x^n\notin B_\infty(0,1)$ for all $n,$ $B_\infty(0,1)$ is not open in $(X,\|\cdot\|_1).$ This proves the identity map is not open.
