Show that $B_1^{d_1}(0) $ is open in $(C[0,1], d_\infty)$ but $B_1^{d_\infty}(0)$ is not open in $(C[0,1],d_1)$
I am clueless on this one.
While proving first part, if I take $f \in $$B_1^{d_1}(0) \rightarrow ||f||_1 < 1$ then I want an $\epsilon$ neighbourhood around $f$ with respect to infinity norm that lies inside this ball
let $r= 1 - ||f||_\infty$
Then $B_{r}^{d_\infty} (f) \subseteq B_1^{d_1}(0)$ because if
$x\in B_{r}^{d_\infty} \rightarrow ||x-f ||_\infty < r( = 1 - ||f||_\infty )\rightarrow ||x||_\infty = ||x-f + f||_\infty \leq ||x-f||_\infty + ||f||_\infty < 1 $
So is this argument good?
for the other part:
$0 \in B_{1}^{\infty}(0)$
Suppose there exists $\epsilon > 0 $ such that $B_{\epsilon}^{d_1}(0)\subseteq B_1^{\infty }(0)$
consider the function $f$ whose graph forms the triangle with vertices $(0,0), (\epsilon,0), (\epsilon/2,1.2)$
then $ ||f||_1$ equals area of this triangle = $0.6\epsilon < \epsilon $ so $f\in B_{\epsilon}^{d_1}(0)$ but $||f||_\infty = 1.2 > 1$ so $f\notin B_1^{\infty}(0)$
so there cannot exists $\epsilon>0$ such that $B_{\epsilon}^{d_1}(0)\subseteq B_1^{\infty }(0)$ so not open
Suppose that $f \in B_1^{d_1}(0)$, which means that $f$ is continuous and $I = \int \vert f \vert < 1$ and denote $r = \frac{1-I}{2}$. For $g \in B_r^{d_\infty}(f)$, we have
$$\int \vert g \vert \le \int \vert f \vert + \int \vert g-f \vert = \int \vert f \vert + \Vert g-f \Vert_\infty \le I + r < 1.$$
That proves that for any $f \in B_1^{d_1}(0)$, we can find an open ball for $d_\infty$ centered on $f$ and included in$f \in B_1^{d_1}(0)$. Therefore $B_1^{d_1}(0)$ is open in $(\mathcal C[0,1], d_\infty)$.
Now consider for all $n \in \mathbb N$
$$g_n(x)= \begin{cases} 1-nx & 0 \le x \le 1/n\\ 0 & 1/n \le x \le 1 \end{cases} $$ We have $d_1(0,g_n) = \int \vert g_n \vert = 1/n$ and $d_\infty(0,g_n) = 1$. As $\lim\limits_{n \to \infty} d_1(0,g_n) = 0$, it is impossible to find an open ball for $d_1$ centered on $0$ and included in $B_1^{d_\infty}(0)$. Proving that $B_1^{d_\infty}(0)$ is not open in $(\mathcal C[0,1], d_1)$.