Open unit ball in $C[0,1]$ 
How to show
(i) the  open unit ball in $(C[0,1],d_\infty)$ is not open in $(C[0,1],d_1)$
(ii) the open unit ball in $(C[0,1],d_1)$ is open in $(C[0,1],d_\infty)$ ?,
where $d_1(f,g)=\int_0^1 \vert f(x)-g(x) \vert dx$ and $d_\infty (f,g)=\text{sup}_x\; \vert f(x)-g(x) \vert$

I know the open unit ball in $(C[0,1],d_\infty)$ is "the set of all functions which are less than the unit distance from the zero function" and the figure look like this:(outline of the graph)
Can I have a hint?
 A: Writing it out explicitly tends to make things clearer. Since the metrics involved are norms, let me write $\|\cdot\|_p$ for the $p$ norm, and let $B_p$ be the open unit ball around 0 in the $p$ norm. For $(i)$, you want to find a function $f$ in $B_\infty$ and a function $g$ with small 1-norm so that $f+g\notin B_\infty$. Do you know of any positive functions that are huge in supremum norm but have small integral? You can take $f=0$.
For $(ii)$, you should again just write everything out. This time, you need to show that if you have a function $f$ in $B_1$, then there is a radius $r$ that can depend on $f$, such that for any function $g$ with  $\|g\|_\infty < r$, that $\|f+g\|_1 < 1$. Try to finish the proof.
A: For 1) define $f_n (x) =1-nx$ for $0\leq x \leq \frac 1 n$ and $0$ for other values of $x$. Note that the zero function $f$  is in the open unit ball for $d_{\infty}$. Now $\int_0^{1}|f_n-f|=\frac 1 {2n} \to 0$ and $\|f_n\|_{\infty}=1$ for all $n$. Hence there is no $r>0$ such that  $d_1(f,g) <r$ implies $g$ is in the open un it ball for $d_{\infty}$. This proves 1).
For 2) suppose $\int_0^{1} |f(x)| \, dx <1$. Let $r$ be a number between $\int_0^{1} |f(x)| \, dx$ and $1$.  Then $\|f-g\|_{\infty} <1-r$ implies $\int_0^{1} |g(x)| \, dx<\int_0^{1} |f(x)| \, dx +1-r <1$. This prove that the ball of radius $1-r$ around $f$ is contained in the open unit ball so $f$ is an interior point. 
