Prove that the identity map $(C[0,1],d_1) \rightarrow (C[0,1],d_\infty)$ is not continuous $$d_\infty = \max|x_i - y_i|$$
$$d_1 = \sum_{i=1}^n |x_i - y_i|$$
The first part of this question was to prove that the identity map $$(C[0,1],d_\infty) \rightarrow (C[0,1],d_1)$$ is continuous, which I did using $\epsilon = n \delta$. I'm not sure how to prove it is not continuous in the opposite direction. I tried using my logic in reverse but it doesn't seem to be working. Can anyone offer any hints or suggestions? 
 A: Is the identity map $C[0,1],d_1\to C[0,1],d_{\infty}$ bounded? $C[0,1],d_1$ is incomplete
$f_n(x)= 2nx; x\in [0,{1\over 2n}]$
$f_n(x)= 2-2nx; x\in [{1\over 2n},{1\over n}]$
$f_n(x)= 0$ else
$f_n$ converges to $0$ in $d_1$, it does not convrges  to $d_{\infty}$ norm,since the value of $\sup f_n(x)=1$ always
A: I will assume that $d_1$ and $d_\infty$ are the distances induced by the one and infinity-norms, respectively, i.e. 
$$
d_1(f,g)=\|f-g\|_1=\int_0^1|f-g|,\ \ \ \ d_\infty(f,g)=\|f-g\|_\infty=\sup\{|f(t)-g(t)|:\ t\in[0,1]\}.
$$. 
It is probably not efficient to think of this problem in terms of the definition of continuity. Rather, since the identity is a linear mapping, one wants to show that it is bounded in one direction and unbounded in the other one. 
On the one hand, you have $d_1(f,g)\leq d_\infty(f,g)$, since
$$
\int_0^1|f|\leq \|f\|_\infty\,\int_0^11=\|f\|_\infty.
$$
To show that the identity is not continuous the other way, we need to find functions with constant $1$-norm, and arbitrarily large infinity-norm. For example, we could take
$$
f_n(t)=\begin{cases}n,&\mbox{ if }t\in[0,1/n),\\ n(2-nt),&\mbox{ if }t\in[1/n,2/n)\\0,&\mbox{ if }t\in[2/n,1]\end{cases}
$$
This $f_n$ is continuous, $\|f_n\|_\infty=n$, $\|f_n\|_1=3/2$.
