Demonstration that two distances aren't equivalent. I'd like to find an example that shows that the following distances aren't equivalent in $X=C([0,1])$:
$$d_\infty(f,g)=\sup_{x\in [0,1]}|f(x)-g(x)|$$
$$d_1(f,g)=\int_0^1|f(x)-g(x)|dx$$
Someone can help me?
 A: Consider two functions that are equal at "most" locations, but are different on a very small set. Then the $d_1$ distance will be small, where the $d_\infty$ distance will be relatively larger - this distance second doesn't care that the functions are close most of the time, it just looks for the biggest difference.
Concretely, you could try $f(x) \equiv 0$ and
$$
g_n(x) = \begin{cases} 1 - n x &: 0 \leq x \leq \frac{1}{n}, \\
0 &: \frac{1}{n} < x \leq 1.
\end{cases}
$$
Then $d_\infty(f,g_n) = 1$, while $d_1(f,g_n) = 1/2n$.
Now, let's assume these metrics are equivalent. Then, among other things$^*$, there needs to be a constant such that
$$
C d_\infty(f,g) \leq d_1(f,g)
$$
for all $f,g$. But, as we have seen
$$
d_\infty(f,g_n) = 1 \\
d_1(f,g_n) = \frac{1}{2n}
$$
and there is no positive constant so that
$$
C \leq \frac{1}{2n}
$$
for all $n$.
*: of course, the other thing that would have to hold is 
$$
\widetilde C d_1(f,g) \leq d_\infty(f,g)
$$
for some other constant $\widetilde C$ and all $f,g$. But, this is true:
$$
d_1(f,g) = \int_0^1 |f(x) - g(x)| \,dx \leq \int_0^1 \sup_{x \in [0,1]} |f(x) - g(x)|\,dx = \int_0^1 d_\infty(f,g) \,dx = d_\infty(f,g)
$$
