Finding difficult to find a counterexample show that two metrics are not equivalent.
Set: $C[0,1] $ of all continuous functions on the interval $[0,1]$.
Metric 1: $d(x,y) = \max\limits_{t \in [0,1]} |x(t) - y(t)|$.
Metric 2: $d^* (x,y) = \int_0^1 |x(t) - y(t)| dt$.
HINT: It suffices to find a sequence $\langle f_n:n\in\Bbb N\rangle$ that converges in one metric but not in the other. In order to keep things easy to visualize, we could try to find a sequence that converges to the zero function $z(t)\equiv 0$ in one of the metrics but not in the other. It’s also worth noting that if $f_n(t)\ge 0$ for all $t$, then $d^*(f_n,z)$ is just the area under $f_n$ between $0$ and $1$, while $d(f_n,z)$ is the maximum value of $f_n(t)$ on $[0,1]$. Can you choose the functions $f_n$ so that either
One of these should seem a much better bet than the other, especially in conjunction with the suggestion in the comments. In case you still don’t see it after some thought, I’ve added a further hint in the spoiler-protected block below.
Triangles of height $1$ can have arbitrarily small areas.
Another way:
Suppose that , $X=C[0,1]$.
If the given two matrices are equivalent then, the identity map $id:(X,d^*)\to (X,d)$ defined by $id(f(t))=f(t)$ , $\forall f(t)\in X$ is continuous.
Not only that $id:(X,d)\to (X,d^*)$ by $id(f(t))=f(t)$ , $\forall f(t)\in X$ is also continuous.
We show by an example that, the identity map $id:(X,d^*)\to (X,d)$ defined by $id(f(t))=f(t)$ , $\forall f(t)\in X$ is NOT continuous.
For it, consider the functions, $f_1(t)=t^n$ & $f_2(t)=0$. Then $f_1,f_2\in X$.
Also, $d^*(f_1,f_2)=\int_0^1|f_1(t)-f_2(t)|\,dt=\int_0^1t^n\,dt=\frac{1}{n+1}\to 0$ as $n\to \infty$.
But, $$d\left(id(f_1(t),f_2(t)\right)=d(f_1(t),f_2(t))=\max_{0\le t\le 1}|f_1(t)-f_2(t)|$$
$$=\max_{0\le t\le 1}t^n=1\not\to0 $$as $n\to \infty$.
So, the identity map $id:(X,d^*)\to (X,d)$ defined by $id(f(t))=f(t)$ , $\forall f(t)\in X$ is NOT continuous.
Hence , $d$ & $d^*$ may not be equivalent.
