1. ## Topology (2)

Show that the metrics d(f,g) = intergral 0 to 1 abs (f(t) - g(t) dt and
d'(f,g) = max {abs f(t) - g(t) : t in [0,1] on the set C[0,1] of continuous real valued functions on [0,1] are not topologically equivalent.

Hint: If d' is topologically equivalent to d, then a sequence {Xn} converges to X in one metric iff it converges to X in the other metric.

Thanks.

2. Originally Posted by taypez
Show that the metrics d(f,g) = intergral 0 to 1 abs (f(t) - g(t) dt and
d'(f,g) = max {abs f(t) - g(t) : t in [0,1] on the set C[0,1] of continuous real valued functions on [0,1] are not topologically equivalent.

Hint: If d' is topologically equivalent to d, then a sequence {Xn} converges to X in one metric iff it converges to X in the other metric.

Thanks.
I never learned topology so I might be way off.

Consider the sequence $x_n = \{ x^n \}$, i.e. $x_1=x,x_2=x^2,x_3=x^3,...$. I claim that the limit of this sequence is $0$. Meaning we need to show $d(x^n,0)<\epsilon$ for sufficiently large $n$. Meaning, $\int_0^1 |x^n - 0 | dx =\int_0^1 x^n dx$ be made sufficiently small. This integral is $\frac{1}{n+1}$ which can be made smaller than any $\epsilon >0$. We have shown this sequence converges to $0$ in the first topology.

Consider the same sequence in the second topology. I claim the limit is $1$. We need to show $d'(x^n,1)<\epsilon$ when $n$ is sufficiently large. But this means $\max_{x\in [0,1]} \{ |x^n - 1| \} = \max_{x\in [0,1]} \{ x^n - 1\}$. But the maximum value is $0$ because $0\leq x^n \leq 1$ on $[0,1]$. This (0) is certainly less than any epsilon thus thus it converges to 1 in the second topology.

Thue, the two fields are not equivalent.