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The vector space of continuous functions on $[0,1]$ can be given the two metrics

$d_\infty(f,g)=\text{sup}_{x\in[0,1]}|f(x)-g(x)|$ and $d_1(f,g)=\int^1_0|f(x)-g(x)|dx$.

Are these two metrics equivalent?

I know how to show two metrics are equivalent in general, so we need to show every set of functions that is open in $([0,1],d_\infty)$ is open in $([0,1],d_1)$ and vice versa.

More specifically, let $X\in[0,1]$ be open in $([0,1],d_\infty)$. Then $X=$ int$X$ and it implies that $\forall f\in X, \exists r_\infty>0$ such that $B_{r_\infty}(x)\subset X$, which is to show that $\forall f\in X, \exists r_\infty>0$ such that $\{g\in X:\text{sup}_{x\in[0,1]}|f(x)-g(x)|<r_\infty\}$. And we need to show that $\forall f\in X, \exists r_1>0$ such that $\{g\in X:\int^1_0|f(x)-g(x)|<r_1\}$.

But I have little clue in showing how can we get from $\{g\in X:\text{sup}_{x\in[0,1]}|f(x)-g(x)|<r_\infty\}$ to $\{g\in X:\int^1_0|f(x)-g(x)|<r_1\}$. I am not very good in working with continuous functions.

Can anyone please give some hints or some direction that I can work on?

Thanks very much!

user71346
  • 4,171

2 Answers2

1

They are not equivalent, $B_1(0,1)$ is not contained in $B_{\infty}(0,n)$for any integer n.

Hint; construct a function $f$ such that $\int_0^1f=1$ and $sup_{[0,1]}f=+\infty$. Take a function which is $0$ on [0,1/n+1] and [1/n,1] on [1/n+1,1/n] you can construct a function $f_n$ such that $f_n(1/n)=f_n(1/(n+1))=0$, on $[1/1+n, {{1/n+1/(n+1)}\over 2}]$ $f_n$ is an increasing linear function, on $[ {{1/n+1/(n+1)}\over 2},1/n]$ $f_n$ is a decreasing linear function to 0 and $\int_{[1/n+1,1/n}f_n=1$. The graph of $f$ looks like a hat

Tsemo Aristide
  • 87,475
0

They're not equivalent. Any $d_1$-neighbourhood of the zero function contains infinitely many functions of the form $x \mapsto x^n$, while all these functions are at $d_\infty$-distance $1$ from the zero function.

timon92
  • 11,329