# Thread: is this norm lipschitz equivalent to uniform norm?

1. ## is this norm lipschitz equivalent to uniform norm?

$\displaystyle \|f\|_{1}=\int_{0}^{1}{|f|} \ \ on \ \ C[0,1]$

uniform norm: $\displaystyle \|f\|_{\infty}=\sup \{|f(x)| : x\in [0,1] \}$

i think this is obvious:

$\displaystyle \|f\|_{1}\leq\|f\|_{\infty}$

but i cant find a positive real number a, $\displaystyle a\|f\|_{\infty}\leq\|f\|_{1}$

2. Originally Posted by szpengchao
$\displaystyle \|f\|_{1}=\int_{0}^{1}{|f|} \ \ on \ \ C[0,1]$

uniform norm: $\displaystyle \|f\|_{\infty}=\sup \{|f(x)| : x\in [0,1] \}$

i think this is obvious:

$\displaystyle \|f\|_{1}\leq\|f\|_{\infty}$

but i cant find a positive real number $\displaystyle a$, $\displaystyle a\|f\|_{\infty}\leq\|f\|_{1}$
This is because there exists no such $\displaystyle a$: for every $\displaystyle \varepsilon>0$ you can easily find $\displaystyle f\in\ C([0,1])$ such that $\displaystyle f(0)=1$ and $\displaystyle \int_0^1 |f|<\varepsilon$: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose $\displaystyle f(x)=x^n$ for $\displaystyle n$ large enough if you prefer to write formulas (then $\displaystyle f(1)=1$))

3. ## ok

ok. so they are not lipschitz equivalent then?

4. ## but "a" can be arbitrarily small too...

Originally Posted by Laurent
This is because there exists no such $\displaystyle a$: for every $\displaystyle \varepsilon>0$ you can easily find $\displaystyle f\in\ C([0,1])$ such that $\displaystyle f(0)=1$ and $\displaystyle \int_0^1 |f|<\varepsilon$: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose $\displaystyle f(x)=x^n$ for $\displaystyle n$ large enough if you prefer to write formulas (then $\displaystyle f(1)=1$))
i don't get it. "a" can be arbitrarily small...

can't you do the following... let a -> 0+ and let A-> positive infinity. then it must be true that (a * uniform norm) <= norm1 <= (A * uniform norm) since 0 <= norm1 <= infinity, for all f.

what's wrong with that?

i understand there's a theory that all norms in Rn are Lipschitz equivalent to the Euclidian norm. then all should be Lipschitz equivalent to each other. seems applicable?