is this norm lipschitz equivalent to uniform norm?

• Nov 1st 2008, 10:26 AM
szpengchao
is this norm lipschitz equivalent to uniform norm?
$\|f\|_{1}=\int_{0}^{1}{|f|} \ \ on \ \ C[0,1]$

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

i think this is obvious:

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

but i cant find a positive real number a, $a\|f\|_{\infty}\leq\|f\|_{1}$
• Nov 1st 2008, 11:31 AM
Laurent
Quote:

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

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

i think this is obvious:

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

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

This is because there exists no such $a$: for every $\varepsilon>0$ you can easily find $f\in\ C([0,1])$ such that $f(0)=1$ and $\int_0^1 |f|<\varepsilon$: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose $f(x)=x^n$ for $n$ large enough if you prefer to write formulas (then $f(1)=1$))
• Nov 1st 2008, 11:32 AM
szpengchao
ok
ok. so they are not lipschitz equivalent then?
• Jul 13th 2010, 07:37 PM
elm
but "a" can be arbitrarily small too...
Quote:

Originally Posted by Laurent
This is because there exists no such $a$: for every $\varepsilon>0$ you can easily find $f\in\ C([0,1])$ such that $f(0)=1$ and $\int_0^1 |f|<\varepsilon$: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose $f(x)=x^n$ for $n$ large enough if you prefer to write formulas (then $f(1)=1$))

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?