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I would like a little help here:

I have two defined norms over $C^{1}([0,1])$ :

  • $\| A(f)\|=|f(0)|+\max_{x\in[0,1]}{|f'(x)|}$
  • $\| B(f)\|=\int_0^1|f(x)|dx+\max_{x\in[0,1]}{|f'(x)|}$

I already proved that $A,B$ are norms over $C^{1}([0,1])$ by showing that the usual axioms hold: zero vector has norm $0$, positive homogeneity and the triangle inequality (if it's not complete tell me please).

The last thing that I need to show is the equivalence of those norms. How am I to do it?

Thanks

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@julien:Topological equivalence . –  abnormal Mar 5 '13 at 21:37
    
what have you tried so far? The topological equivalence follows from the Lipschitz equivalence (if I understand Lipschitz equivalence correctly) –  Quickbeam2k1 Mar 5 '13 at 21:42
    
Forget my previous comment. I was thinking distances, for which the distinction makes sense. Not norms... And they are indeed equivalent here. –  1015 Mar 5 '13 at 22:25

2 Answers 2

By the mean value theorem $$ |f(x)|\leq |f(0)|+\|f'\|_\infty\qquad\forall x\in[0,1]. $$ Hence $$ \int_0^1|f(x)|dx\leq \int_0^1(|f(0)|+\|f'\|_\infty)dx=|f(0)|+\|f'\|_\infty. $$ So $$ \|B(f)\|=\int_0^1|f(x)|dx+\|f'\|_\infty\leq |f(0)|+2\|f'\|_\infty\leq 2\|A(f)\|. $$

Conversely, observe that an integration by parts yields $$ \int_0^1f(x)dx=-f(0) -\int_0^1 (x-1)f'(x)dx. $$ Hence $$ |f(0)|\leq \int_0^1|f(x)|dx+\int_0^1|x-1||f'(x)|dx\leq \int_0^1|f(x)|dx+\|f'\|_\infty. $$ Therefore $$ \|A(f)\|=|f(0)|+\|f'\|_\infty\leq \int_0^1|f(x)|dx+2\|f'\|_\infty\leq 2\|B(f)\|. $$

So we have shown that the two norms are equivalent.

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One part was done before me, we have to carry $|f(0)|$.

So, by the mean value theorem, for any $t\in (0,1]$ we have that $f(t)-f(0)=t\cdot f'(t_1)$ for some $t_1<t$, so $|f(t)|\le |f(0)|+||f'||_\max$ (using also $t\le 1$). This implies $||f||_\max\le |f(0)|+||f'||_\max$.

For the other part, $\int_0^1 |f|dx\le \int_0^1|f(0)|+||f'||_\max dx= |f(0)|+||f'||_\max$, and these altogether show that indeed $$||B(f)|| \le ||A(f)||+||f'||_\max \le 2||A(f)||$$

For the other part, it's a bit trickier: start to draw the function $f$ from $x=0$, wlog on the picture we can assume $f(0)>0$. Then draw the line with the maximal possible slope, $m:=||f'||_\max$, downwards so that it intersects the $x$-axis at $x=f(0)/m$. Then we have that this whole triangle (origin, $(0,f(0))$, $(x,0)$) must be below $f$, so must be contained in $\int_0^1 |f|$. That is, $$ \frac{f(0)^2}{2m}\le \int_0^1|f| \\ f(0)^2\le 2||B(f)||^2 \,,$$ and from this $||A(f)||\le 3||B(f)||$ follows.

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My deleted answer was corrected some time ago. Check the integration by parts. It makes it pretty easy and it yields a $2$ bound. If I did not make another mistake... –  1015 Mar 5 '13 at 23:14
    
Your argument is original (at least for me), +1. –  1015 Mar 5 '13 at 23:23
    
I just drew it on a paper.. :) Probably the bound $2$ would be also available from that, but doesn't really matter.. –  Berci Mar 6 '13 at 9:40

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