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I am a little bit unsure about the following claim. Let $H:[0,1]\times \mathbb{R}\rightarrow \mathbb{R}$ be a mapping of the form $H(x,f(x))$, where $f:[0,1]\rightarrow \mathbb{R}$ is some continuous and nondecreasing function. Suppose there exists a positive constant $K$, independent of $x$ and $f$ such that: $$ \sup_{x\in[0,1]}|H(x,f(x))-H(x,g(x)|\leq K \sup_{x\in [0,1]} |f(x)-g(x)| $$ where $g:[0,1]\rightarrow \mathbb{R}$ is a nondecreasing and continuous function.

The claim is that if the above inequality holds, then $$ |H(x,f(x))-H(x,g(x)|\leq K |f(x)-g(x)| $$ for all $x\in [0,1]$. I want to prove/disprove this claim. I tried a couple of things with no success so far. For instance, I tried to show that $g(x)=\frac{|H(x,f(x))-H(x,g(x))|}{f(x)-g(x)|}$ is bounded above, which would give me the desired result but I am stuck at the point in which I get $$|H(x,f(x))-H(x,g(x))|\leq \sup_{x} g(x) \sup_x|f(x)-g(x)|$$ This looks a lot like the inequality I have, but I do not know how to link $\sup_{x}g(x)$ with $K$.

Any suggestion/comment/help is greatly appreciated it!

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thanks @JavaMan. I see...but is the conclusion of the initial condition still valid? –  Cristian Sep 3 '12 at 23:41
    
Sorry I erased my comment that for $a(x), b(x) \neq 0$, we have $\sup(a/b) \leq \sup(a)/\inf(b)$. I still need to think more carefully about the claim. –  JavaMan Sep 3 '12 at 23:49
    
I don't understand the second sentence. I understand what a mapping $H:[0,1]\times \mathbb R\to \mathbb R$ is. And I understand the meaning of $H(x,f(x))$: we plug $f(x)$ into the second variable. But I don't know what it means for $H$ to be of the form $H(x,f(x))$. For example, is $H(x,y)=x+y$ of such a form or not? –  user31373 Sep 7 '12 at 22:14
    
@LVK Thanks for your comment. It only means what you first say: in the second variable you plug in $f(x)$. In your example, if we let $y=f(x)$, where $f(x)$ is a nondecreasing and continuous function on $[0,1]$, then it is a case of the H function I have in mind. –  Cristian Sep 8 '12 at 22:55
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What prevents you from plugging in two constant functions $f(x)=a$, $g(x)=b$ into the assumption and concluding that $|H(x,a)-H(x,b)|\le K|a-b|$ for all $x\in[0,1], a,b\in\mathbb R$? –  fedja Sep 11 '12 at 10:54

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