Let $f$ be continuous on $\Bbb R$ and $c $ be a positive real number. Let $x \in [0,K]$ for some $K >0$. Then does there exist a constant $C >0$ such that $$ \| f(cx) \|_{L^\infty([0,K])} = C \| f(x) \|_{L^\infty([0,K])} ?$$
About an $L^\infty$ equality
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real-analysis
functional-analysis
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0Should $C$ be the same constant for all $c$? Do you want to show the equality for all $c$? – 2012-12-19
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0@macydanim No, $c$ and $C$ are different. I want to find such $C$ or if exists, then I want to know the dependence . – 2012-12-19
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0Your assumption that $x\in[0,K]$ is meaningless in the context of the question. – 2012-12-19
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0I know that $c$ and $C$ are meant to be different, but do you want $C$ to depend on $c$ i.e. $C=C(c)$ ? – 2012-12-19
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0@Eckhard I think $x \in [0,K]$ guarantees that $f$ is bounded on $\Bbb R$ because $f$ is continuous. – 2012-12-19
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0But if you have $x\in [0,K]$ why do you chose the $||\cdot||_{L^{\infty} R}$ norm and not only on $[0,K]$ – 2012-12-19
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0@macydanim Yes, you are right. That should be replaced by $[0,K]$ I will change. – 2012-12-19
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0If $f$ is continuous, it is bounded on $[0,K]$, but that is not relevant if one considers the $L^\infty(\mathbb{R})$ norm. That being said, the notation $\|f(x)\|_{L^\infty(\mathbb{R})}$ is not well-defined, even less if you restrict $x$ to an interval $[0,K]$. – 2012-12-19
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0@Eckhard In fact $L^\infty$ is just a maximum in this case. – 2012-12-19
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0So does $C$ depend on $c$ ? Because if so, the formulation would be trivial. Or do you want the same $C$ for all $c$ ? – 2012-12-19
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0@macydanim I hope $C$ to be depend on $c$. – 2012-12-19
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0Well then chose C = $||f(cx)||_{L^{\infty}} / ||f(x)||_{L^\infty} $for all $f \not \equiv 0$. – 2012-12-19
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0@macydanim Thank you very much. I was a little confused. – 2012-12-19
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0So that was the answer you wanted ? Because that won't help that much in any application ;) – 2012-12-19
1 Answers
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If $c=1$, then obviously, we can take $C=1$. Otherwise, there can be no such $C$.
If $c\lt1$, let $$ f(x)=\left\{\begin{array}{} 0&\text{if }x\le cK\\ \frac{x-cK}{(1-c)K}&\text{if }cK\lt x\lt K\\ 1&\text{if }x\ge K \end{array}\right. $$ where $\|f(x)\|_{L^\infty([0,K])}=1$, yet $\|f(cx)\|_{L^\infty([0,K])}=\|f(x)\|_{L^\infty([0,cK])}=0$
If $c\gt1$, let $$ f(x)=\left\{\begin{array}{} 0&\text{if }x\le K\\ \frac{x-K}{(c-1)K}&\text{if }K\lt x\lt cK\\ 1&\text{if }x\ge cK \end{array}\right. $$ where $\|f(x)\|_{L^\infty([0,K])}=0$, yet $\|f(cx)\|_{L^\infty([0,K])}=\|f(x)\|_{L^\infty([0,cK])}=1$
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1You should modify your construction of $f$ to satisfy the continuity condition. – 2012-12-19
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0@richard: thanks. I was too busy thinking about $L^\infty$. – 2012-12-19
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0You are welcome! – 2012-12-19
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0@richard @ robjohn Thank you, but how it becomes if $f$ is differentiable and the $L^\infty$ norm of $f(x)$ and $f(cx)$ nonzero? – 2012-12-19
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0@Kamon: you can have $f\in C^\infty$ and have $\|f\|_{L^\infty}\ge1$ as long as you don't have an upper limit on $\|f\|_{L^\infty}$ – 2012-12-19
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0@Kamon: e.g. $$ f(x)=\left\{\begin{array}{} m&\text{if }x\le x_m\\ m+\frac{M-m}{1+\exp\left(\frac1{x-x_m}+\frac1{x-x_M}\right)}&\text{if }x_m\lt x\lt x_M\\ M&\text{if }x\ge x_M \end{array}\right. $$ – 2012-12-19