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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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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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