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What values can $k$ take such that there exists $c<1$ such that for all continuous functions, $F$, defined on the interval $[0,r], r\in \mathbb R^+$ $$\sup\limits_{x\in[0,r]}\left\{e^{kx}\int\limits_0^x|F(s)|\,\,\,ds\right\}\leq {c\over M}\sup\limits_{s\in[0,r]}\left\{e^{ks}|F(s)|\right\}$$

where $M>0$ is fixed?

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    The plural is "suprema"; see e.g. http://english.stackexchange.com/questions/39771/plurals-of-infimum-and-supremum.2011-12-12
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    @joriki: thanks for pointing that out.2011-12-12
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    $m\gt0$, $c\lt1$ doesn't restrict the value of $c/m$ in any way.2011-12-12
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    @joriki: indeed. I have edited the question to remove the redundant info. Thanks.2011-12-12
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    @joriki: just remembered why i added the conditions. M is fixed.2011-12-12
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    $f(s)-g(s)$ can be any continuous function, so you might as well just use one function. Also you need $r\in\mathbb R^+$.2011-12-12
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    So you mean "such that there exists $c\lt1$ such that"?2011-12-12
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    @joriki: thanks! :) I don't quite understand the comment "such that..."?2011-12-12
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    What I mean is, the role of $c$ is unclear. There's no quantifier for it; it just occurs, and then you say "we want". The only way I can make sense of that is if you intended $c$ to be implicitly existentially quantified; my suggested formulation was supposed to express that. It would replace the single "such that" in the first line.2011-12-12
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    @joriki: ah, thanks again. Edited.2011-12-12

1 Answers 1

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Since $F$ can be any continuous function, $G(s)=\mathrm e^{ks}F(s)$ can also be any continuous function. So we want

$$\sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\int\limits_0^x\mathrm e^{-ks}|G(s)|\,\,\,\mathrm ds\right\}\leq {c\over M}\sup\limits_{s\in[0,r]}\left\{|G(s)|\right\}$$

for all continuous functions $G$. Given the supremum on the right-hand side, the left-hand side is maximal for constant $G$, so we can assume constant $G$ without loss of generality. The constant cancels, and thus we want

$$ \begin{eqnarray} \sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\int\limits_0^x\mathrm e^{-ks}\,\,\,\mathrm ds\right\} &=& \sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\frac{1-\mathrm e^{-kx}}{k}\right\} \\ &=& \sup\limits_{x\in[0,r]}\left\{\frac{\mathrm e^{kx}-1}{k}\right\} \\ &\leq& {c\over M}\;. \end{eqnarray} $$

(We can treat the case $k=0$ as a limiting case in which the argument of the supremum becomes simply $x$.) The argument of the supremum monotonically increases with $x$ independent of $k$, so the supremum is attained at $x=r$, and we get

$$\frac{\mathrm e^{kr}-1}{k}\leq {c\over M}\;.$$

The left-hand side monotonically increases with $k$, so you can find $c\lt1$ such that the inequality holds if $k\le k_{\text{max}}$, with $k_{\text{max}}$ determined by the transcendental equation

$$\frac{\mathrm e^{k_{\text{max}}r}-1}{k_{\text{max}}}= {1\over M}\;.$$