# Suprema of continuous functions

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. english.stackexchange.com/questions/39771/…;. – joriki Dec 12 '11 at 12:39
@joriki: thanks for pointing that out. – Mulkey Dec 12 '11 at 13:39
$m\gt0$, $c\lt1$ doesn't restrict the value of $c/m$ in any way. – joriki Dec 12 '11 at 14:03
@joriki: indeed. I have edited the question to remove the redundant info. Thanks. – Mulkey Dec 12 '11 at 15:10
@joriki: just remembered why i added the conditions. M is fixed. – Mulkey Dec 12 '11 at 15:18

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

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

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