Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The plural is "suprema"; see e.g.…;. – 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
up vote 1 down vote accepted

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.