I can't prove that the following inequality true or not: $$ \sup\limits_{t>0}[\frac{g(t)}{\sup\limits_{t<u<\infty}g(u)}]\leq 1 \tag{*}, $$ where $g$ is a positive function. I think it is true if we replace supremum with essential supremum. Because $\text{ess sup}_{t<u<\infty}g(u)=\text{ess sup}_{t\leq u<\infty}g(u)$ so we have $g(u)\leq \text{ess sup}_{t< u<\infty}g(u)$ for all $u\geq t$ and in parcticular for $u=t$. Hence $$ \sup\limits_{t>0}[\frac{g(t)}{\text{ess sup}_{t< u<\infty}g(u)}]\leq 1. \tag{**} $$ What dou think the (*) inequality is true or not and the proof of the (**) inequality is true or not?

  • 2
    $\begingroup$ If $g$ is continuous, then $(\ast)$ is true. If $g$ has a discontinuity, it need not hold. $\endgroup$ – Daniel Fischer Apr 25 '16 at 11:46

Let $g : (0,\infty) \to \mathbb R$ be the function that takes the value $2^{-n}$ on the interval $(n-1,n]$. $g$ is bounded above by $1$, and $$\frac{g(n)}{\sup_{n < u < \infty} g(u)} = 2$$ for all $n$. You can adjust the size of the jumps to make the supremum infinite if you like.

  • 1
    $\begingroup$ To fleshen out the last sentence: Replace $2^{-n}$ with $\frac1{n!}$ $\endgroup$ – Hagen von Eitzen Apr 25 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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