# $\sup\limits_{t>0}[\frac{g(t)}{\sup\limits_{t<u<\infty}g(u)}]\leq 1$

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?

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

## 1 Answer

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.

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