Let define $F(s)=\int\_0^\infty f(u)e^{-su}du$. If $f$ is bounded and $\lim\_{t\to 0}f(t)$ exists. Then we can get $\lim\_{t\to 0}f(t)=\lim\_{s\to\infty}sF(s)$.
Can we use upper limits or lower limits to replace all the above limits? i.e. we only know $\limsup\_{t\to 0}f(t)$ or $\liminf\_{t\to0}f(t)$ exist. Can we get the following equality? $$\limsup\_{t\to 0}f(t)=\limsup\_{s\to\infty}sF(s), $$ $$\liminf\_{t\to 0}f(t)=\liminf\_{s\to\infty}sF(s)? $$
If it is correct, is there someone can give me some reference or if it is incorrect is there someone can give me a counterexample?



Initial and final Theorem for upper and lower limits?
The answer is no.


Let $f(u)$ be defined piecewise. On intervals of the form $(2^{k},2^{k+1}]$, where $k\in \mathbb{Z}$, you set $f(u) = (-1)^k$. Then you have that $\limsup\_{t\to 0} f(t) = +1$ and $\liminf\_{t\to 0} f(t) = -1$.


The integral $s F(s) = \int\_0^\infty f(u/s) e^{-u} ~du$ is:


* Continuous in $s$
* For every $s$ strictly less than 1 in absolute value
* satisfies $s F(s) = 4s F(4s)$.


And hence you must have $|\limsup s F(s)|, |\liminf s F(s)| < 1$.


