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?
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asked Jan 18, 2022 at 18:01
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$\begingroup$ if the limits $\lim_{t\rightarrow 0}$ and $\lim_{s\rightarrow\infty}$ both exist, as you assume, then why would the lim sup or lim inf be different? $\endgroup$– Carlo BeenakkerCommented Jan 18, 2022 at 18:44
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$\begingroup$ I mean we don't know the limit exist or not, we only know upper limit or lower limit exists. Can we get the same results.... $\endgroup$– Fractional analysicsCommented Jan 18, 2022 at 18:52
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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 $|\limsup s F(s)|, |\liminf s F(s)| < 1$.
answered Jan 18, 2022 at 18:53
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$\begingroup$ The $e^{-u}$ is a bit of a red herring. The problem is easier to think about if instead you think about $F(s) = \int_0^{1/s} f(u) ~du$, then $s F(s)$ is the mean of the function $f$ on the interval $[0,1/s]$. The qualitative gist your are looking for is the same. $\endgroup$ Commented Jan 18, 2022 at 18:58