I am reading the book Laplace Transform by Widder (https://archive.org/details/in.ernet.dli.2015.201038). In chapter V (page 181), there is a theorem that states as follows:
If \begin{equation} f(s) = \int_{0}^{\infty} e^{-st}\mathrm{d}{\alpha(t)}\qquad (s>0), \end{equation}
where $\alpha(t)$ is defined as $\displaystyle\alpha(t) = \int_0^t a(x)\rm{d}x$, then for any $\gamma \geq 0$ and for any constant $A$:
\begin{equation} \limsup_{s \to 0^+} \lvert s^{\gamma}f(s) - A \rvert \leq \limsup_{t \to \infty} \lvert \alpha(t)t^{-\gamma} \Gamma(\gamma + 1) - A \rvert, \end{equation}
\begin{equation} \limsup_{s \to \infty} \lvert s^{\gamma}f(s) - A \rvert \leq \limsup_{t \to 0^{+}} \lvert \alpha(t)t^{-\gamma} \Gamma(\gamma + 1) - A \rvert. \end{equation}
I have a question on the equality condition of this theorem. When $\gamma = 0$ and $A = 0$, we obtain the Abel theorem for integral $\displaystyle \limsup_{s \to 0^+} f(s) = \limsup_{t \to \infty} \alpha(t)$. It is very tempting that if $\gamma = 1$ and $A = 0$ we can also obtain an equality $\displaystyle \limsup_{s \to 0^+} sf(s) = \limsup_{t \to \infty} t^{-1}\alpha(t)$ (this is just my guess). May I ask if this guess is correct and is there a way to find the condition for the equality in this theorem?
To simplify things I will assume that $\alpha:[0,\infty)\rightarrow\infty$ is a right continuous function of local finite variation and that $\alpha(0)=0$. This allows for the use of Lebesgue integration and extends the results for a larger class of locally integrable functions $f$ on $[0,\infty)$.
Comment: In Widder's book referenced by the OP, the author works with the Stieltjes integral and the assumption is that $\alpha$ is a normalized function of bounded variation. The funciton $f$ is then implicitly assume to be such that $f$ is Stieltjes-integrable w.r.t $\alpha$ in any interval $[a,b]\subset[0,\infty)$.
Suppose $f\in L^1_{loc}(\mu_\alpha)$ (here $\mu_\alpha$ is the Lebesgue-Stieltjes measure induced by $\alpha$) and that $$f(s)=\int_{(0,\infty)}e^{-st}\alpha(dt:=\lim_{R\rightarrow\infty}\int_{(0,R]}f(t)e^{-st}\,dt<\infty$$ for all $s>0$.
This means that
Having transform the representation of $f$ as an improper Legesgue-Stieltjes integral into an improper Lebesgue integral we have that for any $\gamma>0$ \begin{align} s^\gamma f(s) &=s^{\gamma+1} \int_{(0,T]}e^{-st}\alpha(t)\,dt +s^{\gamma+1}\int_{(T,\infty)}e^{-st}\alpha(t)\,dt\tag{0}\label{zero} \end{align}
For the first integral in the right-hand-side of \eqref{zero} we have \begin{align} s^{\gamma+1} \int_{(0,T]}e^{-st}\alpha(t)\,dt \leq \Big(\sup_{0<t\leq T}\frac{\alpha(t)}{t^{\gamma}}\Big)\int_{(0,sT]}e^{-u}u^{\gamma}\,du \end{align} Hence \begin{align}\lim_{s\rightarrow\infty}s^{\gamma+1} \int_{(0,T]}e^{-st}\alpha(t)\,dt\leq \Big(\sup_{0<t\leq T}\frac{\alpha(t)}{t^{\gamma}}\Big)\Gamma(\gamma+1)\tag{1}\label{one} \end{align} For the second integral on the right-hand-side of \eqref{zero}, fix $\varepsilon>0$. By [1], $\alpha(t)=o(e^{\varepsilon t})$. As $\alpha$ is of local bounded variation, there is $M_\varepsilon>0$ such that $e^{-\varepsilon t}|\alpha(t)|\leq M_\varepsilon$. Consequently, \begin{align} s^{\gamma+1}\int_{(T,\infty)}e^{-st}|\alpha(t)|\,dt\leq M_\varepsilon s^{\gamma+1}\frac{e^{-(s-\varepsilon)T}}{s-\varepsilon}\xrightarrow{s\rightarrow\infty}0\tag{2}\label{two} \end{align} Putting \eqref{zero}, \eqref{one} and\eqref{two} together we obtain $$\limsup_{s\rightarrow\infty}s^{\gamma}f(s)\leq\Big(\sup_{0<t\leq T}\frac{\alpha(t)}{t^{\gamma}}\Big)\Gamma(\gamma+1)$$ Letting $T\rightarrow0$ yields \begin{align}\limsup_{s\rightarrow\infty}s^{\gamma}f(s)\leq\limsup_{t\rightarrow+0}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{3}\label{three} \end{align}
A similar argument shows that $$\liminf_{s\rightarrow\infty}s^{\gamma}f(s)\geq\Big(\inf_{0<t\leq T}\frac{\alpha(t)}{t^{\gamma}}\Big)\Gamma(\gamma+1)$$ Letting $T\rightarrow0$ yields \begin{align} \liminf_{s\rightarrow\infty}s^{\gamma}f(s)\geq\liminf_{t\rightarrow+0}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{4}\label{four} \end{align}
Collecting this limits, we have that \begin{align} \liminf_{t\rightarrow+0}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1) \leq \liminf_{s\rightarrow\infty}s^{\gamma}f(s)\leq \limsup_{s\rightarrow\infty}s^{\gamma}f(s)\leq \limsup_{t\rightarrow+0}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{5}\label{five} \end{align}
We can also treat the first integral in the right-hand-side of \eqref{zero} as \begin{align} s^{\gamma+1} \int_{(0,T]}e^{-st}|\alpha(t)|\,dt &\leq s^{\gamma+1}M\int_{(0,T]}e^{t(\varepsilon-s)}\,dt\\ &=s^{\gamma+1}M\frac{e^{(\varepsilon-s)T} -1}{\varepsilon-s}\xrightarrow{s\rightarrow0+}0\tag{6}\label{six} \end{align} As for the second integral on the right-hand-side of \eqref{zero} \begin{align} s^{\gamma+1}\int_{(T,\infty)}e^{-st}\alpha(t)\,dt&\leq s^{\gamma+1}\Big(\sup_{t>T}\frac{\alpha(t)}{t^\gamma}\Big)\int_{(T,\infty)}e^{-st}t^{\gamma}\,dt\\ &=\Big(\sup_{t>T}\frac{\alpha(t)}{t^\gamma}\Big)\int_{(sT,\infty)}e^{-st}t^{\gamma}\,dt\xrightarrow{s\rightarrow0+}\sup_{t>T}\frac{\alpha(t)}{t^\gamma}\Gamma(\gamma+1)\tag{7}\label{seven} \end{align} Putting \eqref{zero}, \eqref{six} and \eqref{seven} together gives $$ \limsup_{s\rightarrow0+}s^{\gamma}f(s)\leq\Big(\sup_{t>T}\frac{\alpha(t)}{t^{\gamma}}\Big)\Gamma(\gamma+1) $$ Letting $T\rightarrow\infty$ yields \begin{align} \limsup_{s\rightarrow0+}s^{\gamma}f(s)\leq\limsup_{t\rightarrow\infty}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{8}\label{eight} \end{align} A similar argument shows that \begin{align} \liminf_{s\rightarrow0+}s^{\gamma}f(s)\geq\liminf_{t\rightarrow\infty}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{9}\label{nine} \end{align}
Collecting all the limits we obtain
\begin{align} \liminf_{t\rightarrow\infty}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\leq \liminf_{s\rightarrow0+}s^{\gamma}f(s)\leq \limsup_{s\rightarrow0+}s^{\gamma}f(s)\leq \limsup_{t\rightarrow\infty}\frac{\alpha(t)}{t^{\gamma}}\Gamma(\gamma+1)\tag{10}\label{ten} \end{align}