Let $f,g,\alpha:[0,\infty)\to\mathbb{R}$ be continuous nonnegative functions. Suppose $f$ is differentiable on $(0,\infty)$ and satisfies a differential inequality
\begin{align}
\frac{d}{dt}f(t)+g(t)\leq c_1\alpha(t) & & (1)
\end{align}
for some constant $0<c_1<\infty$. Suppose further that $\alpha$ satisfies
\begin{align}
\int^{\infty}_t\alpha(s)ds\leq c_2e^{-\lambda t} & & (2)
\end{align}
for some constants $0<c_2,\lambda<\infty$. Is it true that
an integral inequality of the form
\begin{align}
f(t)+\int^{\infty}_tg(s)ds\leq Ce^{-\lambda t}
\end{align}
holds for some constant $0<C<\infty$?
Naively, I tried to first integrate both sides of (1) on a finite integral $[t,T]$ to obtain
\begin{align}
f(T)-f(t)+\int^T_tg(s)ds\leq c_1\int^T_t\alpha(s)ds\leq c_1c_2e^{-\lambda t} & & (3)
\end{align}
and take limit as $T\to\infty$. However, one really needs to justify that we can take limit of $f(T)$, which does not seem very straightforward to me.
I'm thinking of replacing $lim$ by $liminf$, and note that by (2), $\alpha$ seems to have a similar exponential decay, thus by (1), $f'(t)$ also has a similar exponential decaying bound, so that it grows very slow ultimately. Nevertheless, this is very sketchy and I'm not quite sure if it is entirely correct.
Lastly, even if everything goes well very luckily, in the L.H.S. of (3) there is still an extra minus sign in front of $f(t)$, which is different from the desired.
Any comment or answer are greatly welcomed and appreciated.
Best Answer
I think you only have control on $f'$, not on $f$ : Let $f(t)=1$, $g(t)=\alpha(t)=\exp(-\lambda t)$ then (1) and (2) are true but you have $f(t)+\int_t^{+\infty}g(s)ds\ge1$.