We define the Laplace transform of a non-negative function $f : \mathbb{R_+} \to \mathbb{R_+}$ by $$\mathcal{L}f(q) \triangleq \int_0^{+\infty}f(t)e^{-qt}dt,$$ where $q$ is in the domain of convergence of the integral.

Suppose that $\mathcal{L}f(q) \leq C(q)$ for some $C(q) \in \mathbb{R}_+$ that can depend on $q$. Question : is it possible to find an upper bound on $f(t)$ for any $t > 0$ that involves $C(q)$ ?

One possible answer is to assume that $f$ in non-decreasing and then we obtain, for each $t > 0$ and $q$ in the domain, \begin{equation} \begin{split} q\mathcal{L}f(q) & = \int_0^t q f(s)e^{-qs}ds + \int_t^{+\infty} q f(s)e^{-qs}ds \\ & \geq f(t) \int_t^{+\infty} q e^{-qs}ds = f(t)e^{-qt}. \end{split} \end{equation}

Therefore, $f(t) \leq e^{qt}q\mathcal{L}f(q) \leq e^{qt}qC(q)$.

Now, in my case I know that $f(t) \leq 1$ for all $t > 0$ (which also implies $q\mathcal{L}f(q) \leq 1$, for all $q$, so we can assume that $C(q) < 1$). So, I would like to find an inequality like the one above but where $e^{qt}$ is replaced by something which is smaller or equal to $1$. Does anybody have any ideas ? Thanks a lot !