I have a function $f(t)$ decreasing and a function $g(t)$ nonnegative.
I know for grant that $\int_0^t f(t-\theta)g(\theta)d\theta$ is increasing (I mean, this is part of hypothesys).
Now, I have $h(t)=f(t)k(t)$ where both $f(t) $ and $k(t)$ are decreasing, but $k(t)>1$.
I could say that is $\int_0^t h(t-\theta)g(\theta)d\theta$ increasing?
I know that $\int_0^t h(t-\theta)g(\theta)d\theta>\int_0^t f(t-\theta)g(\theta)d\theta$, but how can I prove that there is not oscilation?
Thank you so much.
It's not true. Here is a counterexample. Let $f(t)=g(t)=e^{-t}$, and $k(t)=1+e^{-t}$. Then $f$ is decreasing, $g$ is non-negative, $k$ is decreasing, and $k>1$. Then,
$$\int_0^t f(t-\theta)g(\theta)d\theta =\int_0^t e^{-t+\theta}e^{-\theta}d\theta = te^{-t} $$
and
$$\int_0^t h(t-\theta)g(\theta)d\theta =\int_0^t (1+e^{-t+\theta})e^{-t+\theta}e^{-\theta}d\theta $$ $$ = \int_0^t (e^{-t}+e^{-2t+\theta}) d\theta $$ $$ = te^{-t} + e^{-2t} \int_0^t e^{\theta} d\theta $$ $$ = te^{-t} + e^{-2t} (e^t-1) $$ $$ = (t+1)e^{-t}-e^{-2t} $$ The derivative wrt $t$ is $$ \frac{d}{dt} \int_0^t h(t-\theta)g(\theta)d\theta = 2e^{-2t}-te^{-t} $$ Then $$ 2e^{-2t}-te^{-t}=0 \iff te^{t}=2 $$ We have that $\int_0^t h(t-\theta)g(\theta)d\theta$ is increasing for $0<t<W(2)$ and decreasing for $t>W(2)$, where $W$ is the Lambert W-function and $W(2)\approx 0.853...$