If the convolution of the exponential function and a function tends to 0 when the time tends to infinity, can we get the following conclusion?

Question

If we know $\lim_{t\rightarrow \infty}\int_{0}^{t}e^{A(t-\tau)}\phi(\tau)\mathrm{d}\tau=0,\lim_{t\rightarrow \infty}e^{At}=0$, can we get: $$\lim_{t\rightarrow \infty}\phi(t)= 0$$ ?

Answer 1

No, it seems to be incorrect.

If you let $\phi(\tau) = \cos(\tau^2)$, then $\lim_{\tau\rightarrow \infty} \phi(\tau) \neq 0$, but the rapid oscillation of $\phi$ makes the convolution go to $0$ anyway. I haven't proven this rigorously, but it shouldn't be too hard. Numerically, with $A = -1$ and $$ F = \int_0^t e^{-(t-\tau)} \cos(\tau^2) d \tau, $$ we get the following values: $$ \begin{align*} F(1) &= \phantom{-}0.55 \\ F(10) &= -0.023 \\ F(100) &= -0.00155 \\ F(1000) &= -0.000175 \\ F(10000) &= \phantom{-}0.000046 \end{align*} $$