Given two positive real functions $h$ and $g$ and,
$$g(t) \simeq \int\limits_0^t h(\tau) \cdot g(t-\tau) d\tau$$
i.e., $g$ asymptotically approximately equals the convolution of itself and another function $h$. Could we say $g$ is an exponential function? My rough thoughts:
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apply laplace transform $\mathcal L$ to both sides and solve $s$:
$$\hat g(s) = \hat h(s) \cdot \hat g(s)$$
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evaluate $g(t) / g(t-1)$
$$\frac{g(t)}{g(t-1)} = \frac{\mathcal L^{-1} \{\hat h(s) \cdot \hat g(s) \}}{\mathcal L^{-1} \{\hat h(s) \cdot \mathcal L \{g(t-1)\} \}}
= \frac{\mathcal L^{-1} \{\hat h(s) \cdot \hat g(s) \}}{\mathcal L^{-1} \{\hat h(s) \cdot e^{-s} \cdot \hat g(s)\}} $$since this only works for the $s$ letting $\hat h(s) = 1$, can we extract $e^{-s}$ out and claim $g(t) / g(t-1) = e^{s}$?
Best Answer
D.B's answer and the one suggested by the OP converts the asymptotic condition $g\sim g\star h$ to an equality $g=g\star h$.
Here an example more in the spirit of the originally stated question: $h=\chi_{[0,1]}$ and $g(x)=x^2$. Then for $x>1$ we have $$ (g\star h)(x) = \int_0^x h(u)g(x-u)du = \int_0^1 (x-u)^2du = x^2-x + 1/3$$ for $x>1$, which is $\sim g$ as $x\to\infty$. But this $g$ is not an exponential function.