Submultiplicative function that converges to $0$ has exponential decay

Question

Let $f : [0,\infty) \to [0,\infty)$ be submultiplicative, i.e. $f(s+t) \leq f(s)f(t)$ for all $t,s\geq 0.$ Further assume $\lim_{t \to \infty} f(t)=0.$ Then show that $f$ decay exponentially.

My attempt: We want to show there exists $r>0,M \geq 1$ such that $f(t)\leq M e^{-rt}$ for all $t \geq 0.$ If not, then for each $n \in \mathbb{N},$ there exists $t_n\geq 0$ such that $$f(t_n)>e^{-nt_n}.$$ Submultiplicativity of $f$ implies $$\left[f\left(\frac{t_n}{n}\right)\right]^n\geq f(t_n)>e^{-nt_n}$$ and hence $$f\left(\frac{t_n}{n}\right)>e^{-t_n}.$$

I'm not sure how to proceed from here.

Answer 1

Fix $\varepsilon\in (0,1)$. Then there is $t_0>0$ be such that $f(t)\leq \varepsilon$ for all $t\geq t_0$. Let $\lambda>0$ small enough so that $e^{-\lambda 2t_0}\geq \varepsilon.$ Then for all $t\in[t_0,2t_0]$ we have $$f(t)\leq \varepsilon \leq e^{-\lambda 2t_0}\leq e^{-\lambda t}.$$ Now for all $t\in [t_0,2t_0]$ and for all $k=1,2,\dots$ we have $$ f(kt)\leq f(t)^k\leq e^{-\lambda kt}$$ Since any element $t\geq t_0$ is in the form $t=kt'$ for $t'\in [t_0,2t_0]$ we deduce that $f(t)\leq e^{-\lambda t}$ for all $t\geq t_0$.

Using the fact that $f$ is bounded on $[0,t_0]$, you should now be able to complete the proof.

Answer 2

First, if there exists $y$ for which $f(y) = 0$, then for all $x > y$, $0 \leqslant f(x) = f((x-y)+y) \leqslant f(x-y)f(y) = 0$ and $f(x) = 0$. The result is then trivial.

Now, consider $f(x) >0$ for all $x$. Take $g = \ln(f)$ which is well defined by assumption. Then, for all $x,y$, $g(x+y) \leqslant g(x)+g(y)$ and $g$ is subadditive. You can show by induction that $\forall n\in \mathbb{N}$, $g(n) \leqslant ng(0)$. You can also show that for all rational $q$, this relation is true : $g(q) \leqslant qg(0)$.

As the limit of $f$ is $0$, there exists $A$ such that if $x > A$, then $f(x) < 1$, thus $g(x) < 0$. Then on $[A,\infty)$, $g$ is decreasing and if $x$ is irrationnal, there exists $q$ rational such that $x < q < x+1$. With all that, you can show that for all $x > A$, $g(x) \leqslant x g(0) + C$ for a constant $C$ that only depends on $A$.

Consequently, on $[A,\infty)$, $f(x) \leqslant e^C \cdot e^{g(0)x}$.