# Submultiplicative function that converges to $0$ has exponential decay

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.

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.

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

• In fact I'm not that sure for the decreasing part I wrote it quickly. Maybe you have to adapt it a bit considering $f(x) < \varepsilon$ and not just $<1$. May 22, 2020 at 13:28