If $f$ is continuous, $f(1) >1$ and $f(x+y)=f(x)f(y)$, then $f$ is increasing.

Question

Consider the function $f$ with the following properties:

$$\lim_{x\rightarrow 0} f(x) =1,$$ $$f(x+y)=f(x)\,f(y),$$ $$f(x) >0,\quad \forall x\in\mathbb{R},$$ $$ -\infty<x,y<\infty.$$

Show that if $\,f(1) = p >1,\,$ then $f$ is increasing on $\mathbb{R}$.

How am I supposed to show this without using continuity?

Edit: Asked my professor and I have now figured it out. Here is the proof.

Assume $f$ is continuous in $\mathbb{Q}$. Suppose $f(1) = p >1$. Define $g(x) = f(1)^x$, where $g(x)$ is continuous on $\mathbb{R}$. From an earlier exercise we found that $f(x)=f(1)^x$ if $x\in\mathbb{Q}$. This shows that $f(x)=g(x)$ for all $x\in\mathbb{Q}$. Since $\mathbb{Q}$ is a dense subset of $\mathbb{R}$, we have that $f(x)=g(x)$ for all $x\in\mathbb{R}$. Further, we have that $f$ is continuous on $\mathbb{R}$. Then, take $x_1<x_2$. This gives $f(x_1) = p^{x_1} > 1$ and $f(x_2) = p^{x_2} > 1$. Thus, algebra dictates that $p^{x_1} < p^{x_2}$ since $p >1$. Therefore, given $x_1<x_2$ we have $f(x_1)<f(x_2)$ which proves that $f$ is increasing.

Answer 1

Hints:

1. Prove continuity at $0$ (i.e. $f(0)=1$) by plugging $x=0$ and considering $y\to 0$ in (2).
2. Using this and (2) again, prove continuity at any $x$.
3. Prove $f(1/n)>1$ for all $n\in\Bbb N$.
4. Then use continuity to prove that $f$ is increasing.

Answer 2

Step 1. $f(x)>0$, for all $x\in \mathbb R$. This is obtained combining the fact that $$ f(x)=\left(f\left(\frac{x}{n}\right)\right)^n\quad\text{and}\quad \lim_{x\to 0}f(0)=1. $$

Step 2. Set $g(x)=\log f(x)$. Then $g(x+y)=g(x)+g(y)$, and $\lim_{x\to 0}g(0)=0$. We shall show that $g$ is continuous. Indeed $$ \lvert g(x+h)-g(x)\rvert=\lvert g(h)\rvert\to 0. $$ Thus $g(x)=xg(1)$. This is due to the fact that $g(n)=ng(1)$, for $n$ integer, and $g(1)=ng(1/n)$, and hence $g(m/n)=mg(1)/n$, and for every $x$ real, let $q_n\to x$, with $q_n$ rational and obtain, due to continuity of $g$, that $$ g(x)=\lim_{n\to\infty}g(q_n)=\lim_{n\to\infty}q_ng(1)=xg(1). $$

and hence $f(x)=\mathrm{e}^{cx}$, where $c=g(1)$, and as $f(1)>1$, then $c>0$, and hence $f$ is increasing.

Answer 3

From $$\lim_{x\to x_0}f(x)\stackrel{(2)}=f(x_0)\lim_{x\to x_0}f(x-x_0)\stackrel{(1)}=f(x_0)$$ we see that $f$ is continuous. Assume $f(x_0)=0$ for some $x_0$. Then $f(x)=f(x-x_0)f(x_0)=0$ for all $x$, contradicting $f(1)>1$. Then $f(x)>0$ for all $x$ by IVT. This allows us to conclude $f(x-y)=\frac{f(x)}{f(y)}$ from $(2)$, so that $f$ is a group homomorphism $\mathbb R\to (0,\infty)$. The kernel is a subgroup of $\mathbb R$, hence is $\{0\}$ or $a\mathbb Z$ for some $a>0$ or dense. In the latter case, $f$ is constant by continuity, contradictimg $f(1)\ne1$. In the second case, $f(\frac12a)^2=1$ and $f(\frac12a)>0$ gives a contradiction. We conclude that $f(x)\ne 1$ for $x\ne0$. Hence $f$ is injective and by continuity strictly monotonic. From $f(0)=1<f(1)$, we see that $f$ must be strictly increasing.