by Masacroso
Last Updated May 16, 2018 14:20 PM

I have a proof for this exercise that seems correct:

Let $g:(0,\infty)\to\Bbb R$ such that $g(s+t)=g(s)+g(t)$ for all $s,t>0$. Then show that if $g$ is bounded in bounded sets then $g(x)=x g(1)$ for all $x>0$.

**My proof:** first note that $g(n\cdot x)=n\cdot g(x)$ for all $n\in\Bbb N$ and all $x\in(0,\infty)$. By the same reason we find that $n\, g(1/n)=g(1)$ for $n\in\Bbb N$ what imply that $n^{-1} g(1)=g(1/n)$. Then both statements together imply that $g(r)=r\, g(1)$ for $r\in\Bbb Q_{>0}$.

Hence if $(x_k)\in\Bbb Q_{>0}^{\Bbb N}$ converge to some $x>0$ we find that $\lim g(x_k)=g(1)\lim x_k=x\, g(1)$. Also note that $$ g(x+y)=g(x)+g(y)\implies g(x+y)-g(y)=g(x),\quad\forall x,y>0\tag1 $$

Now suppose that $g$ is bounded in bounded sets and that $g(s)= s\cdot g(1)+\epsilon$ for some $\epsilon\neq 0$ and some $s>0$, then from the above analysis we can find that $$ g(r(s-q))=r(s-q) g(1)+r\epsilon,\quad\forall r\in\Bbb Q_{>0},\, q\in(0,s)\cap\Bbb Q\tag2 $$ Then let $(q_k)\in\Bbb Q_{>0}^{\Bbb N}$ such that $0<s-q_k<1/k^2$ for each $k\in\Bbb N$. Then we find that $$ \lim_{k\to\infty} g(k(s-q_k))=\lim_{k\to\infty}\big(k(s-q_k)g(1)+k\epsilon)=\infty\tag3 $$ Now, by the definition of $(q_k)$ we find that $0<k(s-q_k)<1/k$ for each $k\in\Bbb N$, hence $g|_{(0,1]}$ is not bounded, thus $g(s)=sg(1)$ for all $s>0$, as desired.$\Box$

Two questions here:

Can someone confirm the correctness of the above proof?

Regardless if the above proof is correct or not it seems that the argument is slightly complicated so I want to know if there is an easier proof.

Sketch of a proof:

You already know that this equation is true for rational numbers. For an irrational number $s$, suppose that $s_n$ is a decreasing sequence of rational numbers approaching $s$. Then since $f$ is bounded in any bounded neighborhood of $s$, $f(s_n-s)$ is a decreasing sequence approaching $f(0^+)$ which can be verified to be 0. But $f(s_n-s)$ = $f(s_n)-f(s)$ so that $f(s_n)$ approaches $f(s)$.

Appendix: $f$ is monotonic. I prove it for the case when f(1) is positive. The other cases are similar. Consider an irrational number $s$ in the domain of the function. We have $f(s)=f(r)+f(s-r)$ for every rational $r$ less than $s$. There must exist a rational $q$ less than $s$ such that $f(s-q)$ is positive (you can easily prove it, find two appropriate irrationals and use the fact that their sum is a rational), so that $f(s)$ as the sum of two positives is positive as well. Hence in this case $f$ is increasing.

It's not clear to me whether what's below is more or less what the OP said; I find the exposition above a little hard to follow. Whether or not it's actually a simpler proof it certainly *looks* simpler...

The result is actually much simpler than I thought. Choose $c$ so that $$|g(t)|<c\quad(|t|<1).$$

It follows that for $n=1,2\dots$ we have $$|g(t)|<c/n\quad(|t|<1/n),$$because if $|t|<1/n$ then $|g(t)|=|g(nt)|/n$ and $|nt|<1$.

So $g$ is continuous at the origin. Since $g(t+\delta)=g(t)+g(\delta)$ this shows that $g$ is continuous. So $g(t)=tg(1)$, since that holds for all rational $t$.

The proof of David C. Ullrich is correct. Let me make it slightly easier to understand.

Step 1. First observe that it suffices to show that $g$ is continuous at $x=0$, and in particularly that $\lim_{x\to 0}g(x)=0$.

Step 2. Set $M=\sup_{x\in [-1,1]}|g(x)|$. We know from the assumptions in the OP that $M<\infty$.

Step 3. Now let $\varepsilon>0$. Then we need to find a $\delta>0$, such that $$|x|<\delta\qquad\Longrightarrow\qquad |g(x)|<\varepsilon.$$ In fact $\delta=\dfrac{\varepsilon}{2M}$ is a suitable candidate, since $$ |x|<\frac{\varepsilon}{2M}\quad\Longrightarrow\quad \left|\frac{2Mx}{\varepsilon}\right|<1 \quad\Longrightarrow\quad \left|\,g\left(\frac{2Mx}{\varepsilon}\right)\right|\le M \quad\Longrightarrow\quad|g(x)|\le\frac{\varepsilon}{2}<\varepsilon. $$

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