calculus - If $f:[0,infty)to [0,infty)$ and $f(x+y)=f(x)+f(y)$ then prove that $f(x)=ax$

Let $\,f:[0,\infty)\to [0,\infty)$ be a function such that $\,f(x+y)=f(x)+f(y),\,$ for all $\,x,y\ge 0$. Prove that $\,f(x)=ax,\,$ for some constant $a$.

My proof :

We have , $\,f(0)=0$. Then ,

$$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(h)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=f'(0)=a\text{(constant)}.$$

Then, $\,f(x)=ax+b$. As, $\,f(0)=0$ so $b=0$ and $f(x)=ax.$

Is my proof correct?

Answer

In your proof you assume that $f$ is differentiable, which is not given.

Let me suggest how to obtain the formula of $f$:

Step I. Show that $\,f(px)=p\,f(x),\,$ when $p$ is a positive rational and $x$ a non-negative real. (At first show this for $p$ integer.) We obtain also that, $\,f(0)=0$.

Step II. Observe that $f$ is increasing, since, for $y>x$, we have
$$
f(y)=f(x)+f(y-x)\ge f(x).
$$

Step III.
Since $f$ is increasing, then the limit $\,\lim_{x\to 0^+}f(x)\,$ exists. However
$$

\lim_{x\to 0^+}f(x)=\lim_{n\to\infty}f\Big(\frac{1}{n}\Big)
=\lim_{n\to\infty}\frac{1}{n}\,f(1)=0.
$$

Step IV. Pick an arbitrary $x\in(0,\infty)$, and a decreasing sequence
$\{q_n\}\subset\mathbb Q$ tending to $x$. Then
$$
f(q_n)=q_n\,f(1)
$$
and

$$
x\,f(1)\longleftarrow q_n\,f(1)=f(q_n)=f(x)+f(q_n-x)\longrightarrow f(x),
$$
since $\,\,q_n-x\to 0^+$, and thus $\,\,\lim_{n\to\infty}f(q_n-x)=0$.

Therefore, $\,f(x)=x\,f(1),\,$ for all $x\in\mathbb [0,\infty)$, and hence $\,f'(x)=f(1)$.