Continuous Cauchy functional equation

We shall prove that continuous solutions $ g :R\rightarrow R  $ of equation

$$g(x+y)=g(x)+g(y)$$

are functions g(x)=ax where a=g(1).
We have g(0)=g(0+0)=g(0)+g(0) so

$$(i)g(0)=0.$$
$$(ii)g(-x)=-g(x)$$

as g(0)=g(x+(-x))=g(x)+g(-x)=0
so g(-x)=-g(x)

$$(iii)g(nx)=ng(x)$$

induction
g(2x)=g(x+x)=g(x)+g(x)=2g(x)
g(nx)=g((n-1)x+x)=g((n-1)x)+g(x)=(n-1)g(x)+g(x)=ng(x);

$$(iv)g(\frac{x}{m})=\frac{1}{m}g(x)$$

as $ g(x)=g(m\frac{x}{m})=mg(\frac{x}{m}) $
so $ g(\frac{x}{m})=\frac{1}{m}g(x) $

$$(v)g(\frac{n}{m}x)=\frac{n}{m}g(x)$$

From (v) we have

$$g(\frac{n}{m})=\frac{n}{m}g(1)=\frac{n}{m}a$$

Now let $ x\in R $ there is a sequence $ x_n\in Q $ such as

$$\lim_{n\to\infty}x_n=x$$

As g is continuous

$$g(x)=g(\lim_{n\to\infty}x_n)=\lim_{n\to\infty}g(x_n)=\lim_{n\to\infty}x_ng(1)=\lim_{n\to\infty}x_na=xa$$

Now we want to solve in the class of continuous function $ f:(0,+\infty)\rightarrow R $ equation

$$f(xy)=f(x)+f(y)$$

let

$$x=e^u  \:y=e^v$$

Then

$$f(xy)=f(e^ue^v)=f(e^{u+v})=f(e^u)+f(e^v)$$

Let

$$g(u)=f(e^u)$$

So

$$g(u+v)=g(u)+g(v)$$

We know that g(u)=au;
so $ f(e^u)=au $
Then

$$f(x)=a\ln x$$

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