QUESTION: Let $A$ be the set of all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(xy)=xf(y)$ for all $x,y \in \mathbb{R}$. $$$$
(a) If $f \in A$ then show that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ $$$$
(b) For $g,h \in A$, define a function $g\circ h$ by $(g \circ h)(x)=g(h(x))$ for $x \in \mathbb{R}$. Prove that $g \circ h$ is in $A$ and is equal to $h \circ g$.
MY ANSWER: Here's what I have done..
Putting $y=1$ we get $f(x)=f(1)x$.
Say $f(1)=k$ for some $k \in \Bbb{R}$
$\therefore f(x)=kx \text{ }\forall x \in \Bbb{R}$
This is the solution to Cauchy's Functional Equation which takes the form $f(x+y)=f(x)+f(y)$. But the problem is the domain of $x$ and $y$ in Cauchy's Equation is $\Bbb{Q}$ while we have $\Bbb{R}$. Therefore we prove $(a)$ as follows –
Say $y=bx$ for some $b \in \Bbb{R}$
$$ax(1+b)=ax+abx$$
$$\implies ax(1+b)=ax+ay$$
This is similar to saying –
$$f(x(1+b))=f(x)+f(y)$$
$$\implies f(x+bx)=f(x)+f(y)$$
$$\implies f(x+y)=f(x)+f(y)$$
Q.E.D. $\square$
Now, I have two questions..
Firstly, is the above proof alright ?
Secondly, how do I solve point $(b)$ ?
$goh(x)$ is simply another way of saying $g(h(x))$, and everyone knows that $goh(x)=g(h(x))$. Then what is the information given based on which we may proceed ?
Thank You for your help in advance..
Best Answer
From your work we know $f(x)=f(1)x=kx$. To show $a)$, note that
$$f(x+y)=k(x+y)=kx+ky=f(x)+f(y)$$
To show $b)$, note that
$$g(h(x))=g(k_hx)=k_gk_hx=(k_g k_h)x$$
Then
$$g(h(xy))=(k_gk_h)xy=x(k_gk_h)y=xg(h(y))$$
Also
$$h(g(x))=h(k_gx)=k_hk_gx=(k_hk_g)x=(k_gk_h)x=g(h(x))$$
We conclude $g(h(x))=h(g(x))\in A$.