Prove that $f(x)=x^\alpha$ for all $x>0$.

Question

Suppose that $f:(0,\infty) \to \Bbb R$ is differentiable and that there exists a constant $\alpha$ belonging to $\Bbb R$ such that $x*f'(x)=\alpha*f(x)$ for all $x>0$ and $f(1)=1$. Prove that $f(x)=x^\alpha$ for all $x>0$.

I tried integrating both sides of the equation, but not sure how the $f(1)=1$ comes into play here. Any hints in the correct direction would be much appreciated.

Just out of curiosity, what is $*$? – goblin GONE Dec 03 '13 at 03:56 — goblin GONE, Dec 03 '13 at 03:56

score 1 · Accepted Answer · answered Dec 03 '13 at 03:27

1

Consider $g(x)=f(x)x^{-\alpha}$. Differentiate. Conclude $g$ is constant over $\{x>0\}$. Find the constant using $f(1)=1$, then find what $f$ is.

answered Dec 03 '13 at 03:27

Pedro

122,002

g'(x)=f(x)[-alphax^(-alpha-1)]+[x^(-alpha)]*f'(x) ??? – user113009 Dec 03 '13 at 03:40
@user113009 Take a common factor of $x^{-\alpha-1}$. Then use $x'f(x)=\alpha f(x)$. – Pedro Dec 03 '13 at 03:41
But xf ' (x)=alphaf(x) – user113009 Dec 03 '13 at 03:44
OK, I see how that gets me that g ' (x) = 0 an thus g is a constant. – user113009 Dec 03 '13 at 03:53
Ah, I think I get the f(1)=1 to find the C. Correct me if I am wrong but g(x)=C implies f(x)/x^alpha = C which implies that f(x)=C*x^alpha. Then using f(1)=1 this implies that C=1, thus concluding that f(x)=x^alpha. – user113009 Dec 03 '13 at 04:00
@user113009 Yep, that is it. – Pedro Dec 03 '13 at 04:02
Thanks for the push in the right direction Pedro! – user113009 Dec 03 '13 at 04:02
@user113009 Any time. =) – Pedro Dec 03 '13 at 04:04

Prove that $f(x)=x^\alpha$ for all $x>0$.

1 Answers1