Friday, March 10, 2017

functional equations - If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t



Let $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$.




I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as $g(xy) = g(x) + g(y)$


I looked up the hint and it says let $g(x) = \log f(a^x) $


The wikipedia page for functional equations only states the form of the solutions without proof.


Attempt Using the hint (which was like pulling a rabbit out of the hat)


Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$ so that we can define the real function $g(x) = \log f(a^x)$ and we have $$g(x+y) = g(x)+ g(y)$$


i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).


Letting $\log_a f(a) = p$ we get $f(a^x) =a^p $. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited


Answer



So, we assume $f$ is continuous. Letting $g(x) = \ln(f(a^x))$, we get $$ \begin{align*} g(x+y) &= \ln(f(a^{x+y})) = \ln(f(a^xa^y)) = \ln(f(a^x)f(a^y))\\ &= \log(f(a^x)) + \ln(f(a^y))\\ &= g(x)+g(y). \end{align*}$$ So $g$ satisfies the Cauchy functional equation; if you assume $f$ is continuous, then so is $g$, hence $g(x) = xg(1)$ for all $x\gt 0$.


Since $g(1) = \ln(f(a))$, we have $$f(a^x) = e^{g(x)} = e^{g(1)x} = (e^{x})^{g(1)}.$$ Given $r\in \mathbb{R}$, $r\gt 0$, we have $r = a^{\log_a(r)}$, hence $$\begin{align*} f(r) &= f\left(a^{\log_a(r)}\right)\\ &= \left(e^{\log_a(r)}\right)^{g(1)}\\ &= \left(e^{\ln(r)/\ln(a)}\right)^{g(1)}\\ &= \left(e^{\ln(r)}\right)^{g(1)/\ln(a)}\\ &= r^{g(1)/\ln(a)}, \end{align*}$$ where we have used the change-of-base formula for the logarithm, $$\log_a(r) = \frac{\ln r}{\ln a}.$$ Finally, since $g(1) = \ln(f(a))$, we have $$f(r) = r^{\ln(f(a))/\ln(a)}.$$ As this works for any positive $a$, $a\neq 1$, taking $a=e$ we get $$f(r) = r^{\ln(f(e))}.$$



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...