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I'm trying to prove that if $g:\mathbb R^+\to \mathbb R$, defined by $g(uv)=g(u)+g(v)$, for all $u, v \in \mathbb R^+$, $g$ continuous, then $g(x)=b\cdot \ln x$

In order to prove this, I'm trying to prove that if there is a function $f:\mathbb R \to \mathbb R$ such that $f(xy)=f(x)f(y)$, then $f(x)=x^b, b\in \mathbb R$.

If I prove the last statement, I prove the question. First, I was trying this by induction on $\mathbb N$, what I found very difficult to prove.

I need help. Thanks!

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  • $\begingroup$ @ThomasAndrews thank you very much! $\endgroup$ – user42912 Feb 12 '13 at 19:37
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    $\begingroup$ It's actually easier to prove that if $h:\mathbb R\to\mathbb R$ is continuous and $h(x+y)=h(x)+h(y)$ then there is an $a\in\mathbb R$ such that $h(x)=ax$ for all $x\in\mathbb R$. Then let $h(x)=g(e^x)$. $\endgroup$ – Thomas Andrews Feb 12 '13 at 19:40
  • $\begingroup$ The problem with your approach is that you have no real way to deal with the case $b$ is not an integer. Then $x^b$ is not necessarily defined for $x<0$, for example, if $b=1/2$ or $b$ is irrational. But $b<0$ is possible for your original function $g$ $\endgroup$ – Thomas Andrews Feb 12 '13 at 19:46
  • $\begingroup$ @ThomasAndrews I got it! thank you again. $\endgroup$ – user42912 Feb 12 '13 at 19:48
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Your approach has problems because, when, say, $b=1/2$, $x^b$ is not defined on all of $\mathbb R$.

The safer approach is to prove that if $h:\mathbb R\to\mathbb R$ is continuous, and for all $x,y\in\mathbb R$, $h(x)+h(y)=h(x+y)$, then there is a $b$ such that $h(x)=bx$ for all $x$.

Setting $h(x)=g(e^x)$ above, from this result it would follow since $$h(x+y)=g(e^{x}e^{y})=g(e^x)+g(e^u)=h(x)+h(y)$$

So for $z>0$, $g(z)=h(\log z)=b\log z$.

To prove this statement about $h$, we proceed by first proving it for natural numbers $x$, then for positive rational numbers $x$, and then for all rationals, and then finally for all real $x$.

We first show $h(0)=0$. Then we show that $h(-x)=-h(x)$ for all $x$. Both of these follow relatively directly from the condition on $h$.

Then we show by induction that if $n$ is a natural number and $x$ any real, then $h(nx)=nh(x)$.

In particular, if $x=1$, this means that $h(n)=nh(1)$. Let $b=h(1)$. So $h(n)=bn$ for all natural numbers $n$.

Then we show that if $x=m/n$ is a positive rational number, then $$bm=h(m)=h(nx)=nh(x)$$ so $h(x)=b\frac{m}{n}=bx$ for $x$ positive rational, as well.

Now, if $f(x)=bx$, then $f(-x)=-f(x)=-bx=b(-x)$. So this means that $f(x)=bx$ for all rationals $x$.

But since the we assumed $h$ was continuous, this means that $h(x)=bx$ for all real number numbers. (This is why you need continuity.)

[I've left out some of the steps - the induction argument, the $h(0)=0$ and $h(-x)=-h(x)$ steps.]

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  • $\begingroup$ Great solution, very complete. Thank you very much!!! $\endgroup$ – user42912 Feb 14 '13 at 6:27
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It seems the question is not right, consider f(x)=0. Is there any condition like f should be positive and continuous?

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    $\begingroup$ $0=0\cdot \ln{x}$ - so it seems right for me. $\endgroup$ – zaarcis Feb 12 '13 at 21:36
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If $u=1$ and $v=1$ then it means $g\left( 1\right)=g\left( 1\right)+g\left( 1\right)$. So $g\left( 1\right)=0$.

Also we will need value of $g^\prime\left(1\right)$:
$g\left(u^{n}\right)=ng\left(u\right)\Rightarrow g\left(\sqrt[n]{u}\right)=\frac{g\left(u\right)}{n}$
$$g^{\prime}\left(1\right)=\lim_{\Delta x\to0}\frac{g\left(1+\Delta x\right)}{\Delta x}=\lim_{w\to1}\frac{g\left(w\right)}{w-1}=\lim_{n\to\infty}\frac{g\left(\sqrt[n]{u}\right)}{\sqrt[n]{u}}=\\=\lim_{n\to\infty}\frac{\frac{1}{n}g\left(u\right)}{\sqrt[n]{u}-1}=g\left(u\right)\cdot\lim_{n\to\infty}\frac{\frac{1}{n}}{\sqrt[n]{u}-1}=g\left(u\right)\cdot\frac{1}{\ln u}=\frac{g\left(u\right)}{\ln\left(u\right)}$$

Now calculate derivative (using knowledge that if $f\left(x\right)=b\ln\left(x\right)$ then $f^{\prime}\left(x\right)=\frac{b}{x}=f^{\prime}\left(1\right)\frac{1}{x}$):

$$g^{\prime}\left(x\right)=lim_{\Delta x\to0}\frac{g\left(x+\Delta x\right)-g\left(x\right)}{\Delta x}=$$ $$=\lim_{\Delta x\to0}\frac{g\left(x\left(1+\frac{\Delta x}{x}\right)\right)-g\left(x\right)}{\Delta x}=\lim_{\Delta x\to0}\frac{g\left(1+\frac{\Delta x}{x}\right)}{\Delta x}=$$ $$=\lim_{\Delta x\to0}\frac{g\left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}x}=\lim_{\Delta x\to0}\frac{g\left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}}\cdot\frac{1}{x}=$$ $$=\lim_{\frac{\Delta x}{x}\to0}\frac{g\left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}}\cdot\frac{1}{x}=\lim_{\frac{\Delta x}{x}\to0}\frac{g\left(1+\frac{\Delta x}{x}\right)-g\left(1\right)}{\frac{\Delta x}{x}}\cdot\frac{1}{x}=$$ $$=g^{\prime}\left(1\right)\cdot\frac{1}{x}=\frac{g\left(u\right)}{\ln\left(u\right)}x$$

So value of $g\left(1\right)$ and $g^\prime\left(x\right)$ is the same as for $f\left(x\right)=b\ln\left(x\right)$. It's enough.

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    $\begingroup$ $g$ is not necessarly differentiable. $\endgroup$ – user42912 Feb 12 '13 at 20:17
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    $\begingroup$ You're right. Will try to fix after few minutes (I'm busy at this moment). $\endgroup$ – zaarcis Feb 12 '13 at 20:20
  • $\begingroup$ @user42912 I think that I fixed it. $\endgroup$ – zaarcis Feb 12 '13 at 22:09

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