Lebesgue measure and characterisation of function $\Phi$ [Rudin-Real&Complex]

Let $m$ be Lebesgue measure on $[0,1]$ and define $||f||_p$ with respect to $m$ as usual. What are all functions $\Phi$ on $[0, \infty)$ such that the relation

$$ \Phi( \lim_{p\to\ 0}||f||_p)= \int_{0}^{1}(\Phi\circ f)dm$$ holds for every bounded, measurable, positive $f$ ?

A hint is provided:

Show first that $$c \Phi(x)+(1-c) \Phi(1)= \Phi(x^c) \ \ (x>0, 0 \le c \le 1) \ \tag 1 $$

I was thinking of using this $$ \lim_{p\to\ 0}||f||_p = \exp\left \{ \int_0^1 \log |f| dm\right \}$$ From here we get that $\log $ satisfies the condition.

Then I tried to show that any $\Phi$ which meets the conditions of $(1)$ must satisfy the relation, $\Phi(xy)= \Phi(x)+ \Phi(y)$ from which I want to conclude that $log$ is the only such function.

Is my line of thinking correct? If not, then how should I proceed?

Solution 1:

First, it's not clear from your post and the comments whether you've seen why $\Phi$ must satisfy (1). Let $f=x\chi_{(0,c)}+\chi_{(c,1)}$. Then $$\int_0^1\Phi(f)=c\Phi(x)+(1-c)\Phi(1),$$while $$||f||_p=\left(cx^p+1-c\right)^{1/p}.$$ So $$\log(||f||_p)=\frac1p\log(1+c(x^p-1))\sim\frac cp\left(e^{p\log(x)}-1\right)\sim c\log(x).$$So (1) holds.

Now if you let $\Psi(s)=\Phi(e^s)-\Phi(1)$ then (1) gives $$c\Psi(t)=\Psi(ct)\quad(0\le c\le 1,\,t\in\Bbb R).$$Setting $t=1$ shows that $$\Psi(c)=c\Psi(1)\quad(0\le c\le 1).$$And $t=2$ shows $\Psi(2c)=2\Psi(c)$. Hence $$\Psi(c)=c\Psi(1)\quad(c\ge0).$$

Unravelling things now shows that $\Phi(t)$ is a linear combination of $1$ and $\log(t)$.

Conversely, any linear combination of $1$ and $\log$ works. This is well known, or at least better known to me than what's above. Proofs exist in various places, for example On the space $L^0$ and $\lim_{p \to 0} \|f\|_p$ . (Ok, all you actually find there is a proof that $\Phi=\log$ works. But it's obvious that $\Phi=1$ works, and that the set of $\Phi$s that work is closed under linear combinations.)