Exercise 3.6 from rudin's real and complex analysis Let $m$ be Lebesgue measure on $[0,1]$, and define $\|f\|_{p}$ with respect to m.Find 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 $f$.Show first that
$$ c\Phi(x)+(1-c)\Phi(1)=\Phi(x^{c})\qquad (x>0,0\leq c\leq 1)$$
My idea:it is easy to check
$$\lim_{p\to 0^{+}}\|f\|_{p}=\exp\left(\int_{[0,1]}\ln|f(x)|dm\right) $$
then we can guess $\Phi(x)=\ln x$.for any $x>0$ use $f(y)=x^{2y}$,then leads
$$ \Phi(x)=\int_{0}^{1}\Phi(x^{2y})dm $$
but I don't know how to going on.
 A: Ok,I find a solution! due to $f$ is bounded on $[0,1]$,then $f\in L^{1}([0,1])$,it is easy to check
$$ \lim_{p\to 0}\|f\|_{p}=\exp\left(\int_{0}^{1}\ln{f}dm \right)$$
then
$$ \Phi\left(\exp\left(\int_{0}^{1}\ln{f}dm \right)\right)=\int_{0}^{1}\Phi\circ f dm $$
holds for every bounded measurable positive $f$.Put
$$ f_{x}(t)=\left\{
              \begin{array}{ll}
                x, & \hbox{$t\in[0,c]$;} \\
                1, & \hbox{$t\in(c,1]$.}
              \end{array}
            \right.$$
where $x>0,0\leq c\leq 1$.then we have
$$ \Phi(\exp(c\ln{x}))=c\Phi(x)+(1-c)\Phi(1) $$
Or
$$ \Phi(x^{c})=c\Phi(x)+(1-c)\Phi(1)\qquad (x>0,0\leq c\leq 1)$$
This relation holds actually for all $x>0,c\geq 0$. In fact,if $c>1,x>0$,let $y=x^c$,then $$\Phi(y^{\frac{1}{c}})=\frac{1}{c}\Phi(x)+\left(1-\frac{1}{c}\right)\Phi(1)$$
,i.e $$c\Phi(x)=c\Phi(x^c)+(c-1)\Phi(1)$$
,hence $\Phi(x^c)=c\Phi(x)+(1-c)\Phi(1)
$.
Let $g(x)=\Phi(x)-\Phi(1)$, then $g(x^c)=cg(x),\forall x>0,c\geq 0$.Taking $x=e,c=\ln{y}$,then $g(y)=g(e)\ln{y}$,so $g(x)=\alpha\ln{x} $for some $\alpha,(x>0)$.then $$\Phi(x)=\alpha\ln{x}+\beta\qquad (\alpha,\beta\in \mathbf{R})$$.
A: Your choice for $\Phi $ seems all right.
We note that
\begin{equation*}
\lim_{p\downarrow 0}\parallel f\parallel _{p}=\exp [\int dx\ln f(x)]
\end{equation*}
We must have
\begin{equation*}
\Phi (\exp [\int dx\ln f(x)])=\int dx\Phi (f((x))
\end{equation*}
Try
\begin{equation*}
\Phi (g)=\ln g
\end{equation*}
Then we must verify
\begin{equation*}
\ln (\exp [\int dx\ln f(x)])=\int dx\ln f(x)
\end{equation*}
which is obviously true. I do not know if other choices for $\Phi $ are
possible
