# What can we say about $\Phi$?

Let $m$ be Lesbegue measure on $[0,1]$ and define $|f|_{p}$ with respect to $m$. Find all functions $\Phi$ on $[0,\infty)$ such that the relation $$\Phi(\lim_{p\rightarrow 0}|f|_{p})=\int^{1}_{0}(\Phi\circ f)dm$$ Show first that $$c\Phi(x)+(1-c)\Phi(1)=\Phi(x^{c}),(x>0, 0\le c \le 1)$$

By previous conclusion we have $$\lim_{p\rightarrow 0}|f|_{p}=e^{\int \log|f|dx}$$ So we only need to find $\Phi$ such that $$\Phi(e^{\int \log|f|dx})=\int^{1}_{0}(\Phi\circ f)dx$$ Let $F$ be a measurable function by being $X$ on $x\in [0,c]$ and $1$ on $[c,1]$. Then the right hand side become $c\Phi(x)+(1-c)\Phi(1)$. The right hand side now become $e^{c\log[x]}=x^{c}$. So we conclude that $$c\Phi(x)+(1-c)\Phi(1)=\Phi(x^{c}),(x>0, 0\le c \le 1)$$ as desired.

Similarly we may let $F$ be a measurable function that is $X$ on $[0,c]$ and $Y$ on $[c,1]$. Then the right hand side is $c\Phi(x)+(1-c)\Phi(y)$. The integral now become $e^{c\log[X]+(1-c)Log[Y]}=X^{c}Y^{1-c}$, and we have the more general identity

$$c\Phi(X)+(1-c)\Phi(Y)=\Phi(X^{c}Y^{1-c})$$ At this point it feels obvious that $\Phi$ can be the $\log[x]$ function or a constant times $\log[x]$ function. But $\Phi(1)\not=0$ is possible; So it is not clear to me how to construct $\Phi$ precisely.

I do not know if this is a duplicate.

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finally solved. –  Bombyx mori Dec 25 '12 at 10:04
Maybe you could post your solution in an answer rather than in the OP. –  Davide Giraudo Apr 7 '13 at 14:46
@DavideGiraudo: I don't remember the solution over the top of my head, but it should be trivial. –  Bombyx mori Apr 8 '13 at 3:43