Need help with problem 6 in chapter 3 of rudin's real and complex analysis

Question

Let $m$ be the Lebesgue measure on $[0,1]$ and define $\lVert f\rVert_p$ with respect to $m$. Find all functions $\Phi$ on $(0,\infty)$ such that the relation $$\Phi\left(\lim_{p \to 0} \lVert f\rVert_p\right) = \int_0^1 \Phi(f)\,dm$$ holds for every bounded, measurable positive $f$. Show first that $$c\Phi(x) + (1-c)\Phi(1) =\Phi(x^c)$$

I can't even prove the inequality he gave as a hint, and even if I assume it is true I don't know how to proceed, help would be appreciated.

Answer 1

The $c$ and $1-c$ suggests you should consider the function $f(t):=x1_{[0,c]}(t)+1_{(c,1]}(t)$. I'll let you check this gives the hint.

Then let $\Psi(t):=\Phi(t)-\Phi(1)$. Deduce a functional equation for $\Psi$, solve it, and hence the general form of $\Phi$.

Answer 2

It was proven in the previous exercise (I assume you could do it), that for all $f:[0,1]\to ]0,\infty[$ that is measurable and bounded, we have $\lim_{p\to 0} ||f||_p = \exp(\int_0^1 \log fdm)$. Composing with $\Phi$ on the left, we get $\Phi\circ \exp(\int_0^1 \log f(x)dx)=\int_0^1 \Phi\circ fdm$. Now if $g:[0,1]\to R$ is any bounded and measurable function, the function $f=e^g$ is bounded, measurable and >0, so we get $\Phi\circ\exp(\int_0^1 gdm)=\int_0^1 \Phi\circ e^g dm = \int_0^1 \Phi\circ\exp\circ g dm$.

Let $H=\Phi\circ\exp$, and take for $g$ the function equal to $x$ on $[0,c]$ and to $y$ on $]c,1]$, where $x,y\in R$ and $0\le c\le 1$. We get the relation $H(cx+(1-c)y)=\int_0^c H(x)dm +\int_c^1 H(y)dm=cH(x)+(1-c)H(y)$.

This means that $H$ is affine on every interval $[x,y]\subset R$, which implies (I will let you prove this) that $H$ is in fact affine on $R$, i.e. $H(x)=ax+b$ for some $a,b\in R$. It follows that for all $x\in R$ we have $\Phi\circ \exp(x)=H(0)+(H(1)-H(0))x=\Phi(1)+(\Phi(e)-\Phi(0))x$. Therefore, for $y\in ]0,\infty[$ we have $\Phi(y)=\Phi(1)+(\Phi(e)-\Phi(1))\log(y)$.

Conversely, it is immediate that any function $\Phi:]0,\infty[\to R$ of the form $\Phi(y)=A+B\log(y)$ is such that $\Phi\circ \exp(\int_0^1 \log fdm) = \int_0^1 \Phi\circ f dm$, for all $f:[0,1]\to ]0,\infty[$ measurable and bounded.