All the functions $\phi$ such that $\phi (\lim_{p\to 0}||f||_p)=\int_0^1 (\phi of)dm$

Question

I was working through the third chapter of big Rudin and I am a bit stuck on the sixth question which asks us to find all the functions $\phi$ on [0,$\infty$) such that the relation $$\phi(\lim_{p\to 0}||f||_p)=\int_0^1 (\phi of)dm$$ (m is a lebesgue measure on [0,1]) for all bounded, measurable and positive f. It also asks us to first show $$c\phi (x)+(1-c)\phi(1)=\phi(x^c)$$(x $\gt$0, 0$\le c\le 1)$ I have already done the question preceding this one and it would make sense that log would be such a function once the second equality is proved. I would highly appreciate if I could get a hint as to how to prove that. I was thinking of using contradiction and then using that x to make a delta type function but that didn't take me anywhere. I may be missing something very silly.Thanks for any help. $$Edit$$ I think I have a solution to the problem. Please have a look. Consider f such that f(y)=x when $y \lt c$ and f(y)=1 otherwise. Now $\int_0^1 (\phi of)dm=c\phi (x)+(1-c)\phi(1)$. Note that from problem 5 ch 3 of rudin, $\lim_{p\to 0}||f||_p=exp(\int_0^1 (\log f)dm)$. Therefore $\phi(\lim_{p\to 0}||f||_p)=\phi(exp(clog(x)+(1-c)log(1))=\phi(x^c)$. This proves the second equality.

Answer 1

$$cϕ(x)+(1−c)ϕ(1)$$ is a convex combination of $ϕ(x)$ and $ϕ(1)$ and thus situated inside the interval $[ϕ(x),ϕ(1)]$. By the intermediate value theorem, there is some $x^c\in[x,1]$ with the proscribed value $ϕ(x^c)$.

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For your framing question, I do not understand the notation. What is $\lim:\lim$, why is the limes in $x$ when the only other scalar variable is $p$, ...