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Where I am in the Problem

I am now trying to solve the smaller problem:

Find all functions $f$ on $[0,\infty)$ satisfying the relation $cf(x)+(1-c)f(1)=f(x^c)$ where $0\leq c\leq 1$

I know constant functions work, and logrithmic functions would work if $0$ was omitted from the domain, but I am having a hard time approaching this problem when I can't assume that the function $\Phi$ is at least continuous.

For some context this problem came out of a bigger problem from rudin 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 $$\Phi(\lim\limits_{p\to 0}||f||_p)=\int\limits_{0}^{1}(\Phi \circ f)d\mu$$ holds for every bounded, measurable, positive $f$. Show first thtat $$c\Phi(x)+(1-c)\Phi(1)=\Phi(x^c)\quad (x>0,0\leq c\leq 1)$$

My Work In total:

Let $f=a\chi_A+b\chi_B$ be measurable, where $a,b>0$, $A\cup B=[0,1]$, and $A\cap B=\emptyset$. Then $$||f||_p=\bigg\{\int\limits_{X}|f|^pd\mu\bigg\}^{1/p}$$ so $$||a\chi_A+b\chi_B||_p=\bigg\{\int\limits_{X}|a\chi_A+b\chi_B|^pd\mu\bigg\}^{1/p}=\bigg\{\int\limits_{X}|a|^o\chi_A+|b|^p\chi_Bd\mu\bigg\}^{1/p} \quad (*)$$

$$=\big\{a^p\mu(A)+b^{p}\mu(B)\big\}^{1/p}$$

Let $a=x$, $b=1$, and $\mu(A)=c$.

$$||f||_p=\big\{cx^p+(1-c)\big\}^{1/p}$$

Thus

$$\Phi(\lim\limits_{p\to 0} ||f||_p)=\Phi(\lim\limits_{p\to 0}\big\{cx^p+(1-c)\big\}^{1/p})=\Phi(x^c)\quad (**)$$

$$\int\limits_{0}^{1}(\Phi \circ f)d\mu= c\Phi(x)+(1-c)\Phi(1)$$

Thus

$$c\Phi(x)+(1-c)\Phi(1)=\Phi(x^c)\quad (x>0,0\leq c\leq 1)$$

Footnotes

(*) While it is normally true that $(a+b)^p\neq a^p+b^p$, since $a$ and $b$ are being multiplied by characteristic functions it means that $$|a\chi_A(x)+b\chi_B(x)|^p=\begin{cases}|a|^p & x\in A\\ |b|^p & x\in B\end{cases}$$. Given that the proceding equality holds.

(**) The limit on the inside is achived by letting $\big\{cx^p+(1-c)\big\}^{1/p}=e^{\frac{\ln(cx^p+(1-c))}{p}}$, applying l'hopital's rule to the exponent once gives the desired equality.

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Ignore the point $x = 0$ for the moment. Rudin asserts

$$c\Phi(x) + (1-c)\Phi(1) = \Phi(x^c)$$

only for $x > 0$. Then for fixed $x > 0$ consider $g_x \colon [0,1] \to \mathbb{R}$ given by $g_x(c) = f(x^c)$. Since by assumption

$$f(x^c) = cf(x) + (1-c)f(1),\tag{1}$$

we know that $g_x$ is continuous, and that in turn implies that $f$ is continuous on $(0,x)$. Since $x > 0$ was arbitrary, we thus have the continuity of $f$ on $(0,+\infty)$.

Now let $f_0(x) = f(x) - f(1)$. By $(1)$, we have

$$f_0(x^c) = c\cdot f_0(x)\tag{2}$$

for all $x > 0$ and $c\in [0,1]$. Since for $t > 1$ we have

$$f_0(e) = f_0\bigl((e^t)^{1/t}\bigr) = \frac{1}{t}f_0(e^t)\qquad\text{and}\qquad f_0(e^{-1}) = f_0\bigl((e^{-t})^{1/t}\bigr) = \frac{1}{t}f_0(e^{-t})$$

by $(2)$, it follows that

$$f_0(x) = \begin{cases}\quad f_0(e)\cdot \log x &, x \geqslant 1 \\ -f_0(e^{-1})\cdot \log x &, x < 1.\end{cases}$$

Note that it need not be the case that $f_0(e^{-1}) = -f_0(e)$.

If $(1)$ is supposed to hold also for $x = 0$, then inserting $c\in [0,1)$ shows

$$f(0) = c\cdot \bigl(f(0) - f(1)\bigr) + f(1),$$

whence $f(0) = f(1)$, or, if we allow it, $f(0) = \pm\infty$.

Clearly, adding constants to $\Phi$ doesn't influence whether

$$\Phi\Bigl(\lim\limits_{p\to 0} \lVert f\rVert_p\Bigr) = \int_0^1 \Phi\circ f\,dm\tag{$\ast$}$$

holds, so we can assume $\Phi(1) = 0$ and just need to check for which choices of $A,B,C$ the function

$$\Phi_{A,B,C} \colon x \mapsto \begin{cases} A\log x &, x \geqslant 1 \\ B\log x &, 0 < x < 1 \\ \quad C &, x = 0\end{cases}$$

satisfies $(\ast)$.

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