Functions that Calculate their $L_p$ Norm are there any examples of functions $f:x\in\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$ and intervals $(a,b), 0\le a \lt b \le \infty$ , for which $$\Big(\int_a^b{|f(x)|^p dx}\Big)^\frac{1}{p} = f(p)$$ $$\forall p\in(a,b)$$
This question came up when thinking about $L_p$ norms as functions of $p$ and thus as a mapping of a real function to another one. 
In view of the comment of Mark Meckes, I would like to know, if there are also examples of non-trivial functions, i.e. that are not identical to $1$
 A: I'm reactivating this, but it's an extended comment at best. (It started out as what I thought was a complete answer, until I realized that I hadn't read the question carefully enough.) The interesting part of the question remains open.
There are no such functions if $b-a>1$: Indeed, $f(x)=\|f\|_x\ge (b-a)^{1/x}\min f$,
which leads to a contradiction if we take a point $x$ for which the min is assumed. Similarly, if $b-a=1$ and $f$ is not constant, then we obtain strict inequality here, so this isn't possible either. The case $b=\infty$ is also easily ruled out because then $f(x)\ge \min_{a+1\le t\le a+2}f(t)>0$, so $f\notin L^p$. Similarly, $f\in L^p$ for $p\in [a,b]$ and $b-a<1$ is ruled out in the same way, by estimating $f$ at its maximum.
This leaves us with the following slightly more focused version of the question: Is there an $f\in \bigcap_{a\le p<b}L^p$
with $f(x)=\|f\|_x$ for all $a\le x<b$? Here necessarily $b-a<1$ and $f\notin L^b$.
A: Based on Christian Remling's answer, we assume $b-a<1$. We know that $f$ has a pole on $b$. I don't know whether such an $f$ exists but I would at least like to understand the order of growth at this pole.
$f(x)$ is an increasing function. For simplicity of notation let's change variables to $g(x)=f(b-x)$, then $f(x)$ is decreasing. So for any $x$, the $L^p$ norm is at least $g(x) x^{1/p}$. If we plug in $p=(b+1/\log x)$, we get
$g(-1/\log x) \geq g(x) x^{ 1/ (b+ 1/\log x ) } =g(x) x^{1/b- 1/(b \log x) + 1/(b\log x^2) \dots) }= g(x) x^{1/b} e^{1/b + o(1) } $
$g(x) \leq g(-1/\log x) x^{-1/b} e^{1/b + o(1) } $
Let $h(x)= g(1/x)$, then we further simplify:
$h(x) \leq h( \log x) x^{1/b} e^{1/b+o(1)}$
This gives $h(x) \leq (x \cdot \log x  \cdot \log \log x \cdot  \dots )^{1/b} $
I believe this asymptotic upper bound is fairly sharp, that is, if we have a bound of the form
$h(x) \leq (x \cdot \log x\cdot  \log \log x \cdot  \dots  \log^n x)^{1/b}$
then by taking the $L^p$ norm $n$ times we get a bound of the form $h(x)=O(1)$ and a contradiction, but I haven't checked this.
The method of proof suggests that in the relevant regime, the operator that sends a function $f$ to the set of $L^p$ norms of $f$ is unstable, so iterating it might be a bad idea for finding a fixed point. However, some form of inverse of it might be stable? 
