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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 $1$

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    $\begingroup$ How about $a=0$, $b=1$, $f(x) = 1$? $\endgroup$
    – Mark Meckes
    Commented Jul 2, 2014 at 12:55
  • $\begingroup$ @MarkMeckes your example is correct; the question would be whether also non-constant functions exist, that calculate their $L_p$ norm $\endgroup$
    – Manfred Weis
    Commented Jul 2, 2014 at 13:12
  • $\begingroup$ It's easy to see that non-constant examples are only possible if $b-a<1$ and $f\notin L^b$. (I had this posted as what I thought was a complete answer when I discovered that I hadn't read your question carefully enough.) $\endgroup$
    – Christian Remling
    Commented Jul 3, 2014 at 3:42
  • $\begingroup$ In particular, if $b-a>1$, then there are no solutions, and if $b-a=1$, then the solutions are exactly the constants $f=c>0$. $\endgroup$
    – Christian Remling
    Commented Jul 3, 2014 at 3:46
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    $\begingroup$ Not quite the same, but there is a function $f$ defined on some interval such that $f(p) = \|f\|_p^p$ on that interval. Namely, $1/(1-x)$ on $[0,1]$. $\endgroup$
    – Sean Eberhard
    Commented Jul 4, 2014 at 9:55

2 Answers 2

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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 $f\notin L^b$.

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  • $\begingroup$ Considering continuous functions on $(a,b)$ (as they must be), you excluded all cases, $b-a>1$, $b-a=1$ and $b-a<1$. The interval being bounded, can't you remove the last case by just considering $(c,d)\subset(a,b)$? $\endgroup$
    – username
    Commented Jul 3, 2014 at 10:46
  • $\begingroup$ @AthanagorWurlitzer: I don't think restricting to a subinterval will work since $\|f\|_{L^p(c,d)}<\|f\|_{L^p(a,b)}$, so the restricted $f$ will no longer satisfy the equation you're trying to solve. $\endgroup$
    – Christian Remling
    Commented Jul 3, 2014 at 16:48
  • $\begingroup$ @AthanagorWurlitzer: No problem, of course. As for the remaining case, I don't really have a good idea right now, but I have a feeling that one has to study $f$ near $b$ rather carefully; just Holder feels too "soft." $\endgroup$
    – Christian Remling
    Commented Jul 4, 2014 at 4:50
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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?

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  • $\begingroup$ I think you lost an (inessential) $b$ in the very first line: it's really $e^{1/b^2}$ throughout. Also, could you please clarify the meaning of "$\ldots$" in your bound on $h$. Are you saying that you continue until $\ln\ln\ldots\ln x\approx 1$ and then bound $h$ by the corresponding product (the number of factors will then depend on $x$, of course)? (What happened to the $e^{1/b^2}$ factors?) $\endgroup$
    – Christian Remling
    Commented Jul 9, 2014 at 18:40
  • $\begingroup$ I made that intentionally vague because I didn't want to write exactly what the bound is. Yes, it's essentially like what you said. You stop when $\log x < 1/ (a-b)$, say, and you know you have some $O(1)$ upper bound on $h$ in that range. You do have to include the $e^{1/b^2}$ factors, which becomes $e^{ log^* (x) / b^2}$, I guess. The lower terms just become $O(1)$. $\endgroup$
    – Will Sawin
    Commented Jul 9, 2014 at 20:44

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