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Motivation for this question: If the operators $B_i'$ satisfy an inequality, prove that $B_1'+\dots B_n'$ also satisfies the same inequality

Let $A$ be a non-constant operator acting on $C^\infty(\mathbb{R})$, and let $u(t,x)=e^{-tA}f(x)$, where $f\in C^\infty(\mathbb{R})$. Also, let $s(t)$ be a function of $t$. Then it is easy to see that $$\frac{\partial}{\partial_t} \|u\|_{s(t)}=\|u\|_s^{1-s}\left[\frac{s'}{s}\int u^s\log\left(\frac{u}{\|u\|_s}\right)d\mu(x)-\langle Au,u^{s-1}\rangle\right]$$ Here $\langle Au,u^{s-1}\rangle=\int (Au)u^{s-1}d\mu(x)$.

Short proof: Let $$y=\|u\|_s=\left(\int u^s\right)^{1/s}d\mu(x)$$ Now take $\log$ on both sides and differentiate with respect to $t$.

  1. I now want to find an closed-form expression $X(u,t)$ such that $$\frac{\partial}{\partial_t} X(u,t)=\|u\|_s^{1-s}\left[\frac{s'}{s}\int u^s\left[\log\left(\frac{u}{\|u\|_s}\right)\right]^2 d\mu(x)-\langle Au,u^{s-1}\rangle\right]$$ Note that the only different is that $\log$ has been squared.

  2. I want to find a differential operator $L$, perhaps comprising of $\frac{\partial}{\partial_t}$, $\frac{\partial^2}{\partial_t^2}$, etc, such that $$L \|u\|_{s(t)}=\|u\|_s^{1-s}\left[\frac{s'}{s}\int u^s\left[\log\left(\frac{u}{\|u\|_s}\right)\right]^2 d\mu(x)-\langle Au,u^{s-1}\rangle\right]$$

I have been trying to find $X(u,t)$ and/or $L$ for at least a month, without success.

EDIT: Note that $X(u,t)$ can be thought of as $$\int_0^t \|u\|_s^{1-s}\left[\frac{s'}{s}\int u^s\left[\log\left(\frac{u}{\|u\|_s}\right)\right]^2 d\mu(x)-\langle Au,u^{s-1}\rangle\right]dt$$ However, I am looking for an algebraic expression that doesn't involve integrals. Ideally, $X(u,t)$ should be a norm of some kind.

Things I have tried:

  1. Integration by parts, studying examples, etc. Nothing solves the problem completely.
  2. $X(u,t)=\|u\|_s(1+\log \frac{u}{\|u\|_s})$ when $u$ is a constant function with respect to $x$. However, I have not been able to make this work for non-constant functions.
  3. It doesn't matter if you don't get $\frac{s'}{s}$ exactly. The only thing that matters is that the $\log u$ term should be squared and within the integral.

Thank you for your time.

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  • $\begingroup$ can you clarify what properties you want from $A(u,t)$? Because clearly by integrating $A(u,t)=\int^t...$. $\endgroup$
    – Thomas Kojar
    Commented Jan 5 at 20:11
  • $\begingroup$ @ThomasKojar- I would ideally like a closed-form expression for $A(u,t)$. $\endgroup$
    – matilda
    Commented Jan 5 at 21:32
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    $\begingroup$ closed-form in terms of what? $A(u,t)=\int^{t}||u||^{1-s}... dt$ is a closed form since $u=e^{-tA}f$. $\endgroup$
    – Thomas Kojar
    Commented Jan 5 at 21:56
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    $\begingroup$ The integral expressed here as $$ \int u^s\log\left(\frac{u}{\|u\|_s}\right)-\langle Au,u^{s-1} \rangle$$ could easily be mistaken for $$ \int \left( u^s\log\left(\frac u {\|u\|_s}\right)-\langle Au,u^{s-1} \rangle \right) \, ds, $$ or for $$ \int u^s \left( \log\left(\frac u{\|u\|_s} \right) - \langle Au,u^{s-1}\rangle \right) \, dt. $$ But it appears that (maybe?) you intended $$ \int u^s \left( \log\left(\frac u{\|u\|_s} \right) - \langle Au,u^{s-1}\rangle \right) \, dx. $$ Explicitness about this could avoid confusion. $\endgroup$
    – Michael Hardy
    Commented Jan 6 at 0:19
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    $\begingroup$ I prefer to keep it here in MO because there are many experts here that can contribute. If you are concerned with sharing preliminary work, perhaps try to only mention the very particular technical estimate you think might be true but are having trouble proving. There is likely no need to mention which conjecture you are working on. $\endgroup$
    – Thomas Kojar
    Commented Jan 21 at 7:51

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