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Let $\nu$ be a finite Borel measure on $\mathbb{R}^n$ and define the shift operator $T_a$ on $L^p_{\nu}(\mathbb{R}^n)$ by $f\to f(x+a)$ for some fixed $a\in \mathbb{R}^n-\{0\}$. Suppose moreover that $\nu$ is absolutely continuous wrt the Lebesgue measure $m$ and let $ \frac{d \nu}{dm}(x)= h(x). $

In this case, can we obtain a bound on $\|T_{a}\|_{\mathrm{op}}$ in terms of $h$ and of $a$?

Usually when $\nu$ is the Lebesgue measure then this is commonly known to be $1$, but here, in the finite and dominated case I can't seem to find such a result...

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Well, for large $a$ the norm goes to infinity. Find a ball $B$ such that $\nu(B) > \nu(\mathbb{R}^n) - \epsilon$ and consider the characteristic function of $B$ shifted by $-a$, for any $a$ greater than the radius of $B$. Its $L^2$ norm is at most $\sqrt{\epsilon}$, but after shifting by $a$ its norm is $> \sqrt{\nu(B)}$.

For general $a$ it's just a matter of comparing $h$ and its shift by $a$. The issue is if $f$ is the characteristic function of a tiny ball $B_1$ (tiny compared to $a$), and $B_2$ is the shift of this ball by $a$, then the ratio of the square roots of $s = \int_{B_2} h$ to $r = \int_{B_1} h$ gives a lower bound on the norm of the shift. Tiny balls are all we need to look at by a short argument using Lebesgue density. So the norm of the shift will be $\sqrt{\left\|\frac{h_a}{h}\right\|_\infty}$, where $h_a$ is the shift of $h$ by $a$. (Note that this could be infinite.)

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  • $\begingroup$ As a lower bound (when $\alpha \neq 0$) was always have 1, since the translation operator is hypercyclic no? $\endgroup$ – Zorn's Lama Jun 5 at 14:18
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    $\begingroup$ By $\alpha$ do you mean $a$? Then yes, 1 is always a lower bound. $\endgroup$ – Nik Weaver Jun 5 at 14:22
  • $\begingroup$ I've been thinking about it but is even possible to have a probability measure $\nu$ and some $a \in \mathbb{R}^n-\{0\}$ for which $\sqrt{\frac{\|h_a\|}{\|h\|}_{\infty}}$ achieves value $1$? I think it's not possible.. $\endgroup$ – Zorn's Lama Jun 5 at 17:04
  • $\begingroup$ No, not possible. Whatever $a$ is, you can find a positive measure ball which is small enough that it is disjoint from its translation by $a$. So if you translate it by $na$ for all $n \in \mathbb{Z}$ you get an infinite sequence of disjoint sets and finiteness of $\nu$ implies that they can't all have the same measure. $\endgroup$ – Nik Weaver Jun 5 at 19:48
  • $\begingroup$ But it's easy enough to come up with an $h$ such that the norms tend to $1$ as $a \to 0$. $\endgroup$ – Nik Weaver Jun 5 at 19:48
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You get a rather obvious bound for $\|T_a\|_{op}$ from $$ \int|f(x+a)|^p h(x)dx =\int |f(y)^p|h(y-a)dy = \int |f(y)|^ph(y) \left|h(y-a)/h(y)\right|dy \le c\int|f(y)|^ph(y)dy$$ with $c=\|h(y-a)/h(y)\|_\infty$.

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  • $\begingroup$ I accepted Nik's result only because it has a couple more details but both a great! :) $\endgroup$ – Zorn's Lama Jun 5 at 13:58
  • $\begingroup$ Moreover, Nick was faster than me. $\endgroup$ – Jochen Wengenroth Jun 5 at 14:04
  • $\begingroup$ Right, but you were both faster than me. $\endgroup$ – Zorn's Lama Jun 5 at 14:05
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This was intended as an extended comment but started to have too many formulas, so I thought that it would be more legible if posted as an answer. (You don't actually state if there are $\nu$-null sets which are not Lebesgue null, so I'm going to build an example which is mutually absolutely continuous wrt Lebesgue measure.)

If $h$ is unbounded then it seems to me that your shift operator could be unbounded. You don't specify whether you want $p$ to be in the reflexive range, so let me take $p=2$ just to be sure, and take $n=1$, $a=1$ for simplicity. Take $h(x)=|x|^{-3/4}$ for $x\in [-1,1]$ and $h(x)=e^{-|x|}$ outside that interval, so that $h \in L^1_m({\bf R})$. Put $d\nu = h\ dm$, so that $\nu$ is a finite measure on ${\bf R}$.

Consider $$ f(x) = \begin{cases}(x-1)^{-1/3} & \hbox{if $x\in (1,2]$} \\ 0 & \hbox{otherwise} \end{cases}$$ This belongs to $L^2_\nu({\bf R})$ since $$ \int_{\bf R} |f(x)|^2 h(x) \,dx = \int_1^2 (x-1)^{-2/3} e^{-x}\,dx <\infty $$ On the other hand, $$ (T_1f)(x) =f(x+1) = \begin{cases} x^{-1/3} & \hbox{if $x\in (0,1]$} \\ 0 & \hbox{otherwise} \end{cases}$$ so $$ \int_{\bf R} |T_1f(x)|^2 h(x) \,dx = \int_0^1 x^{-2/3} x^{-3/4}\,dx = +\infty $$

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  • $\begingroup$ This is a very interesting point Yemon. I didn't notice it for example... $\endgroup$ – Zorn's Lama Jun 5 at 14:02
  • $\begingroup$ I could have said this in a similar style to Nik Weaver's answer: the basic issue is that translation is good for Lebesgue measure because you don't change the "mass" assigned to a given region when you translate it. But as soon as some regions have "a lot more $\nu$-mass than Lebesgue mass" you are going to run into trouble by shifting an $f$-shaped pile of soil from regions with small $\nu$-mass to a place which has large $\nu$-mass $\endgroup$ – Yemon Choi Jun 5 at 14:02
  • $\begingroup$ Actually, your example was very illustrative since I mistakenly has a Gaussian measure visualized (which of-course has no blowups). Thanks a lot! :) $\endgroup$ – Zorn's Lama Jun 5 at 14:03

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