# Operator norm of shift operator for finite measure spaces

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...

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.)

• As a lower bound (when $\alpha \neq 0$) was always have 1, since the translation operator is hypercyclic no? – Zorn's Lama Jun 5 at 14:18
• By $\alpha$ do you mean $a$? Then yes, 1 is always a lower bound. – Nik Weaver Jun 5 at 14:22
• 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.. – Zorn's Lama Jun 5 at 17:04
• 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. – Nik Weaver Jun 5 at 19:48
• But it's easy enough to come up with an $h$ such that the norms tend to $1$ as $a \to 0$. – Nik Weaver Jun 5 at 19:48

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 $$c=\|h(y-a)/h(y)\|_\infty$$.

• I accepted Nik's result only because it has a couple more details but both a great! :) – Zorn's Lama Jun 5 at 13:58
• Moreover, Nick was faster than me. – Jochen Wengenroth Jun 5 at 14:04
• Right, but you were both faster than me. – Zorn's Lama Jun 5 at 14:05

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 $$\int_{\bf R} |T_1f(x)|^2 h(x) \,dx = \int_0^1 x^{-2/3} x^{-3/4}\,dx = +\infty$$

• This is a very interesting point Yemon. I didn't notice it for example... – Zorn's Lama Jun 5 at 14:02
• 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 – Yemon Choi Jun 5 at 14:02
• Actually, your example was very illustrative since I mistakenly has a Gaussian measure visualized (which of-course has no blowups). Thanks a lot! :) – Zorn's Lama Jun 5 at 14:03