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...



Operator norm of shift operator for finite measure spaces
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.)


