Q:

Operator Norm of Translation Operator

Define the operator $T_a(f)\to f(\cdot +a)$ from $L^1_{\nu}\to L^1_{\nu}$ for a Borel probability measure $\nu$ on $\mathbb{R}^n$.  Is the operator norm of $T_a$ well-known?   Ie, the quantity:
$$
\sup_{\|f\|\leq 1, f\neq 0}\,
\frac{\|T_a(f)\|}{\|f\|},
$$
where $\|\cdot\|$ is the $L^1_{\nu}$-norm and $\nu$ is absolutely continuous with respect to the Lebesgue measure.  

A:

It is easy to construct examples where $\|T_a\|$ is as big as you want.  Let $\mu$ be Lebesgue measure, $b\in(0,1)$, $g\in L^1(\mu)$ positive, with $\|g\|_1=1$, $\int_{B_1(0)}g\,d\mu=b$, and $g$ supported in $B_1(0)\cup B_1(2)$. Define a Borel probability measure on $\mathbb R^n$ by 
$$
\nu(E)=\int_Eg\,d\mu.  
$$
Let $f=\tfrac1{\mu(B_1(2))}\,1_{B_1(2)}$. Let $a=2$. We have 
$$
\int_{\mathbb R^n}f\,d\nu=\frac{1-b}{\mu(B_1(2))}
$$
and
$$
\int_{\mathbb R^n}T_{2}f\,d\nu=\frac{b}{\mu(B_1(0))}. 
$$
Thus
$$
\|T_2\|\geq\frac{\|T_2f\|}{\|f\|}=\frac{b}{1-b}. 
$$
Taking for instance $b=\frac{m}{m+1}$ for $m\in\mathbb N$, we get 
$$
\|T_a\|\geq m. 
$$

