Lower bounds on translates of a function over a compact set Let $f\in L^p(\mathbb{R})$ and define $f_\theta(x)=f(x-\theta)$. Let $K\subset\mathbb{R}$ be a compact set. I would like to compute (or at least lower bound) the following:
$$
\inf_{\theta\ne\theta'\in K}\frac{\Vert f_\theta - f_{\theta'}\Vert_p}{|\theta-\theta'|}.
$$
In particular, I want to understand how this depends on $f$, and would like a bound that depends explicitly on $f$. This is also where the properties of $f$ come in: The weaker the assumptions the better, but e.g. if there a nice bound that depends (say) on the deriviatives of $f$, then we can assume the needed regularity.
My suspicion is that there is an easy counterexample to show this can be rather poorly behaved even for smooth functions, but I have not been creative enough so far.
 A: Here "is an easy counterexample to show this can be rather poorly behaved even for smooth functions". The idea is to make $f$ almost periodic with a quasi-period $T$ (but with a slowly decreasing amplitude near $\pm\infty$, to satisfy the condition $f\in L^p(\mathbb R)$), and then shift $f$ by $T$. Then the shifted function $f_T$ will differ little from $f_0=f$.
For instance, let
$$f(x):=c(1+\sin x)e^{-c^2 (x-\pi)^2},$$
where $c>0$ is small. Then $\|f\|_1=\sqrt\pi$ for all real $c>0$. However,
$$\frac{\|f_{2\pi}-f_0\|_1}{2\pi}=\frac{\text{erf}(c\pi)}{\sqrt\pi}\to0$$
as $c\downarrow0$. So, there is no nontrivial (that is, nonzero) lower bound here -- unless $f$ is fixed or, more, generally, other conditions on $f$ are imposed, in addition to smoothness; see below.

Here is the graph $\{(x,f(x))\colon|x|<11\pi\}$ for $c=1/100$:



$\newcommand\de\delta$Now a simple result in the positive direction. Suppose that for some real $c>0$ and all $x\in[a,b]$ we have $f'(x)\ge c$. Then for any real $\de>0$ and all $x\in[a,b-\de]$ we have $f(x+\de)-f(x)\ge c\de>0$ and hence $\|f_\de-f_0\|_p\ge c\de(b-a-\de)_+^{1/p}$, where $u_+:=\max(0,u)$.
Suppose now that
(i) for some real $c_1>0$ and all $x\in[a_1,b_1]$ we have $f'(x)\ge c_1$ and (ii) for some real $c_2>0$ and all $x\in[a_2,b_2]$ we have $f'(x)\le -c_2$. Then, similarly, for any real $\de>0$ we have $\|f_\de-f_0\|_p^p\ge c_1^p\de^p(b_1-a_1-\de)_+ + c_2^p\de^p(b_2-a_2-\de)_+$. So, if $K-K\subset[-\de_*,\de_*]$ for some real $\de_*>0$, then the infimum in question is
$$\ge(c_1^p(b_1-a_1-2\de_*)_+ + c_2^p(b_2-a_2-2\de_*)_+)^{1/p}.$$
