TITLE: $L^p$ implies polynomial decay?
QUESTION [6 upvotes]: Suppose $f$ is uniformly continuous on $\mathbb{R}^n$.
Question: If $f \in L^p(\mathbb{R}^n)$, must there exist a $q > 0$ such that $|f(x)| \leq C|x|^{-q}$ for all large x (or perhaps a.e. large x)?
I know $f \in L^p(\mathbb{R}^n)$ implies $f(x) \to 0$ as $|x| \to \infty$. 
And I know $|f(x)| \leq C|x|^{-q}$ implies $f \in L^p(\mathbb{R}^n)$ when $pq > n$. 
Remark: One could ask a similar question about infinite series.

REPLY [7 votes]: First we can look at the series version. 
Let $(a_k)_{k\in \mathbb{N}}$ be the series given by $a_k = 0$ if $k$ is not a power of $2$ and $(\log_{2} k)^{-\gamma}$ for some $\gamma > 1$ if $k$ is a power of $2$. Then clearly there is no uniform bound of $(a_k)$ by $k^{-q}$ for any $q \geq 0$. But the series $\sum a_k$ is absolutely summable. 
For the function version, you can build little triangular tents centered at $2^{\ell}$ with width 1 and height $\ell^{-\gamma}$. This function is uniformly continuous, decays to zero, but at rate no better than $(\log \ell)^{-\gamma}$ which is far from polynomial.