If $x^\alpha f$ is in $L^1$, for some $\alpha>0$, then $f$ vanishes at infinity. How I got to the question:
In Folland p.249, Theorem 8.22, there is a list of properties about the Fourier Transform on $\mathbb{R}^n$. Letter $d$ property says
If $f$ and $x^\alpha f$ are in $L^1$ for $|\alpha|\leq k$, then $\hat{f}\in C^k$ and $\partial^\alpha\hat{f} = {[(-2\pi i x)^\alpha f]}\verb!^!$.
Then he goes on to say that this property above tells us about a fundamental property of the Fourier transform: that smoothness properties of $f$ are reflected in the rate of decay of $\hat{f}$ at infinity, and vice-versa.
Question:
This led me to look at the condition $x^\alpha f \in L^1$. Intuitively it tells us that $f$ vanishes at infinity. So my question is, if $x^\alpha f \in L^1$, then is $f\in C_0$?
Reasoning:
I am particularly confused with dealing with the fact that $f$ may not be continuous so I could have random values very far from the origin which are not bounded. These could make a set of finite measure, nonetheless.
To clarify, my definition of vanishing at infinity is that the set $\{x: |f(x)|\geq \epsilon\}$ is bounded for every $\epsilon >0$ (generally we say compact but in $\mathbb{R}^n$ it's equivalent to saying bounded). The definition of $f$ being in $L^1$ is that the integral $\int_{\mathbb{R}^n}|f(x)|dx$ is finite.
 A: Your intuition works for sequences - if a sequence $\{x_n\}$ satisfies $\sum_{n=1}^\infty n^\alpha |x_n|<\infty$, then $x_n\to0$ as $n\to\infty$ - but fails for functions. This is because a function can be integral without being bounded. For instance, think of a function which is usually zero, but occasionally "spikes" up to very large values. As long as the "spikes" have negligible area, there is no reason why this function should not be integrable. Explicitly, consider
$$ g(x) = \begin{cases}
2^n-2^nn^2|2^n-x|,& x\in[2^n-\frac1{n^2},2^n+\frac1{n^2}]\text{ for some }n,\\
0,&\text{otherwise.}\end{cases}$$
There are spikes around $x=2^n$, $n=1,2,\ldots,$ and the spike at $x=2^n$ has area $\frac1{n^2}$, thus the total integral is $\sum_{n=1}^\infty\frac1{n^2} = \frac{\pi^2}6<\infty$. Note also that $g$ is continuous; you could modify the spikes slightly so that the resulting function is smooth.
Now take $f(x) = x^{-\alpha}g(x)$. Then $x^\alpha f(x) = g(x) \in L^1(\mathbb R)$, but $f$ does not vanish at infinity for $\alpha\le1$.
