If $f \in L^1(\mathbb{R})$ and $\hat f \geq 0$, is $f$ continuous? Suppose $f \in L^1(\mathbb{R})$.  I am wondering what conditions on $\hat f = \left[ s \mapsto \int e^{its} f(t) \ dt \right] \in C_0(\mathbb{R})$ suffice to make $f$ continuous (or, more accurately, equal almost everywhere to some unique continuous function). 

Example: If $\hat f \in L^1(\mathbb{R})$, then $f$ is, almost everywhere, equal to the inverse transform of $\hat f$ (Fourier inversion) so $f$ is continuous. 

However, it is possible for $f$ to be continuous, even to have $f \in C_c(\mathbb{R})$, and still not have $\hat f$ integrable. 

Examples: In his answer here, robjohn says that function 
  $$f(t) = \begin{cases}
\frac{-1}{\log(t) + \log(1-t)} & \text{ if } 0 < t < 1 \\
0 & \text{ otherwise } \\
\end{cases}$$ (see this plot) has $\hat f \notin L^1(\mathbb{R})$. The problem of finding an $f \in C_c(\mathbb{R})$ such that $\hat f \notin L^1(\mathbb{R})$ is also discussed here.

So, I'm wondering about other ways in which $\hat f$ can "see" continuity of $f$. Specifically:

Question: If $\hat f \geq 0$, is $f$ continuous?

If this is false, or even if it's true, I am also interested in the more general question: 

Question: In what ways can continuity of $f$ manifest itself as some property of the Fourier transform $\hat f$?

Thanks.
 A: I think your assertion is probably false.  The "triangle" $f(x) = \max(1-|x|,0)$ has positive Fourier transform.  If you take some linear combination of these:
$$ g(x) = \sum a_n f(x/b_n) $$
where $a_n,b_n>0$, you will end up with a function satisfying $f(x) = f(-x)$, and the function is convex and decreasing on $[0,\infty)$ (and probably any such function can be written as a limit of these).  I think in this way you should be able to construct a function in $L^1$ whose Fourier transform is positive, but the function is discontinuous (that is, unbounded) at $x=0$.

Details: Let $f$ be the above triangle wave. Notice $\| f\|_1 =  1$ and $\hat f \geq 0$. For each nonnegative integer $n$, define $f_n(x) = 2^n f(2^n x)$. Notice
$$ \|f_n \|_1 = \int |f(2^n x)| d(2^n x) = \|f\|_1 = 1$$
and $\hat f_n(x) = f(x/2^n) \geq 0$. 
The series $\sum_{n=0}^\infty \frac{1}{2^n} f_n$ converges $\| \cdot \|_1$-absolutely a function $g \in L^1(\mathbb{R})$. Since the Fourier transform is a contractive linear mapping from $L^1(\mathbb{R}) \to C_0(\mathbb{R})$ we get that $\sum_{n=0}^\infty \frac{1}{2^n} \hat f_n$ converges absolutely in the uniform norm to $\hat g$. Since $\hat f_n \geq 0$, it follows that $\hat g \geq 0$. 
Lastly, we must argue that $g$ is not equal almost everywhere to a continuous function. By monotonicity, the sum defining $g$ converges pointwise. On the other hand, since the series converges in $L^1$, a subsequence of the partial sums must converge almost everywhere. Thus $g$ equals, almost everywhere, the pointwise limit of its series.  For any $N > 0$, there is an $\epsilon > 0$ such that  $f_n(y) > 2^{n-1}$ whenever $|y| < \epsilon$ and $n<N$.  It follows that $g$ is, unbounded on a neighbourhood of $0$, so $g$ cannot be equal, almost everwhere, to a continuous function.
A: One kind of very clear answer to this sort of question is in terms of Sobolev spaces and the Sobolev imbedding theorem: on $\mathbb R^n$, if $\hat{f}(\xi)\cdot(1+|\xi|^2)^{k+{n\over 2}+\epsilon}$ is in $L^2$ for some $\epsilon>0$, then $f$ is continuously $k$ times differentiable. 
