Given a lattice $L$ and a subset $P\subset \mathbb R^d$, we define for each positive integer $t$ $$f_P(L,t)=|(tP\cap L)|$$ the number of lattice points in $tP$. Let's say $P$ is nice if $f_P(L,t)$ is a polynomial. We know that if $P$ is a convex polytope with vertices in $L$ then $P$ is nice and $f_P(L,t)$ is its Ehrhart polynomial. My question is about some converse of this statement.

Are there some mild assumptions (for example convexity etc.) on $P$, under which if $f_P(L,t)$ is a polynomial with respect to at least some lattice $L$ then $P$ must be a convex polytope? Or a weaker question: Is any polynomial arising this way also the Ehrhart polynomial of some polytope?

P.S. I haven't thought much about this question so I apologize if it is well-known or it has an obvious negative answer. Also feel free to retag.

Richard Stanley suggested the following in the comments (edited to take into account a trivial family of counter-examples):

Could the following be true? It seems more in line with the question. Let $P$ be a compact convex $n$-dimensional set in $\mathbb R^n$. Suppose that the Ehrhart function $f_P(t)$ is a polynomial for positive integers $t$. Then $P$ is a translation of a rational polytope.

**Edit**: I would also be interested in a slightly weaker statement: Suppose a convex set has positive curvature almost everywhere, must the Ehrhart function necessarily be non-polynomial?

For example given an arbitrary lattice, what would be the easiest way to see that a circle doesnt have a polynomial Ehrhart function?