Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function and denote the part of the plane under the graph of $f$
$$\mathcal{G} = \{(x,y)\in\mathbb{R}^2 \space : \space y\leq f(x)\}$$ The collection of all closed balls inside $\mathcal{G}$ $$\mathcal{B} = \{B \space \space \text{is a closed ball, of positive radius, contained in} \space \mathcal{G} \}$$ And the union of all these closed balls $$A = \bigcup_{B\in\mathcal{B}}B$$
The question is: when is it that $A=\mathcal{G}$ ?
My thougts:
There shouldn't be 'upward cusps' in the graph of the function, since however close to the tip of the cusp and however small a ball ( contained in $\mathcal{G}$ ) you take, it will never contain the tip. Downward cusps are OK, but if we also consider the set $\{(x,y)\in\mathbb{R}^2 \space : \space y\geq f(x)\}$ (or also think about $-f$), these aren't allowed either. Maybe then the answer could be something like $f\in C^1$, but $f'$ can still be $\infty$ at some point.
Clearly $A\subset \mathcal{G}$ and on the other hand the interior points of $\mathcal{G}$ are contained in $A$. So it becomes a question of whether the points on the boundary of $\mathcal{G}$ (the graph of $f$) can be captured by a circle that is below the graph. This is related to the curvature of the curve $(x, f(x))$, right? So somehow second derivatives also appear to matter, but I think that $C^1$ (or even letting the derivative jump upwards if we don't make the requirement for the upper part also) should be enough and also there can be points where the derivative becomes infinity.