Let $f:[a,b]\rightarrow \mathbb{R}$, and suppose $f \in W^{2,p}$ for some real $p \geq 1$. Does that imply $f \in C^1([a,b])$?
My attempt: since $f$ is twice (weakly) derivable, and continuity is a necessary condition for weak derivability, then I suppose that $f$ has to belong to $C^1((a,b))$. But can we deduce that $f$ is continuous at the extremes of the interval?
Strictly speaking, the answer is no. $f \in W^{2,p}$ is really an equivalence class of functions which are equal a.e. It doesn't even exactly make sense to speak of their pointwise regularity. But even if you "open up" the equivalence class and look at the functions inside, not all of them have pointwise regularity. (For instance, you can do whatever you want at all rational points, and nothing changes in terms of $W^{2,p}$.)
Morally, the answer is yes. This is because of Morrey's inequality. In one dimension this tells you that $W^{k,p}$ embeds continuously into a space that I like to think of as "$C^{k-\frac{1}{p}}$", but which is more commonly denoted by $C^{k-1,1-\frac{1}{p}}$. That is, there are $k-1$ continuous derivatives, and the last one is $1-\frac{1}{p}$-Holder continuous. Back in terms of the equivalence classes, this means each equivalence class has a representative with this much pointwise regularity. Taking $k=2$ and any $p \geq 1$, you get $C^1$ regularity. Since these are $C^1$ functions with finite $C^1$ norm, they have a continuous extension to the boundary. (Speaking of pointwise derivatives at the boundary is not generally wise.)