Is the statement "$f$ is continuously differentiable over a closed set" mathematically incorrect?

Question

For example, I am currently reading a sentence that writes:

"let $f$ be a continuously differentiable function over a closed and convex set $C$."

Sentence of this type appears everywhere in analyis, optimization, etc.

So let's imagine $f(x) = x^2$ over a closed set $C$ that is symmetric around $x = 0$.

Then we immediately have a problem in that $f(x) = x^2$ is not differentiable at the corners of $C$, hence $f$ could not have be continuously differentiable and this statement does not make sense.

Can someone check my reasoning?

Answer 1

I would agree that it is an "incorrect" statement, since it's mathematically imprecise. There are multiple (perfectly well-defined, but distinct) interpretations of what this phrase means. For instance, as pointed out in the comments, this could mean either (A) the existence of directional derivatives on the boundary of $C$, or (B) the existence of a differentiable extension of $f$ on an open superset of $C$.

Oftentimes in practice, when this appears in articles, one has to do a bit of guesswork. In some communities, one notion is more common/accepted than the other.