Question on Apostol necessary and sufficient conditions for a field to be a gradient

Question

I am reading Apostol Calculus II (2nd edition) and in section 10.15 it is stated that:

THEOREM 10.5. NECESSARY AND SUFFICIENT CONDITIONS FOR A VECTOR FIELD TO BE A GRADIENT. Let $f$ be a vector field continuous on an open connected set $S$ in $R^K$.

Then the following three statements are equivalent.

(a) $f$ is the gradient of some potential function in $S$.

(b) The line integral off is independent of the path in $S$.

(c) The line integral of $f$ is zero around every piecewise smooth closed path in S.

however later in the chapter, section 10.16 the author describes a test to determine whether a given field $f$ is a gradient:

10.16 Necessary conditions for a vector field to be a gradient

The first fundamental theorem can be used to determine whether or not a given vector field is a gradient on an open connected set $S$. If the line integral of $f$ is independent of the path in $S$, we simply define a scalar field $\phi$ by integrating $f$ from some fixed point to an arbitrary point $x$ in $S$ along a convenient path in $S$. Then we compute the partial derivatives of $\phi$ and compare $D_k\phi$ with $f_k$, the k th component of $f$. If $D_k\phi(x) =f_k(x)$ for every $x$ in $S$ and every $k$, then $f$ is a gradient on $S$ and $\phi$ is a potential. If $D_k\phi(x) \neq f_k(x)$ for some $k$ and some $x$, then $\phi$ is not a gradient on $S$.[...]

I don't understand how the sentence in bold can be possible, given that at the beginning of the paragraph we assumed that the line integral of $f$ was independent of the path in $S$ and for theorem 10.5 this was a necessary condition for $f$ to be a gradient. Any clues?

Thanks

Answer 1

The sentence

If the line integral of $f$ is independent of the path in $S$, we simply define a scalar field $\phi$ by integrating $f$ from some fixed point to an arbitrary point $x$ in $S$ along a convenient path in $S$.

could be followed by "If the line integral of $f$ is not independent of the path in $S$, then the given vector field is not a gradient on $S$." This uses the theorem you state. As you note, this is a straightforward obstruction to the vector field being a gradient, so Apostol doesn't explicitly write it.

However, there is more to do. The theorem you cite requires that the vector field be continuous. For some discontinuous vector fields, the given prescription works and this $\phi$ is well defined. In such cases, there is more work to do to reject discontinuous vector fields -- such fields will have an obstruction in one of their derivatives.

(What's the problem with a discontinuous vector field? Either the field is undefined at one or more points, so is not a vector field on all of $S$, or it is nondifferentiable in at least one direction at at least one point, which prevents the gradient existing at that point. Either way, the gradient cannot exist on all of $S$, so the field cannot be a gradient on $S$.)