I tried to answer the following exercise:
The answer says:
And theorem 11 is:
I answered by other methods but am curious (and very confused): How does that contradicts theorem 11 in a way that makes the existence of $G$ impossible? I guess we are using the contrapositive but then, we have:
$$A\wedge B\implies C$$
And then:
$$¬C \implies ¬(A \wedge B) $$
But I guess this means that we have three possibilities:
- G exists but the second order partial derivatives do not exist.
- G doesn't exists but the second order partial derivatives exist.
- G doesn't exists and the second order partial derivatives do not exist.
If $G$ doesn't exist, we're done. We need only to refute the first case but this means it is possible that $G$ exists and yet, the second order partial derivatives do not exist but we have the curl of $G$, which means that first order partial derivatives could exist. And we only need them to compute $curl \;G$.