f:Rn→Rn is defined by f(x) =(f1(x), f2(x), ... , fn(x)), where x = (x1, x2, ... , xn) and the n functions fi:Rn→R have continuous 2nd order partial derivatives and satisfy ∂fi/∂xj - ∂fj/∂xi = cij (for all 1 ≤ i, j ≤ n) for some constants cij. Prove that there is a function g:Rn→R such that fi + ∂g/∂xi is linear (for all 1 ≤ i ≤ n).
Well, it is easy, provided one can assume the standard result of the vector calculus that if ∂fi/∂xj = ∂fj/∂xi for all i, j, then there is a function g:Rn→R such that f = ∇g.
But can one assume this result? Well, it is pure bookwork, and relatively time-consuming to prove, so presumably yes. Of course, it is a bad question, because it is unclear that you can assume it. The question does not tell you that you can assume it, because that would make it trivial - the only difficulty of the question is that it is slightly inverted, apparently asking about some linear expression rather than about g.
Let hi(x) = (ci1x1 + ci2x2 + ... + cinxn)/2 - f(x). Then ∂hi/∂xj - ∂hj/∂xi = cij/2 - cji/2 - cij = 0, since it follows directly from the definition that cij = - cji.
47th Putnam 1986
© John Scholes
30 Sep 1999