Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it true that any smooth function $f\colon \mathbb{R}^n \to \mathbb{R}^n$ can be represented as $$ f(x) = \nabla U(x) + g(x) $$ where $U(x)$ is a scalar function and $\langle g(x), f(x) \rangle \equiv 0$? Is this representation unique?

share|cite|improve this question
up vote 2 down vote accepted

Let me summarize my comments as an answer to you original question. Given a smooth map $f$, consider a problem of existence of a pair $(U,g)$ such that $U$ is a smooth scalar function and $g$ is a smooth map and such that $$ f = \nabla U+g, \tag{1} $$ $$ f\cdot g = 0. \tag{2} $$

Multiplying both sides in $(1)$ by $f$, we obtain $$ \|f\|^2 - f\cdot \nabla U = f\cdot g = 0. $$ Hence, problem $(1)+(2)$ can be reduced to the problem of existence of $U$ such that $$ f\cdot \nabla U = \|f\|^2 \tag{3} $$ and if the solution of the latter 1st order linear PDE exists, then $g = f - \nabla U$ is the function needed in $(1),(2)$. Unfortunately, I cannot say anything about the existence of solution of this PDE in the general case (i.e. when $f$ is smooth).

Uniqueness does not hold: take $f = 0$, then we are looking for $(U,g)$ such that $$ g = - \nabla U. $$ Clearly, any $U$ smooth solves the problem.

share|cite|improve this answer

This is not true as Ilya said in his comment, however if we modify the statement:

Smooth vector field with compact support $f: \mathbb{R}^n \to \mathbb{R}^n$ can be represented as $f = \nabla U + g$, where $\langle g, \nabla U \rangle = 0$.

And this is Helmholtz decomposition for smooth vector fields in $\mathbb{R}^n$, we can set up a variational problem as follows: Find $U\in H^1_c(\mathbb{R}^n)$, such that $$ \langle \nabla U, \nabla v \rangle = \langle f, \nabla v \rangle = -\langle \mathrm{div}f, v \rangle \quad \text{ for any } v\in H^1_c(\mathbb{R}^n) $$ This is an elliptic problem with a unique solution $U$, letting $g = f- \nabla U$ will give what you want, and $g$ is orthogonal to $\nabla U$, then the decomposition is unique, you won't have a problem like Ilya mentioned in the first comment, set $f=0$ would get you $g = -\nabla U$, and the orthogonality implies $\nabla U = 0$ and $U$ is a constant, by compact supportedness, $U=0$.

share|cite|improve this answer
Sounds interesting – Nimza Jun 27 '12 at 15:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.