Determining the set of functions that share a minimum in a vector flow Let $U = \{x \in \Bbb R^n : ||x-p|| < \varepsilon\}$ be the open ball with center p and $V: U \to \Bbb R^n$ a vector field with the property that starting at any point in $U$ it will eventually flow towards the point $p$.
I'm conjecturing that for some convex function $f:U \to \Bbb R$ with the property that $p$ is the global minimum, $V= - \nabla f $.
My question is, would all vector fields be of this form? Just a subset? Are there vector fields out there that are an exception to this?
 A: Not to put too fine a point on it, but the conjecture advanced in this question is not true.  Here is a simple, linear system on $\Bbb R^3$ which serves as a counterexample:
By a simple translation of coordinates we can assume that $p = 0$.
Consider the vector field $V$ on $\Bbb R^3$ given by $V = (V_x, V_y,  V_z)^T$, where
$V_x = \alpha x + \beta y, \tag 1$
$V_y = -\beta x + \alpha y, \tag 2$
$V_z = \gamma z, \tag 3$
where $\alpha, \gamma < 0$ and $\beta \ne 0$; the corresponding differential equation may be expressed in matrix-vector form as
$\dot {\mathbf r} = A\mathbf r, \tag 3$
where
$\mathbf r = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \tag 4$
$A = \begin{bmatrix} \alpha & \beta & 0 \\ -\beta & \alpha & 0 \\ 0 & 0 & \gamma \end{bmatrix}; \tag 5$ 
it is easily seen that
$\dot{\Vert \mathbf r(t) \Vert^2} = \dot{\langle \mathbf r(t), \mathbf r(t) \rangle} = 2\langle \mathbf r(t), \dot{\mathbf r(t)} \rangle = 2\langle \mathbf r(t), A\mathbf r(t) \rangle, \tag 6$
and 
$\langle \mathbf r(t), A\mathbf r(t) \rangle = x\dot x + y\dot y + z\dot z = x(\alpha x + \beta y) + y(-\beta x + \alpha y) + z(\gamma z)$
$= \alpha x^2 + \beta xy - \beta yx + \alpha y^2 + \gamma z^2 = \alpha x^2 + \alpha y^2 + \gamma z^2; \tag 7$
setting
$\mu = \sup \{ \alpha, \gamma \}, \tag 8$
we have
$\mu < 0, \tag 9$
hence from (6) and (7),
$\dot{\Vert \mathbf r(t) \Vert^2} = 2(\alpha x^2 + \alpha y^2 + \gamma z^2) \le 2 \mu \Vert \mathbf r(t) \Vert^2; \tag{10}$
now if $\mathbf r(0) \ne 0$, uniqueness of the solution $\mathbf r(t) = 0$ implies we must have $\mathbf r(t) \ne 0$ for all $t$, whence we may divide by $\Vert \mathbf r(t) \Vert$:
$\dfrac{\dot{\Vert \mathbf r(t) \Vert^2}}{\Vert \mathbf r(t) \Vert^2} \le 2\mu, \tag{11}$
or
$\dfrac{d \ln \Vert \mathbf r(t) \Vert^2}{dt} \le 2\mu, \tag{12}$
so that if we integrate 'twixt $0$ and $t > 0$ we find
$\displaystyle \int_0^t \dfrac{d \ln \Vert \mathbf r(s) \Vert^2}{ds} ds \le 2\int_0^t \mu ds; \tag{13}$
thus
$2\ln \Vert \mathbf r(t) \Vert - 2\Vert \ln \mathbf r(0) \Vert = \ln \Vert \mathbf r(t) \Vert^2 - \ln \mathbf \Vert r(0) \Vert^2 \le 2\mu t, \tag{14}$
or
$\ln \dfrac{\Vert \mathbf r(t) \Vert}{\Vert \mathbf r(0) \Vert} \le \mu t, \tag{15}$
so that
$\dfrac{\Vert \mathbf r(t) \Vert}{\Vert \mathbf r(0) \Vert} \le e^{\mu t}, \tag{16}$
and at last
$\Vert \mathbf r(t) \Vert \le \Vert \mathbf r(0)\Vert e^{\mu t}; \tag{17}$
since $\mu < 0$, this implies that $\Vert \mathbf r(t) \Vert$ is monotonically decreasing and that
$\lim_{t \to \infty} \mathbf r(t) = 0. \tag{18}$
We have in fact shown that if we take
$U = \{ \mathbf r \in \Bbb R^3 \mid \Vert \mathbf r \Vert < \varepsilon \}, \tag{19}$
and
$\mathbf r(0) \in U, \tag{20}$
then
$\mathbf r(t) \in U \tag{21}$
under the flow of $V$ for all $t > 0$. 
There is, however, no function $f:U \to \Bbb R$, convex or not, with $V = - \nabla f$; if there were, then we would have
$\nabla \times V = -\nabla \times \nabla f = 0, \tag{22}$
but it follows from (1)-(3) that
$\nabla \times V = \begin{pmatrix} 0 \\ 0 \\ -2\beta \end{pmatrix} \ne 0; \tag{23}$
hence $V$ cannot be a gradient.
Note Added in Edit; Saturday 2 September 2017 9:07 PM PST  I would like to add a few words about our OP Alex Reed's closing questions, and also to address his comment.  It should be clear from this answer that by no means "all vector fields be of this form"; indeed, the example $V$ presented here was constructed to not be a gradient vector field.  So, yes, just a subset of all vector fields are gradients; this is manifested in the fact that $V = \nabla f$ for some function $f$ if and only if $\nabla \times V = 0$.  The fact that gradient fields $V$ are constrained to satisfy $\nabla \times V = 0$ constrains them to lie in some subspace of the linear space of vector fields on $U$; for example, if we consider the space of $k$-times differentiable fields $C^k(U, \Bbb R^3)$, then the fields $V$ such that $\nabla \times V = 0$ form a closed subspace in the $\Vert \cdot \Vert_k$ topology; also, the fields with $\nabla \times V \ne 0$ form an open subset of $C^k(U, \Bbb R^3)$; these facts, which follow from the observation that $\nabla \times$ is a bounded linear map from $C^k(U, \Bbb R^3) \to C^{k - 1}(U, \Bbb R^3)$, show that in fact most vector fields in $C^k(U, \Bbb R^3)$ are not gradients. End of Note.
