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Denote $\mathcal F$ as the function class consisting of gradients of all real-valued convex functions in $\mathbb R^d$, that is, $\mathcal F = \{ \nabla \phi ~|~ \phi: \mathbb R^d \to \mathbb R \text{ and $\phi$ is convex}\}$. Note that every element of $\mathcal F$ is a function from $\mathbb R^d$ to $\mathbb R^d$. Then is $\mathcal F$ closed under composition operation? That is, suppose $f \in \mathcal F$ and $ g\in \mathcal F$, do we have $f\circ g \in \mathcal F$ where $\circ$ denotes function composition?

Note: the statement should be true for $d = 1$ since:

  1. Gradient of a univariate real-valued convex function is non-decreasing;
  2. Composition of two non-decreasing functions is still non-decreasing;
  3. Non-decreasing gradient corresponds to a convex function.
Ethan
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  • So, the question boils down to whether $f \circ g$ can be written as $\nabla \phi$ for some real-valued, convex, (probably want $C^1$ too) function $\phi: \mathbb{R}^d \rightarrow \mathbb{R}$. – Selrach Dunbar May 19 '22 at 04:19
  • So $\mathcal{F}$ is a subset of the set of conservative vector fields on $\mathbb{R}^d$ (to use some lingo from multivariable calculus). (Most of the time folks talk about these the discussion immediately goes to line integrals.) Not to take over this post, but can anyone quickly just confirm that if we (temporarily) ignore the convexity requirement, the set of conservative vector fields is closed under composition? – Selrach Dunbar May 19 '22 at 04:45
  • So, just FYI, your question prompted me to ask the related question (in the preceding comment) concerning what happens if we drop the convexity requirement. Here's a link to that question. The short answer is no, the set $\mathcal{F}'$ (your $\mathcal{F}$ with the convexity requirement dropped) is not closed under composition. That said, perhaps the convexity requirement saves it? (Restricts it enough that the restricted set is closed under composition?) – Selrach Dunbar May 23 '22 at 18:24
  • Although his description is a bit sparse, @max_zorn has offered a valid counterexample to show $\mathcal{F}$ is not closed under composition.

    In max_zorn 's answer below he identifies:

    • $f(x,y)= x^2+xy+\frac{1}{2}y^2=\frac{1}{2}[x,y]\begin{bmatrix} 2 & 1\ 1 & 1\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$ and
    • $g(x,y) = \frac{5}{2}x^2+2xy+\frac{1}{2}y^2=\frac{1}{2}[x,y]\begin{bmatrix} 5 & 2\ 2 & 1\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$.
     – Selrach Dunbar May 24 '22 at 03:31
  • Hence, $F := \nabla f = \left<2x+y,x+y \right> =\begin{bmatrix} 2 & 1\ 1 & 1\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$ and $G := \nabla g = \left<5x+2y,2x+y \right> = \begin{bmatrix} 5 & 2\ 2 & 1\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$.

    Thus, $F \circ G: \langle x,y \rangle\mapsto \left<12x+5y,7x+3y\right>=\begin{bmatrix} 2 & 1\ 1 & 1\end{bmatrix} \begin{bmatrix} 5 & 2\ 2 & 1\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}=\begin{bmatrix} 12 & 5\ 7 & 3\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$ which is not the gradient of any scalar function $\phi$.

     – Selrach Dunbar May 24 '22 at 03:32

1 Answers1

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Set
$$A = \begin{pmatrix} 2 & 1\\ 1 & 1\end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix} 5 & 2\\ 2 & 1\end{pmatrix}.$$ Both matrices are positive definite because their diagonal entries are positive as are their determinants. Hence, they are gradients of convex functions, namely $\tfrac{1}{2}\langle x,Ax\rangle$ and $\tfrac{1}{2}\langle x,Bx\rangle$, respectively.

However, $$AB = \begin{pmatrix} 12 & 5 \\ 7 & 3\end{pmatrix}$$ is not even symmetric, so it cannot be a gradient of a convex function.

Note: The matrices were already used in a sister question: Product of any two arbitrary positive definite matrices is positive definite or NOT?

max_zorn
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  • This is NOT a counter example (the function class $\mathcal F$ consists of GRADIENTS of all real-valued convex functions). $\nabla g = 2x$ and $\nabla f = -1$, so $\nabla g \circ \nabla f = -2$ which corresponds to a convex function $x \mapsto -2x + const.$. Also, $\nabla f \circ \nabla g = -1$ corresponding to convex function $x \mapsto -x + const.$ – Ethan May 20 '22 at 15:11
  • @TrueRealEason I misread. Working on a better example. – max_zorn May 21 '22 at 01:53
  • Unfortunately, I don't think this second example is going to scratch the itch either. The questions asks about gradients of real-valued functions. These appear to be Jacobians of vector-valued functions.

    That is, $A = D \overrightarrow{ F ;}= \left<\nabla f_1, \nabla f_2\right>$ where $\overrightarrow{ F ;}(x,y) = \left<f_1 (x,y), f_2(x,y)\right> = \left<2x+y, x+y\right>$.

     – Selrach Dunbar May 21 '22 at 16:24
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    Please read the “Hence” sentence where I give the functions. – max_zorn May 23 '22 at 01:26
  • Nice! I see it now. Didn't understand what you were saying at first. – Selrach Dunbar May 24 '22 at 03:37
  • @max_zorn This follow-up question might be irrelevant, but in your example, is it possible that $x \mapsto AB x$ belongs to the subgradient set of a real-valued convex function (I am not assuming continuous differentiability here)? – Ethan Jun 10 '22 at 02:36
  • @Selrach Dunbar IF that is the case (see my previous comment please), even though $AB$ is not positive semidefinite, the argument in the counterexample won't work because no continuous differentiability is assumed there. – Ethan Jun 10 '22 at 02:40
  • @TrueRealEason The answer is NO. $AB$ is actually maximally monotone. If it was a selection of $\partial f$, then it must coincide with $\partial f$ because that guy is also maximally monotone. So we would have that $AB = \nabla f$ but $AB$ is not positive semidefinite. – max_zorn Jun 10 '22 at 06:36
  • @max_zorn Thanks, that is helpful! So this one can also be served as a COUNTEREXAMPLE for the statement "the composition of two cyclically monotone maps is cyclically monotone", am I right? – Ethan Jun 10 '22 at 16:59
  • You are correct! – max_zorn Jun 11 '22 at 17:26