Let's say I have a differentiable function $F: \mathbb{R}^n \to \mathbb{R}$ with Lipschitz continuous gradient $\nabla F$. Then I take a function $\phi:\mathbb{R} \to \mathbb{R}$, and consider the vector where $\phi$ has been applied to each of the components of $\nabla F$, i.e., $$ \left( \phi \left(\frac{\partial F(x)}{\partial x_1} \right), \dots, \phi \left(\frac{\partial F(x)}{\partial x_n} \right) \right).$$ Are there any requirements (except trivial like $\phi$ being linear) that I can put on $\phi$ for there to exist a function $F_{\phi}: \mathbb{R}^n \to \mathbb{R}$ such that the partial derivatives of $F_{\phi}$ are given by $$ \frac{\partial F_{\phi}(x)}{\partial x_n} = \phi \left(\frac{\partial F(x)}{\partial x_n} \right) ? $$


1 Answer 1


This is an interesting question. Unfortunately, I think the answer is "no". In the one dimensional case, any $\phi$ which has an integral should work. You mention that $\nabla F$ is Lipschitz continuous. This means that at least the weak second derivatives exist. In particular this means that

$$\frac{\partial^2 F_\phi(\mathbf{x})}{\partial x_1 \partial x_2} = \phi'\left(\frac{\partial F(\mathbf{x})}{\partial x_1}\right) \frac{\partial^2 F(\mathbf{x})}{\partial x_1 \partial x_2} = \phi'\left(\frac{\partial F(\mathbf{x})}{\partial x_2}\right) \frac{\partial^2 F(\mathbf{x})}{\partial x_2 \partial x_1}$$

Now, in particular if $F$ and $F_\phi$ are well behaved such that we can interchange mixed partials, then unless we wish to say each mixed partial of $F$ is identically zero, we must have

$$\phi'\left(\frac{\partial F(\mathbf{x})}{\partial x_1}\right) = \phi'\left(\frac{\partial F(\mathbf{x})}{\partial x_2}\right) = ... = \phi'\left(\frac{\partial F(\mathbf{x})}{\partial x_n}\right)$$

From here it seems pretty clear that at least sometimes $\phi'$ must be constant, and $\phi$ linear, at least on the union of the ranges of the partial derivatives. For example take the $2D$ example

$$ F(x,y) = x^2y$$

Now we have

$$\phi'\left(2xy\right) = \phi'\left(x^2\right)$$

If we take $y=0$ we immediately get that $\phi(s) = \phi(0)$ for any positive $s$. But then since $x^2$ is nonnegative, $\phi'(2xy) = \phi'(x^2) = \phi'(0)$. Thus $\phi$ is linear.


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