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Let $\zeta: \mathbb{R^m} \times \mathbb{R^n} \mapsto \mathbb{R}$ be a smooth function and define $\phi(x) = \inf_{y \in C} \zeta(x,y)$ where $C \subset \mathbb{R^n}$ is compact. Suppose that for every $x \in X \subset \mathbb{R^m}$ there is a unique $y(x) \in C$ such that $\phi(x) = \zeta(x,y(x))$.

Now, according to a book a have, $$\frac{\partial \phi}{\partial x}(x_0) = \frac{\partial \zeta}{\partial x} (x_0, y(x_0)) \quad \forall x_0 \in X$$ (this is let as a exercise). The problem is: 1) this looks simple but I don't where to start to prove this, 2) even if I knew how to deal with the first part, I also need the second differential (the Hessian) of $\phi$ (given in term of $\zeta$, $y$ and their derivatives).

I tried to do something like $$\frac{\partial \phi}{\partial x}(x_0) = \frac{\partial \zeta}{\partial x} (x_0, y(x_0)) + \frac{\partial \zeta}{\partial y} (x_0, y(x_0)) \frac{\partial y}{\partial x}(x_0)$$ (this suggests that $\frac{\partial y}{\partial x}(x_0) = 0$) and find the derivative of $y$ by the implicit function theorem, but I can only get the tautology $\frac{\partial \phi}{\partial x}(x_0) =\frac{\partial \phi}{\partial x}(x_0)$, so I suspect this isn't the good way.

Note: this question is related to this one (Infimum is a continuous function, compact set).

Community
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  • It isn't necessarily true that $\partial_xy(x_0) = 0$, but it should be true that $\partial_y \zeta (x_0,y(x_0)) = 0$, because it is a critical point (actually a minimum, so you also have a sign on $\zeta_{yy}$ at this point) of $\zeta$ in the $y$-coordinate. – Glen Wheeler Mar 30 '11 at 18:19
  • What is the name of the book where you found this exercise question? – user6952886 Apr 29 '22 at 00:54

3 Answers3

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I think that the point is the following: is $y(x)$ sufficiently regular to apply chain rule?

In fact, you know by hypothesis that, for fixed $x$, the partial function $y\mapsto \zeta (x,y)$ takes its minimum in $y(x)$ (which actually is $\phi (x)=\zeta (x,y(x))$), hence you have $\frac{\partial \zeta}{\partial y} (x,y(x))=0$; now if $x\mapsto y(x)$ is sufficiently regular, you can apply the chain rule to get:

$$\frac{\partial \phi}{\partial x} (x) =\frac{\partial}{\partial x} \zeta (x,y(x)) = \frac{\partial \zeta}{\partial x}(x,y(x))$$

as your book claimed... Therefore, I think the real problem here is the regularity of $y(x)$ rather than the computation of the derivative of $\phi(x)$.

Pacciu
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Exactly as Pacciu said, regularity of $y(x)$ is the real problem. I think the key is to apply Implicit Function Theorem.

Fix $x_0\in \mathbb{R}^m$, and let $y_0=y(x_0)$. Since $y_0$ is a strict minimum, $D_{yy}\zeta(x_0,y_0)<0$ (strictly negative definite Hessian, but for simplicity, we can consider 1-dim case: $m=n=1$.) Then, in a neighborhood of $(x_0,y_0)$, $D_{yy}\zeta<0$. Within this neighborhood, $D_{y}\zeta(\hat x, \hat y)=0$ if and only if $\hat y=y(\hat x)$ (where minimum attained). Now, by the Implicit Function Theorem, because $D_{yy}\zeta<0$ locally, $D_y\zeta(\hat x, \hat y)=0$ can be solved locally as $\hat y=\hat y(x)$ with a unique continuous differentiable function $\hat y$. This function also coincides with $y(x)$. So $y(x)$ is $C^1$.

GWu
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  • This should all make sense in the context of a separable Hilbert space too. No? – ABIM Oct 23 '18 at 00:04
  • You do not necessarily have $_y \zeta(_0,_0)=0$. Recall that $C$ is a compact set in $\mathbb{R}^d$. – user6952886 Apr 29 '22 at 01:00
  • A strict minimum only requires $D_{yy} \zeta(x_0, y_0) \le 0$. It is sufficient but not necessary that the inequality be strict. This can be seen with $f(x) = x^4$, which has a unique global minimum at $x = 0$, but has $f''(0) = 0$. – infinitylord Jul 21 '22 at 10:16
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Can you guys provide me with the reference in which this theorem is mentioned? I need to apply this theorem to my work but don't know the book to cite.

user300836
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