Suppose $\phi (x,z)$ is a continuous function of two arguments,
$$\phi :{\mathbb {R} }^{n}\times Z\rightarrow {\mathbb {R} }$$
where $Z\subset {\mathbb {R} }^{m}$is a compact set. Further assume that $ \phi (x,z)$is convex in $x$ for every $z\in Z$.
Define$$f(x)=\max _{z\in Z}\phi (x,z)$$
and
$$Z_{0}(x)=\left\{{\overline {z}}:\phi (x,{\overline {z}})=\max _{z\in Z}\phi (x,z)\right\}.$$
Danskin's theorem: $f(x)$ is differentiable at $x$ if $Z_{0}(x)$ consists of a single element $\overline {z}$. Furthermore, the derivative of $f(x)$ is given by$$ {\frac {\partial f}{\partial x}}={\frac {\partial \phi (x,{\overline {z}})}{\partial x}}.$$
Taken from wikipedia
My question: When evaluating $\frac {\partial \phi (x,{\overline {z}})}{\partial x}$ at a given $x$, is $\overline z$ is to be treated as a constant or as a function of $x$?
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$\begingroup$$\bar{z}$ is a constant for the purposes of the partial derivative. In fact, the points $\bar{z}$ are just the points where $\phi(x,z)$ attains its maximum on the set $Z$ so you could, if it makes things clearer, write the partial derivative as$$ \frac{\partial f}{\partial x} = \frac{\partial \phi(x,z)}{\partial x}{\left| _{z\in Z_0}\right.} $$
The set $Z_0$ does not depend on $x$ in particular.
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