1

Suppose there is a problem

$$\min\limits_v\max\limits_x E(v,x).$$

$E$ is a concave function w.r.t. $x$. But w.r.t. $v$, $E$ is a convex function plus a concave function.

I can get $x^*=\arg\max\limits_x E(v,x)=\phi(v)$.

Since $E$ is a convex function plus a concave function, it is hard for me to find the minimal even all stationary points.

If I fix $x$, then I calculate the partial derivative $\frac{\partial E}{\partial v}=f(x,v)$. Then I plug $x^*=\phi(v)$ into $f(\phi(v),v)$.

But another case is first I plug $x^*=\phi(v)$ into $E(v,x)=E(v,\phi(v))=G(v)$. (This is more complex. ) Then I calculate the derivative of $\frac{dG}{dv}$.

My question is should $f(\phi(v),v)$ be equal to $\frac{dG}{dv}$?

Heather
  • 163

1 Answers1

1

No. Note that by the chain rule $$ G'(v) = \partial_x E\bigl(\phi(v), v\bigr)\phi'(v) + \partial_v E\bigl(\phi(v),v\bigr) = \partial_x E\bigl(\phi(v), v\bigr)\phi'(v) + f\bigl(\phi(v),v\bigr) $$ So $G' = f\bigl(\phi(\cdot), \cdot\bigr)$ only if $\partial_x E(\phi(\cdot),\cdot) \phi' = 0$.

martini
  • 84,101
  • Thank you very much for your answer. $x^=\phi(v), {\partial _xE(v,x^)}=0={\partial_x E(v,\phi(v))}=0$. So $\partial \limits_x E(\phi(v),v)= 0$ always be true? – Heather Jul 11 '14 at 08:31
  • So $$\partial _x E(\phi(v),v)= 0$$ always be true? – Heather Jul 11 '14 at 08:38