Condition for Differentiable Functional to have Extremum

Theorem
Let $ S $ be a set of mappings.

Let $ y, h \in S $.

Let $ J \left [ { y } \right ] : S \to \R $ be a differentiable functional.

Then a necessary condition for the differentiable functional $ J \left [ { y; h } \right ] $ to have an extremum for $ y = \hat { y } $ is:


 * $ \displaystyle \delta J \left [ { y; h } \right ] \bigg \rvert_{ y = \hat { y } } = 0 $

Proof
Suppose $J \left [ { y; h } \right ] $ acquires a minimum for $ y = \hat { y } $.

Then:


 * $ \displaystyle \Delta J \left [ { \hat { y }; h } \right ] \ge 0 $

By definition of the differentiable functional,


 * $ \Delta J \left [ { y; h } \right ] = \delta J \left [ { y; h } \right ] + \epsilon \left\vert { h } \right \vert $

where:
 * $ \displaystyle \lim_{ \left\vert{h}\right\vert\to 0 } \epsilon = 0 $.

Hence, there exists $ \left \vert { h } \right \vert $ small enough, that signs of $ \Delta J \left [ { y; h } \right ] $ and $ \delta J \left [ { y; h } \right ] $ match.

Therefore, for $ \left \vert { h } \right \vert $ small enough it holds that $ \delta J \left [ { \hat { y }; h } \right ] \ge 0 $.

Suppose, $ \delta J \left [ { y; h_0 } \right ] \ne 0 $ for some $ h_0 \in S $.

Then:


 * $ \displaystyle \forall \alpha > 0 : \delta J \left [ { y; - \alpha h_0 } \right ] = - \delta J \left [ { y; \alpha h_0 } \right ] $

Therefore, for $\left\vert{h}\right\vert$ however small $\Delta J[y;h]$ can be made to have any sign.

However, by assumption that $ J \left [ { y; h } \right ] $ has a minimum, for sufficiently small $ \left \vert { h } \right \vert $ it holds that:


 * $\displaystyle \Delta J \left [ { \hat { y }; h } \right ] = J \left [ { \hat { y } + h } \right ] - J \left [ { \hat { y } } \right ] \ge 0 $

which automatically fixes the sign of $ \delta J \left [ { y; h } \right ] $.

This is a contradiction.

Hence, for all $ h $ it holds that:


 * $ \displaystyle \delta J \left [ { y; h } \right ] \bigg \rvert_{ y = \hat { y } } = 0 $