Condition for Differentiable Functional to have Extremum

Theorem
Let $S$ be a set of mappings.

Let $y, h \in S: \R \to \R$ be real functions.

Let $J \sqbrk y: S \to \R$ be a differentiable functional.

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


 * $\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$

Proof
Suppose $J \sqbrk {y; h}$ attains a minimum for $y = \hat y$.

Then:


 * $\Delta J \sqbrk {\hat y; h} \ge 0$

By definition of the differentiable functional:


 * $\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \epsilon \size h$

where:
 * $\ds \lim_{\size h \mathop \to 0} \epsilon = 0$

Hence, there exists $\size h$ small enough, that signs of $\Delta J \sqbrk {y; h}$ and $\delta J \sqbrk {y; h}$ match.

Therefore, for $\size h$ small enough it holds that $\delta J \sqbrk {\hat y; h}\ge 0$.

$\delta J \sqbrk {y; h_0} \ne 0$ for some $h_0 \in S$.

Then:


 * $\forall \alpha > 0: \delta J \sqbrk {y; -\alpha h_0} = -\delta J \sqbrk {y; \alpha h_0}$

Therefore, for $\size h$ however small, $\Delta J \sqbrk {y; h}$ can be made to have any sign.

However, by assumption that $J \sqbrk {y; h}$ has a minimum, for sufficiently small $\size h$ it holds that:


 * $\Delta J \sqbrk {\hat y; h} = J \sqbrk {\hat y + h} - J \sqbrk {\hat y} \ge 0$

This automatically fixes the sign of $\delta J \sqbrk {y; h}$.

This is in contradiction with the previous statement.

Hence for all $h$ it holds that:


 * $\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$