Condition for Differentiable Functional to have Extremum

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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 $\delta J \sqbrk {\hat y; h}\ge 0$.

Aiming for a contradiction, suppose $\delta J \sqbrk {y; h_0} \ne 0$ 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.


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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:

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

$\blacksquare$


Sources

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