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:

$\delta J\sqbrk{y;h}\bigg\rvert_{y=\hat y}=0$


Proof

Suppose $J\sqbrk{y;h}$ acquires 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:

$\displaystyle\lim_{\size h \to 0}\epsilon=0$.

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

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.



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:

$\delta J\sqbrk{y;h}\bigg\rvert_{y=\hat y}=0$


$\blacksquare$


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