Necessary Condition for Twice Differentiable Functional to have Minimum

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Theorem

Let $J\sqbrk y$ be a twice differentiable functional.

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

Suppose, for $y=\hat y$ and all admissible $h$

$\delta^2 J\sqbrk{y;h}\ge 0$


Then $J$ has a minimum for $y=\hat y$ if

Proof

By definition, $\Delta J\sqbrk y$ can be expressed as:

$\Delta J\sqbrk{y;h}=\delta J\sqbrk{y;h}+\delta^2 J\sqbrk{y;h}+\epsilon\size {h}^2$

By assumption:

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

Hence:

$\Delta J \sqbrk{\hat y;h}=\delta^2 J\sqbrk{\hat y;h}+\epsilon\size {h}^2 $

Therefore, for sufficiently small $\size h$ both $\Delta J\sqbrk{\hat y;h}$ and $\delta^2 J\sqbrk{\hat y;h}$ will have the same sign.


$\Box$

Suppose, there exists $h=h_0$ such that:

$\delta^2 J\sqbrk{\hat y;h_0}<0$

Then, for any $\alpha\ne 0$

\(\displaystyle \delta^2 J\sqbrk{\hat y;\alpha h_0}\) \(=\) \(\displaystyle \alpha^2\delta^2 J\sqbrk{\hat y;h_0}\)
\(\displaystyle \) \(<\) \(\displaystyle 0\)

Therefore, $\Delta J\sqbrk{\hat y;h}$ can be made negative for arbitrary small $\size h$.

However, by assumption $\Delta J\sqbrk{\hat y;h}$ is a minimum of $\Delta J\sqbrk{y;h}$ for all sufficiently small $\size h$.

This is a contradiction.

Thus, a function $h_0:\delta^2 J\sqbrk{\hat y;h_0}<0$ does not exist.

In other words:

$\delta^2 J\sqbrk{\hat y;h}\ge 0$

for all $h$.

$\blacksquare$

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