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


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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.


$\Box$


Aiming for a contradiction, suppose there exists $h = h_0$ such that:

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

Then, for any $\alpha \ne 0$:

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

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 $h$.

$\blacksquare$


Sources

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