# Necessary Condition for Twice Differentiable Functional to have Minimum

## 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

{{explain|if what?}

## 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 $\alpha \ne 0$:

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

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 $\size h$.

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