Sufficient Condition for Twice Differentiable Functional to have Minimum

Theorem
Let $J$ be a twice differentiable functional.

Let $J$ have an extremum for $y=\hat y$.

Let the second variation $\delta^2 J\sqbrk{\hat y;h}$ be strongly positive $h$.

Then $J$ acquires the minimum for $y=\hat y$.

Proof
By assumption, $J$ has an extremum for $y=\hat y$:


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

The increment is expressible then as:


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

where $\epsilon\to 0$ as $\size h\to 0$.

By assumption, the second variation is strongly positive:


 * $\delta^2 J\sqbrk{\hat y;h}\ge k\size {h}^2,\quad k\in\R_{>0}$

Hence,


 * $\Delta J\sqbrk{\hat y;h}\ge\paren {k+\epsilon} \size {h}^2$

What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that is always positive.

Since $\epsilon\to 0$ as $\size h\to 0$, there exist $c\in\R_{>0}$, such that


 * $\size h\frac 1 2 k\size {h}^2 $

For $k\in\R_{>0}$ and $\size h\ne 0$ is always positive.

Thus, there exists a neighbourhood around $y=\hat y$ where the increment is always positive:


 * $\exists c\in\R_{>0}:\size h0$

and $J$ has a minimum for $y=\hat y$.