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:


 * $\size h < c \implies \size \epsilon < \dfrac 1 2 k$

Choose $h$ such that this inequality holds.

Then

Therefore:


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


 * $\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$

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