Sufficient Condition for Twice Differentiable Functional to have Minimum

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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 with respect to $h$.


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 right hand side is always positive.

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

$\size h<c\implies\size \epsilon<\frac 1 2 k$

Choose $h$ such that this inequality holds.

Then

\(\displaystyle \frac 1 2 k\) \(>\) \(\displaystyle \epsilon>-\frac 1 2 k\) $ \big \vert+k$, by Membership is Left Compatible with Ordinal Addition
\(\displaystyle \frac 3 2 k\) \(>\) \(\displaystyle k+\epsilon>\frac 1 2 k\) $\big\vert\cdot\size {h}^2$, by Membership is Left Compatible with Ordinal Multiplication
\(\displaystyle \frac 3 2 k\size {h}^2\) \(>\) \(\displaystyle \paren {k+\epsilon} \size {h}^2>\frac 1 2 k\size {h}^2\)

Therefore:

$\Delta J\sqbrk {\hat y;h}\ge\paren{k+\epsilon}\size {h}^2>\frac 1 2 k\size {h}^2 $

For $k\in\R_{>0}$ and $\size h\ne 0$ right hand side 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 $y=\hat y$.

$\blacksquare$


Sources