Sufficient Condition for Twice Differentiable Functional to have Minimum

This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

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:

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

Choose $h$ such that this inequality holds.

Then

\(\ds \frac 1 2 k\)

\(>\)

\(\ds \epsilon > -\frac 1 2 k\)

$\big \vert + k$, by Membership is Left Compatible with Ordinal Addition

\(\ds \frac 3 2 k\)

\(>\)

\(\ds k + \epsilon > \frac 1 2 k\)

$\big \vert \cdot \size h^2$, by Membership is Left Compatible with Ordinal Multiplication

\(\ds \frac 3 2 k \size h^2\)

\(>\)

\(\ds \paren {k + \epsilon} \size h^2 > \frac 1 2 k \size h^2\)

This article, or a section of it, needs explaining. In particular: What does this mean? : $\big \vert + k$ and $\big \vert \cdot \size h^2$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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 $\size h \ne 0$ 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

• 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.24$: Quadratic Functionals. The Second Variation of a Functional