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 \left[ { \hat { y }; h } \right ]$ 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 \left [ { \hat { y }; h } \right ]=0$

The increment is expressible then as:


 * $ \Delta J \left [ { \hat { y }; h } \right ]= \delta^2 J \left [ { \hat { y }; h  } \right ]+ \epsilon \left \vert h \right \vert^2$

where $ \epsilon \to 0$ as $ \left \vert h \right \vert \to 0$.

By assumption, the second variation is strongly positive:


 * $ \delta^2 J \left [ { \hat { y }; h } \right ] \ge k \left \vert h \right \vert^2, \quad k \in \R_{ > 0 }$

Hence,


 * $ \Delta J \left [ { \hat { y }; h } \right ] \ge \left ( { k + \epsilon } \right ) \left \vert h \right \vert^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 $ \left \vert h \right \vert \to 0$, there exist $c \in \R_{ > 0 }$, such that


 * $ \left \vert h \right \vert < c  \implies \left \vert \epsilon \right \vert < \frac{ 1 }{ 2 } k $

Choose $ h $ such that this inequality holds.

Then

Therefore:


 * $ \Delta J \left [ { \hat { y }; h } \right ] \ge \left ( { k + \epsilon } \right ) \left \vert h \right \vert^2 >   \frac{ 1 }{ 2 }k  \left \vert h \right \vert^2 $

For $ k \in \R_{ > 0}$ and $ \left \vert h \right \vert \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} : \left \vert h \right \vert < c \implies \Delta J \left [ { \hat { y }; h } \right ]>0$

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