Necessary Condition for Twice Differentiable Functional to have Minimum

Theorem
Let $ J \left[ { y } \right]$ be a twice differentiable functional.

Then $ J$ has a minimum for $ y= \hat { y }$ if


 * $ \delta^2 J \left [ { y; h } \right] \le 0$

for $ y= \hat { y }$ and all admissible $ h$.

Proof
By definition, $ \Delta J \left [ { y } \right] $ can be expressed as:


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

According to theorem


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

Hence:


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

and $ \Delta J \left [ { \hat { y }; h } \right ] $ and $ \delta^2 J \left [ { \hat { y } } \right ] $ will have the same sign for sufficiently small $ \epsilon$.

Suppose:


 * $ \delta^2 J \left [ { \hat { y }; h_0 } \right ] <0$

Then, for any $ \alpha \ne 0$

Therefore, $ \Delta J \left [ { \hat { y }; h } \right ]$ can be made negative for arbitrary $ \left \vert h \right \vert$.

However, $ \Delta J \left [ { \hat { y }; h } \right ]$ is a minimum of $ \Delta J \left [ { y ; h } \right ] $ for all sufficiently small $ \left \vert h \right \vert$.

This is a contradiction.

Thus, a function $ h_0 : \delta^2 J \left [ { \hat { y }; h_0 } \right ] <0$ does not exist.

In other words:


 * $ \delta^2 J \left [ { \hat { y }; h } \right ] \ge 0$

for all $ h$.