Legendre's Condition/Lemma 2

Lemma 2
Let $ h$ be a real function such that:


 * $ h \in C^1 \left ( { a, b } \right ), \quad h \left ( { a } \right )= 0, \quad h \left ( { b } \right ) = 0 $

Let


 * $ \displaystyle \delta^2 J \left [ { y; h } \right ] = \int_a^b \left ( { P \left ( { x, y \left ( { x } \right ) } \right ) h'^2+Q \left ( { x, y \left ( { x } \right ) } \right ) h^2 } \right ) \mathrm d x $

where $ P \in C^0 \left [ { a \,. \,. \, b } \right ] $.

Then a necessary condition for


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

is


 * $ P \left ( { x, y \left ( { x } \right ) } \right ) \ge 0 \quad \forall x \in \left [ { a \,. \,. \, b } \right ]$

Proof
Assume that above is not true.

Then


 * $ \left( \exists x_0 \in \left [ { a \,. \,. \, b } \right ] \right ) \land \left( \exists \beta \in \R_{ < 0 } \right ) : P \left ( { x_0 } \right ) = - 2 \beta $

$ P $ is continuous.

Thus


 * $ \exists \alpha \in \R_{ > 0 } : \left ( {a \ge x_0 - \alpha } \right ) \land \left ( {x_0 + \alpha \ge b } \right )$

and


 * $ P \left ( { x } \right ) < - \beta \quad \forall x \in \left ( { { x_0 - \alpha } \,. \,. \, { x_0 + \alpha } } \right ) $

In other words:

$ P \left ( { x } \right ) \begin{cases} = 0 \quad \forall x \in \left [ { a \,. \,. \, {x_0 - \alpha} } \right ] \lor \left [ { { x_0 + \alpha } \,. \,. \, b } \right ] \\ < 0 \quad \forall x \in \left [ { { x_0 - \alpha } \,. \,. \, { x_0 + \alpha } } \right ] \end{cases} $

Let


 * $ h = \begin{cases}

\sin^2 \left [ { \frac{ \pi \left ( { x-x_0 } \right)  }{ \alpha  } } \right ] & x_0 - \alpha \ge x \ge x_0 + \alpha \\ 0 & otherwise \end{cases}$

It belongs to $ C^1 \left ( { a, b } \right ) $ because:


 * $ \displaystyle \lim_{ x_0 - \alpha + 0^- } h^{ \left ( { k } \right ) } = 0 \quad \forall k \in \N_{ \ge 0 } $


 * $ \displaystyle \lim_{ x_0 + \alpha + 0^+ } h^{ \left ( { k } \right ) } = 0 \quad \forall k \in \N_{ \ge 0 } $

In other words, only $ h $ and $ h' $ are continuous in $ \left [ {a \,. \,. \,b } \right ] $

Then

where


 * $ \displaystyle M = \max_{ a \le x \le b } \left \vert Q \left ( { x } \right ) \right \vert $

For sufficiently small $ \alpha $ the is negative.

Hence, $ \delta^2 J $ is negative for the corresponding $ h$.

To conclude, it has been shown that


 * $ P \ge 0 \quad \neg \forall x \in \left [ { a \,. \,. \, b } \right ] \implies \delta^2 J <0$

Then, by contrapositive statement this is equivalent to


 * $ \delta^2 J \ge 0 \implies P \ge 0 \quad \forall x \in \left [ { a \,. \,. \, b } \right ]$