Legendre's Condition/Lemma 2

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


 * $h \in C^1 \openint a b$
 * $\map h a = 0$
 * $\map h b = 0$

Let:


 * $\displaystyle \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$

where $P \in C^0 \closedint a b$.

Then a necessary condition for:
 * $\delta^2 J \sqbrk {y; h} \ge 0$

is:
 * $\forall x\in\closedint a b: \map P {x, \map y x} \ge 0$

Proof
Assume that above is not true.

Then:


 * $\paren {\exists x_0 \in \closedint a b} \land \paren {\exists \beta \in \R_{<0} }: \map P {x_0} = -2 \beta$

$P$ is continuous.

Thus:


 * $\exists \alpha \in \R_{>0}: \paren {a \ge x_0 - \alpha} \land \paren {x_0 + \alpha \ge b}$

and:


 * $\forall x \in \openint {x_0 - \alpha} {x_0 + \alpha}: \map P x < -\beta$

In other words:

$\map P x \begin {cases} = 0 & : x \in \closedint a {x_0 - \alpha} \lor \closedint {x_0 + \alpha} b \\ < 0 & : x \in \closedint {x_0 - \alpha} {x_0 + \alpha} \end {cases}$

Let


 * $h = \begin {cases}

\sin^2 \sqbrk {\dfrac {\map \pi {x - x_0} } \alpha} & : x_0 - \alpha \ge x \ge x_0 + \alpha \\ 0 & : \text {otherwise} \end{cases}$

It belongs to $C^1 \openint a b$ because:


 * $\displaystyle \forall k \in \N_{\ge 0}: \lim_{x_0 - \alpha + 0^-} h^{\paren k} = 0$


 * $\displaystyle \forall k \in \N_{\ge 0}: \lim_{x_0 + \alpha + 0^+} h^{\paren k} = 0$

In other words, only $h$ and $h'$ are continuous in $\closedint a b$

Then:

where:


 * $\displaystyle M = \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$

For sufficiently small $\alpha$ the is negative.

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

To conclude, it has been shown that


 * $P \ge 0 \quad \neg \forall x \in \closedint a b \implies \delta^2 J<0$

Then, by contrapositive statement this is equivalent to:


 * $\forall x \in \closedint a b: \delta^2 J \ge 0 \implies P \ge 0$