Legendre's Condition/Lemma 2

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


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

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


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

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


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

In other words:

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

Let


 * $h=\begin{cases}

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

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


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


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

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

Then

where


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

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


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