[proofplan]
The proof is the compactness argument behind the coercive leading term. Define the scalar coefficient obtained by evaluating $L_{pp}$ along the curve $y$. The hypotheses make this coefficient continuous on the compact interval $[a,b]$, and the strengthened Legendre condition makes it pointwise positive. The extreme value theorem then gives a positive minimum, which is the required uniform lower bound.
[/proofplan]
[step:Define the leading coefficient along the extremal]
Define $g:[a,b]\to\mathbb{R}$ by
\begin{align*}
g(x)=L_{pp}(x,y(x),y'(x)).
\end{align*}
Here $L_{pp}$ denotes the second derivative of $L$ with respect to its third variable.
Since $L\in C^2([a,b]\times\mathbb{R}\times\mathbb{R})$, the function $L_{pp}$ is continuous on $[a,b]\times\mathbb{R}\times\mathbb{R}$. Since $y\in C^2([a,b])$, both $y$ and $y'$ are continuous on $[a,b]$. Therefore the map
\begin{align*}
x\mapsto (x,y(x),y'(x))
\end{align*}
is continuous from $[a,b]$ to $[a,b]\times\mathbb{R}\times\mathbb{R}$, and hence $g$ is continuous on $[a,b]$.
[guided]
Define $g:[a,b]\to\mathbb{R}$ by
\begin{align*}
g(x)=L_{pp}(x,y(x),y'(x)).
\end{align*}
The notation $L_{pp}$ denotes the second derivative of $L$ with respect to its third variable. Since $L\in C^2([a,b]\times\mathbb{R}\times\mathbb{R})$, the function $L_{pp}$ is continuous on $[a,b]\times\mathbb{R}\times\mathbb{R}$. Since $y\in C^2([a,b])$, both $y$ and $y'$ are continuous on $[a,b]$. Therefore the map
\begin{align*}
x\mapsto (x,y(x),y'(x))
\end{align*}
is continuous from $[a,b]$ to $[a,b]\times\mathbb{R}\times\mathbb{R}$. Composing this map with $L_{pp}$ gives the function $g$, so $g$ is continuous on $[a,b]$.
[/guided]
[/step]
[step:Use compactness to obtain a positive minimum]
By the strengthened Legendre condition along $y$ on $[a,b]$,
\begin{align*}
g(x)>0
\end{align*}
for every $x\in[a,b]$. The interval $[a,b]$ is compact, and $g$ is continuous on $[a,b]$. By the extreme value theorem, there exists $x_0\in[a,b]$ such that
\begin{align*}
g(x_0)=\min_{x\in[a,b]} g(x).
\end{align*}
Set
\begin{align*}
\theta=g(x_0).
\end{align*}
Since $g(x_0)>0$, we have $\theta>0$. For every $x\in[a,b]$, the definition of $x_0$ gives
\begin{align*}
g(x)\ge \theta.
\end{align*}
Substituting the definition of $g$ gives
\begin{align*}
L_{pp}(x,y(x),y'(x))\ge \theta
\end{align*}
for every $x\in[a,b]$, as required.
[guided]
The point of this step is to turn pointwise strict positivity into a single uniform constant. The strengthened Legendre condition says that the [continuous function](/page/Continuous%20Function)
\begin{align*}
g(x)=L_{pp}(x,y(x),y'(x))
\end{align*}
is positive at each point of the closed interval $[a,b]$. Pointwise positivity alone would not give a uniform lower bound on a noncompact set, so the compactness of $[a,b]$ is the essential input.
Because $[a,b]$ is compact and $g$ is continuous, the extreme value theorem applies to $g:[a,b]\to\mathbb{R}$. Therefore there is a point $x_0\in[a,b]$ such that
\begin{align*}
g(x_0)=\min_{x\in[a,b]} g(x).
\end{align*}
Define
\begin{align*}
\theta=g(x_0).
\end{align*}
The strengthened Legendre condition applies at the point $x_0$, so $g(x_0)>0$ and hence $\theta>0$. Since $x_0$ is a minimizer, every $x\in[a,b]$ satisfies
\begin{align*}
g(x)\ge \theta.
\end{align*}
Finally, substituting back the definition of $g$ gives
\begin{align*}
L_{pp}(x,y(x),y'(x))\ge \theta
\end{align*}
for every $x\in[a,b]$.
[/guided]
[/step]
