Theorem: Nonnegative Quadratic Functional implies no Interior Conjugate Points

If the quadratic functional 
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x$
where:
:$\forall x \in \closedint a b: \map P x > 0$
is nonnegative for all $\map h x$:
:$\map h a = \map h b = 0$
then the closed interval $\closedint a b$ contains no inside points conjugate to $a$.
In other words, the open interval $\openint a b$ contains no points conjugate to $a$.
{{explain|Rewrite the above so it makes better sense. For example, should the "nonnegative" comment be above the condition on $\map P x$?}}

Proof:
Consider the functional:
:$\forall t \in \closedint 0 1: \ds \int_a^b \paren {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$
By assumption:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x \ge 0$
For $t = 1$, Euler's Equation reads:
:$\map {h''} x = 0$
which, along with condition $\map h a = 0$, is solved by:
:$\map h x = x - a$ 
for which there are no conjugate points in $\closedint a b$.
In other words:
:$\forall x \in \openint a b: \map h x > 0$
Hence:
:$\forall t \in \closedint 0 1: \ds \int_a^b \paren {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x \ge 0$
The corresponding Euler's Equation is:
:$2 Q h t - \map {\dfrac \d {\d x} } {2 t P h' + 2 h' \paren {1 - t} } = 0$
which is equivalent to:
:$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$
Let $\map h {x, t}$ be a solution to this such that:
:$\forall t \in \closedint 0 1: \map h {a, t} = 0$
:$\map {h_x} {a, t} = 1$
{{explain|What is $h_x$?}}
Suppose that for $\map h {x, t}$ there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$.
In other words:
:$\exists \tilde a \in \closedint a b: \map h {\tilde a, 1} = 0$
By definition, $a \ne \tilde a$.
Suppose $\tilde a = b$.
Then by lemma 1 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$
This agrees with the assumption.
Therefore, it is allowed that $\tilde a = b$.
For $t = 1$, any other conjugate point of $\map h {x, t}$ may reside only in $\openint a b$.
Consider the following set of all points $\tuple {x, t}$:
:$\set {\tuple {x, t}: \paren {\forall x \in \closedint a b} \paren {\forall t \in \closedint 0 1} \paren {\map h {x, t} = 0} }$
If it is non-empty, it represents a curve in $x - t$ plane, such that $h_x \left({x, t}\right) \ne 0$.
{{explain|Rather than using "it", give it a name and reference that name.}}
By the Implicit Function Theorem, $\map x t$ is continuous.
By hypothesis, $\tuple {\tilde a, 1}$ lies on this curve.
Suppose that the curve starts at this point.
The curve can terminate either inside the rectangle or its boundary.
If it terminates inside the rectangle $\closedint a b \times \closedint 0 1$, it implies that there is a discontinuous jump in the value of $h$.
{{explain|Again, rather than using "it", refer to the object in question directly, so it is clear what is being referred to.}}
:Therefore, it contradicts the continuity of $\map h {x, t}$ in the interval $t \in \closedint 0 1$.
{{explain|Specify what contradicts what, by invoking the AimForCont construct}}
If it intersects the line segment $x = b, 0 \le t \le 1$, then by lemma 2 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite it vanishes.
{{explain|"it" again}}
:This contradicts positive-definiteness of the functional for all $t$.
{{explain|See above note on the above contradiction proof}}
If it intersects the line segment $a \le x \le b, t = 1$, then $\exists t_0: \paren {\map h {x, t_0} = 0} \land \paren {\map {h_x} {x, t_0} = 0}$.
{{explain|$h_x$ again}}
If it intersects $a \le x \le b, t = 0$, then Euler's equation reduces to $h'' = 0$ with solution $h = x - a$, which vanishes only for $x = a$.
If it intersects $x = a, 0 \le t \le 1$, then $\exists t_0: \map {h_x} {a, t_0} = 0$
{{explain|"it" again}}
{{Help|explain cases $t = 1$ and $x = a$}}
By Proof by Cases, no such curve exists.
Thus, the point $\tuple {\tilde a, 1}$ does not exist, since it belongs to this curve.
Hence there are no conjugate points of $\map h {x, 1} = \map h x$ in the interval $\openint a b$.
{{qed}}