If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes

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Theorem

Let $\map \alpha x$ be a continuous real function on the closed real interval $\closedint a b$.

Let $\displaystyle \int_a^b \map \alpha x \map h x \rd x = 0$ for every real function $\map h x \in C^0 \closedint a b$ such that $\map h a = 0$ and $\map h b = 0$.



Then $\map \alpha x = 0$ for all $x \in \closedint a b$.


Proof

Aiming for a contradiction, suppose the real function $\map \alpha x$ is nonzero at some point in $\closedint a b$ for some arbitrary $\map h x$.

Due to belonging to $C^0$ it is also nonzero in some interval $\closedint {x_1} {x_2}$ contained in $\closedint a b$.


Let us choose $\map h x$ to be of a specific form, while still satisfying the requirements in the statement of the theorem:

$\map h x = \begin {cases} \map \sgn {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} & : x \in \closedint {x_1} {x_2} \\ 0 & : x \notin \closedint {x_1} {x_2} \end {cases}$

Then:

$\displaystyle \int_a^b \map \alpha x \map h x \rd x = \int_{x_1}^{x_2} \size {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} \rd x$

where we used the fact that:

$\map \alpha x = \map \sgn {\map \alpha x} \size {\map \alpha x}$

as well as:

$\map {\sgn^2} x = 1$ if $x \ne 0$ and $x \in \R$.

The integrand is positive for $x \in \closedint {x_1} {x_2}$, whence the integral is positive.

However, that contradicts the condition on the integral in the statement of the theorem.

Thus, with the provided assumption the condition for the integral does not hold for all $\map h x$ with aforementioned conditions.

Hence the result by Proof by Contradiction.

$\blacksquare$


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