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

Theorem
If $\alpha\left(x\right)$ is continous in $\left[{{a}\,.\,.\,{b}}\right]$, and if $\displaystyle \int_{a}^{b}\alpha\left(x\right)h\left(x\right)\mathrm{d}{x}=0$ for every function $h\left(x\right)\in C^0\left[{{a}\,.\,.\,{b}}\right]$ such that $h\left(a\right)=0$ and $h\left(b\right)=0$,

then $\alpha\left(x\right)=0$ for all $x\in \left[{{a}\,.\,.\,{b}}\right]$.

Proof
Suppose the function $\alpha\left(x\right)$ is nonzero at some point in $\left[{{a}\,.\,.\,{b}}\right]$. Due to belonging to $C^0$ it is also nonzero in some interval $\left[{{x_1}\,.\,.\,{x_2}}\right]$ contained in $\left[{{a}\,.\,.\,{b}}\right]$.

Now we choose $h\left(x\right)$ to be of a specific form, which satisfies the requirements:


 * $h\left(x\right) = \begin{cases}

\operatorname{sgn}\left(\alpha\left(x\right)\right)(x-x_1)(x_2-x) & : x \in \left[{{x_1}\,.\,.\,{x_2}}\right] \\ 0 & : x \not\in \left[{{x_1}\,.\,.\,{x_2}}\right] \end{cases}$

Then


 * $\displaystyle \int_{a}^{b}\alpha\left(x\right)h\left(x\right)\mathrm{d}{x}=\int_{x_1}^{x_2}\left\vert\alpha\left(x\right)\right\vert (x-x_1)(x_2-x)\mathrm{d}{x}$

where we used the fact that


 * $\alpha\left(x\right)=\operatorname{sgn}\left(\alpha\left(x\right)\right)\left\vert\alpha\left(x\right)\right\vert$

as well as


 * $\displaystyle \operatorname{sgn^2}\left(x\right)=1$ if $x\ne 0$ and $x\in\R$.

The integrand is positive for $x\in\left[{{x_1}\,.\,.\,{x_2}}\right]$, whence the integral is positive.

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

Hence, by contradiction, the theorem holds.