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

If $\alpha\left(x\right)$ is continuous 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 $\left[{{a}\,.\,.\,{b}}\right]$ for any $h\left(x\right)$.

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, while still satisfying the requirements in the statement of theorem:

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

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.

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

Hence, by contradiction, the theorem holds.


$\blacksquare$

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