Zero Definite Integral of Nowhere Negative Function implies Zero Function

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Theorem

Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval.

Let $h: \left[{a \,.\,.\, b}\right] \to \R$ be a continuous real function such that:

$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) \ge 0$

Let:

$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$


Then:

$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) = 0$


Proof

Aiming for a contradiction, suppose that:

$\exists c \in \left[{a \,.\,.\, b}\right]: h \left({c}\right) > 0$

As $h$ is continuous, there exists some closed real interval $\left[{r \,.\,.\, s}\right] \subseteq \left[{a \,.\,.\, b}\right]$ where $r < s$ such that:

$\exists \epsilon \in \R_{>0}: \forall x \in \left[{r \,.\,.\, s}\right]: h \left({x}\right) > \dfrac {h \left({c}\right)} 2$

From Sign of Function Matches Sign of Definite Integral:

$\displaystyle \int_r^s h \left({x}\right) \rd x > 0$

This makes a strictly positive contribution to the integral.

$h$ is still continuous on $\left[{a \,.\,.\, r}\right]$ and $\left[{s \,.\,.\, b}\right]$.


So, again from Sign of Function Matches Sign of Definite Integral:

$\forall x \in \left [{a \,.\,.\, r} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_a^r f \left({x}\right) \rd x \ge 0$
$\forall x \in \left [{s \,.\,.\, b} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_s^b f \left({x}\right) \rd x \ge 0$


Thus as:

$\displaystyle \int_a^b h \left({x}\right) \rd x = \int_a^r f \left({x}\right) \rd x + \int_r^s f \left({x}\right) \rd x + \int_s^b f \left({x}\right) \rd x$

it follows that:

$\displaystyle \int_a^b h \left({x}\right) \rd x > 0$


But by hypothesis:

$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$

Thus by contradiction, there can be no $c \in \left[{a \,.\,.\, b}\right]$ such that $h \left({c}\right) > 0$.

Hence the result.

$\blacksquare$


Sources