Zero Definite Integral of Nowhere Negative Function implies Zero Function
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Theorem
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $h: \closedint a b \to \R$ be a continuous real function such that:
- $\forall x \in \closedint a b: \map h x \ge 0$
Let:
- $\ds \int_a^b \map h x \rd x = 0$
Then:
- $\forall x \in \closedint a b: \map h x = 0$
Proof
Aiming for a contradiction, suppose that:
- $\exists c \in \closedint a b: \map h c > 0$
As $h$ is continuous, there exists some closed real interval $\closedint r s \subseteq \closedint a b$ where $r < s$ such that:
- $\exists \epsilon \in \R_{>0}: \forall x \in \closedint r s: \map h x > \dfrac {\map h c} 2$
From Sign of Function Matches Sign of Definite Integral:
- $\ds \int_r^s \map h x \rd x > 0$
This makes a strictly positive contribution to the integral.
$h$ is still continuous on $\closedint a r$ and $\closedint s b$.
So, again from Sign of Function Matches Sign of Definite Integral:
- $\forall x \in \closedint a r: \map f x \ge 0 \implies \ds \int_a^r \map f x \rd x \ge 0$
- $\forall x \in \closedint s b: \map f x \ge 0 \implies \ds \int_s^b \map f x \rd x \ge 0$
Thus as:
- $\ds \int_a^b \map h x \rd x = \int_a^r \map f x \rd x + \int_r^s \map f x \rd x + \int_s^b \map f x \rd x$
it follows that:
- $\ds \int_a^b \map h x \rd x > 0$
But by hypothesis:
- $\ds \int_a^b \map h x \rd x = 0$
Thus by contradiction, there can be no $c \in \closedint a b$ such that $\map h c > 0$.
Hence the result.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Lemma $2.2.10$