Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 2
Theorem
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let:
- $\map f x \ge 0$
for all $x \in \closedint a b$.
Let:
- $\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$.
Proof
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
Let $F : \closedint a b \to \R$ be a real function defined by:
- $\ds \map F x = \int_a^x \map f x \rd x$
We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$.
From Fundamental Theorem of Calculus: First Part, we have:
- $F$ is continuous on $\closedint a b$
- $F$ is differentiable on $\openint a b$ with derivative $f$
Note that:
| \(\ds \map {F'} x\) | \(=\) | \(\ds \map f x\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds 0\) |
for all $x \in \openint a b$.
We therefore have, by Real Function with Positive Derivative is Increasing:
- $F$ is increasing on $\closedint a b$.
However, by hypothesis:
| \(\ds \map F b\) | \(=\) | \(\ds \int_a^b \map f x \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_a^a \map f x \rd x\) | Definite Integral on Zero Interval | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map F a\) |
So, it must be the case that:
- $\map F x = 0$ for all $x \in \closedint a b$.
We therefore have, from Derivative of Constant:
- $\map {F'} x = \map f x = 0$ for all $x \in \closedint a b$
as required.
$\blacksquare$