Prove a property of the integral of the product of continuous functions

Let $f$ be a continuous function on the interval $[a,b]$ and assume

$\int_a^b f(x) g(x) \, dx = 0$

for every function $g$ which is continuous on the interval $[a,b]$ . Prove that $f(x) = 0$ for all $x \in [a,b]$ .

Proof. Since $\int_a^b f(x) g(x) \, dx = 0$ must hold for every function $g$ that is continuous on $[a,b]$ , it must hold for $f$ itself (since $f$ is continuous on $[a,b]$ by hypothesis). Therefore we must have,

$\int_a^b f(x) f(x) \, dx = \int_a^b (f(x))^2 \, dx = 0$

However, $(f(x))^2 \geq 0$ for all $x \in [a,b]$ . We know from the previous exercise (Section 3.20, #7) that a non-negative function whose integral is zero on an interval must be zero at every point at which it is continuous. By hypothesis $f$ is continuous at every point of $[a,b]$ ; hence, $f^2$ is also continuous at every point of $[a,b]$ (since the product of continuous functions is continuous). Therefore,

$(f(x))^2 = 0 \qquad \text{for all } x \in [a,b].$

Which implies,

$f(x) = 0 \qquad \text{for all } x \in [a,b]. \qquad \blacksquare$

2 comments

August 2, 2017 at 8:10 am

Anonymous says:

It’s easier to choose g(x) = 1 for every x then by the previous exercice the answer holds immediately

- September 28, 2019 at 7:06 am
  
  Anonymous says:
  
  Note that we cannot choose g(x) = 1, because f may change sign on the interval. In that case we are integrating both negative and positive areas which may sum to zero. For example, suppose f(x) = cos(x) and g(x) = 1. Then if we integrate from 0 to 2pi we get zero, even though cos(x) != 0 on every point in the interval.

Stumbling Robot

Prove a property of the integral of the product of continuous functions

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