-2
$\begingroup$

Let $I:=[a,b]$, let $f:I\to\Bbb R$ be continuous, and let $f(x)$ greater than or equal to $0$ for all $x \in I$. Prove that if $L(f)=0$, then $f(x) = 0$ for all $x \in I$.

Should I show by contradiction?

  • 0
    What is $L(f)$ in this question?2012-11-29

1 Answers 1

2

Assume by contradiction that $f$ is not identically zero.

Then $f(x_0)>0$ for some $x_0$. By continuity you get that there exists some $\delta$ so that

$f(x) > \frac{f(x_0)}{2} \,;\, \forall x \in [x_0-\delta, x_0+\delta] \,.$

Can you see how does this contradict the fact that $L(f)=0$?

  • 0
    @JacksonHart What is $\int_{x_0-\delta}^{x_0+\delta} \frac{f(x_0)}{2}dx$?2012-11-14