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?
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?
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 $L(f)=0$?