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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?

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What is $L(f)$ in this question? –  anegligibleperson Nov 29 '12 at 3:14
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up vote 2 down vote accepted

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 $L(f)=0$?

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I do not see it right off the bat, how does that show a contradiction? –  Jackson Hart Nov 14 '12 at 0:06
    
@JacksonHart What is $\int_{x_0-\delta}^{x_0+\delta} \frac{f(x_0)}{2}dx$? –  N. S. Nov 14 '12 at 3:12
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