Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What is $L(f)$ in this question? – anegligibleperson Nov 29 '12 at 3:14
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 $L(f)=0$?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.