If $f$ is continuous and positive and $\int_a^b f(x)dx=0$ therefore $f(x)=0$ for all $x\in[a,b]$.

Asked Nov 26 '15 at 08:45

Active Nov 26 '15 at 08:49

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If $f$ is continuous and non-negative and $\int_a^b f(x)dx=0$ therefore $f(x)=0$ for all $x\in[a,b]$.

My attempts

Suppose $f\neq 0$, i.e. there is an $x\in ]a,b[$ s.t. $f(x)\neq 0$. Since $f$ is continuous, there is a $\delta>0$ s.t. $f|_{]x-\delta,x+\delta[}> 0$ (we take $\delta$ small enough to have $]x-\delta,x+\delta[\subset ]a,b[$. Therefore $$\int_a^b f(t)dt\geq \int_{x-\delta}^{x+\delta}f(t)dt>0$$ contradiction. The cas $x=a$ or $x=b$ goes the same but with $f|_{]a,a+\delta[}>0$ or $f|_{]b-\delta,b[}>0$.

edited Nov 26 '15 at 08:49

asked Nov 26 '15 at 08:45

Rick

1,707

2

the question makes more sense if the condition is continuous and non-negative – David Holden Nov 26 '15 at 08:49
@DavidHolden: to me positive is $\geq 0$. For $>0$ I would say strictly positive. – Rick Nov 26 '15 at 08:51
It's not duplicate since my proof is different. – Rick Nov 26 '15 at 08:51
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What's your question, @Rick? – user37238 Nov 26 '15 at 08:52
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You did not strictly show that $f > 0$ on an interval $I$ implies that $\int_I f(t)dt > 0$. The answers to the proposed duplicate show how to handle that. – Martin R Nov 26 '15 at 08:57
It's a result I can use :-) @MartinR – Rick Nov 26 '15 at 09:12
@Rick Nice proof, what are you asking? – Eric S. Nov 26 '15 at 09:52
Just if way way to prove it is correct :-) Thanks. But by your last comment, I can say it is :-) – Rick Nov 26 '15 at 12:55

If $f$ is continuous and positive and $\int_a^b f(x)dx=0$ therefore $f(x)=0$ for all $x\in[a,b]$.

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