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Let $f:[a, b]\to \mathbb{R}$ be continuous. Assume $f(x)\geq 0$ for all $x\in[a, b]$, and that $$\int_a^b f\text{ }\mathrm{ d}x=0$$ Prove that $f(x)=0$ for all $x\in[a, b]$.

Would you recommend using a direct proof or a proof by contradiction by assuming there exists $y\in[a, b]$ such that $f(y)\geq 0$?

Shannon
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  • You should try proving the contrapositive. That will be easier. – Vinod Mar 11 '15 at 00:05
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    if $f\geq g$ then $\int f\geq \int g$. assume f was not equal to $0$ everywhere and then find $g\leq f$ for which you can explicitely tell that the integral is larger than $0$ – Bananach Mar 11 '15 at 00:06

3 Answers3

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Sketch of proof:

Assume the opposite, so $f(x)>0$ for some $x$. use continuity to prove $f$ is larger than a positive $l$ in a closed interval $[c,d]$ around $x$.

Use $\int_a^b\geq L\{a,c,d,b\}\geq (d-c)l>0$ to conclude if $f$ is strictly positive in a point the integral is greater than zero.

Therefore if $\int_a^bf=0$ we have $f$ is never strictly positive.

Asinomás
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Suppose that $f(c) > 0$, for some $c \in [a,b]$. Let $m = \frac{f(c)}{2}$ then there exists $\delta > 0$ such that $f(x) > m$, for all $x \in [c- \delta, c+ \delta]$ then considering $P$ a partition of $[a,b]$ we have $$L(f; P) > 2m \delta \implies \int_a^b f(x)dx \geq L(f;P) > 2m\delta > 0$$

Aaron Maroja
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    Could you further explain how you used partitions? – Shannon Mar 11 '15 at 02:57
  • Sure, how do you mean? – Aaron Maroja Mar 11 '15 at 02:58
  • Well first, by L(f; P) are you referring to the infimum. My professor does not use U(f, P) and L(f, P). Just the summation S(f, P) as a whole. The older definition. – Shannon Mar 11 '15 at 03:44
  • I'm referring to the lower sum, and we have $L(f;P) \leq \int_a^b f(x)dx \leq U(f;P)$. As $[c-\delta, c + \delta]$ has length $2\delta$ then $L (f;P) = \sum m_i (t_i - t_{i-1}) > m 2\delta.$ – Aaron Maroja Mar 11 '15 at 13:18
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Let $F'(x) = f(x)$. Then $F(b) - F(a) = 0$. The differential is defined in $b$ and $a$, and thus $(F(b) - F(a))' = 0'$, so $f(b) = f(a)$.

Since the function $f$ is continuous, there is $c$ between $[a,b]$ such that $f(c) = f(a)$. We also have $F(b) - F(a) = F(b) - F(c) + F(c) - F(a)$, and as $f(x) \ge 0$, $F(c) = F(a)$ (because the integral is crescent) for every $c$ between $a and b$, so $f(c) = f(b) = f(a)= k$ for every $c$ in the interval and the integral reduces itself to $k(b-a)=0$, so $f(c)=k=0$.

Raphael Venturelli
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