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If $f$ is continuous on $[a,b]$, $f(x)≥0$ on $[a,b]$ and $$\int_{a}^{b} f(x) =0$$

then prove that $f(x)=0$ for all $x \in [a,b]$.

I tried with Riemann's definite integral definition but couldn't proceed

user26857
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TheDragonReborn
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  • Use the contra-reciprocal for demonstrating that. – Wmmoreno Oct 27 '13 at 06:58
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    Since $f$ is continuous, and $f(x_0)>0$, there is an interval $I$ containing $x_0$ where $f$ is positive (elementary exercise- see Spivak's Caclulus). On this interval, integral of $f$ is positive. – Groups Oct 31 '14 at 09:35

2 Answers2

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Something more easy. If $f$ is continuous in $[a,b]$ then we can define a map $F:[a,b]\to\Bbb R$ such that $F(x)=\int_a^x f(x) dx$ and $F'=f$.

We have $F(x)≥0$ for every $x \in [a,b]$ because $f(x)\geq0$ for every $x \in [a,b]$.

Also $F$ is increasing because $f(x)≥0$ and we have $F(a)=F(b)=0$, thus $F$ is constant and especially $F(x)=0$.

So $f(x)=0$ for every $x \in [a,b]$.

Haha
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    If $f$ is only integrable, then you can't say $F'=f$. The Second Fundamental Theorem requires that $f$ be continuous. For example, if $f$ was nonzero at a single point, then it would still integrate to 0. – kbolino Jun 16 '16 at 20:52
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    How does F being increasing and nonnegative make F(b) and F(a) equal to 0 – angelo086 Nov 05 '16 at 18:14
  • Haha, I have the same question as @angelo086 (wasn't stalking you. just recognised your username hahaha) – BCLC Jan 11 '17 at 08:37
  • cc @kbolino.... – BCLC Jan 11 '17 at 08:37
  • @Haha Can u help me in this question? If $f$ belongs to $C[0, 1]$. Determine the cases where the given condition implies that $f$ is identically zero.

    Case 1 $\int\ x^nf(x) =0$ for all non negative n

    Case 2 $\int\ f(x)Cosnx =0$ for all non negative n

    Case 3 $\int\ f(x)Sinnx =0$ for all positive n

     – anonymous May 28 '17 at 01:46
  • @sani Asking random questions in comments is not a reasonable way how to deal with the question ban. (Neither is asking questions on other sites, where they are off-topic.) You have enough reputation to talk in chat, so you can try to ask there. For example Calculus and analysis chat room or the main chat room might be a reasonable place for the question from your comment. – Martin Sleziak May 28 '17 at 08:30
  • But this is not a random question...This is very much relevant to the main question..However If u know the answer U could have helped me with lesser effort..@MartinSleziak – anonymous May 28 '17 at 09:03
  • hiii@Martin Sleziak Do u have any Idea How can I proceed????? – anonymous May 28 '17 at 10:42
  • This proof is wrong. We can't ensure the derivative with continuity. – Hackerman Jul 20 '21 at 03:49
  • @RudikRobertoRompich you can check that if f is continuous, F so defined is always differentiable. – Arctic Char Jul 20 '21 at 03:56
  • @ArticChar, counterexample: Tagaki, and Weierstrass function. Continuity does not imply being differentiable. There is even a book called "Continuous Nowhere Differentiable Functions" from Springer. – Hackerman Jul 20 '21 at 04:23
  • @RudikRobertoRompich of course there are functions $f$ that is continuous but not differentiable. But for any continuous $f$, $F$ is differentiable. – Arctic Char Jul 20 '21 at 10:06
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Hint: Suppose that $f$ were not identically zero, and choose $x_0 \in (a, b)$ for which $f(x_0) > 0$. Then there exists an $\epsilon > 0$ such that

$$|x - x_0| < \epsilon \implies f(x) > \frac{1}{2} f(x_0)$$

Try considering a partition of the interval containing the interval $[x_0 - \epsilon, x_0 + \epsilon]$, and compute a lower sum.

  • Is there a reason why it was chosen $f(x)>\dfrac12f(x_0)$ instead of $f(x)>f(x_0)$? –  May 14 '20 at 04:24
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    @User1618 He has chosen an $\epsilon$ so that $f(x)$ lies in the $f(x_0)/2$ radius interval around $f(x_0)$. If, for example, $f(x_0)$ was an upper bound, then no such $\epsilon$ exists. – ZSMJ Nov 26 '21 at 07:55