1
$\begingroup$

Let $f$ be a continuous function on $[0,1]$. Suppose $$\int_0 ^x f(t)=0$$ for all x in [0,1] Then $f(t)=0$ for all $t\in[0,1]$. Is this statement true? If false can u give me a counter example.

$\endgroup$
  • 2
    $\begingroup$ It is as false as you can expect. Just observe that $$\int_0^1\cos\pi x\,dx=0$$ For the claim to be true there is lacking one very important condition on $\;f\;$ ... $\endgroup$ – DonAntonio May 12 '18 at 18:23
  • $\begingroup$ No, a function symmetric around the point $(x/2,0)$ would also do the job. And there are more counterexamples $\endgroup$ – Peter May 12 '18 at 18:24
  • 8
    $\begingroup$ Is $x$ fixed, or is the integral 0 for all $x\in [0,1]$? $\endgroup$ – Alex R. May 12 '18 at 18:24
  • 1
    $\begingroup$ If you mean this to hold for all $x$ in the interval, then it is true. Sketch: Suppose $f(x_0)>0$. By continuity suppose we have $\epsilon>0$ with $f(x)>0$ for $x\in [x_0-\epsilon, x_0]$. Then $\int_0^{x_0-\epsilon}f(t)\,dt$ and $\int_0^{x_0}f(t)\,dt$ can't both be $0$. But, really, your question is not clear. $\endgroup$ – lulu May 12 '18 at 18:29
  • $\begingroup$ I apologize ...yes this holds for all x $\endgroup$ – Normal May 12 '18 at 18:56
2
$\begingroup$

Suppose $F(x)=\int_0^x f(t)\,dt= 0$ for all $x\in [0,1].$ Then obviously $F'(x)=0$ for all $x.$ But since $f$ is continuous, the FTC gives $F'(x) = f(x)$ for all $x$. Thus $0=F'(x)=f(x)$ for all $x.$

$\endgroup$
1
$\begingroup$

If you meant that $\forall x\in[0,1] \int\limits_{0}^x f(t)dt$, then yes, it follows that $f(x)=0\ \forall x\in[0,1]$.

Let's assume that $f(y)>0$ for some $y\in[0,1]$. Then $f$ is strictly positive in some neighbourhood of $y$. Then there is $\delta$ such that $\int\limits_{y-\delta}^{y+\delta} f(t)dt>0$, so $\int\limits_{0}^{y+\delta} f(t)dt=\int\limits_{y-\delta}^{y+\delta} f(t)dt+\int\limits_{0}^{y-\delta} f(t)dt>0$, which leads to a contradiction. Analogously for $f(y)<0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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