4
$\begingroup$

True/False test: Let $f:[0,1]\to[0,1]$ be continuous then $f$ assumes the value $\int_0^1 f^2(t)dt$ somewhere in $[0, 1].$

$$f:[0,1]\to[0,1]\implies f^2:[0,1]\to[0,1]\implies 0\le\int_0^1 f^2(t)dt\le1$$

So it's true.

but the paper says the statement is false.

Please help.

$\endgroup$
  • 1
    $\begingroup$ your final statement would imply the statement you want if the function $f$ were surjective, so that's a place to start looking for a counterexample $\endgroup$ – citedcorpse Jul 17 '13 at 12:32
  • $\begingroup$ Please clarify. I didn't get it. $\endgroup$ – Sriti Mallick Jul 17 '13 at 12:34
  • $\begingroup$ I thought it as multiplication i.e. $f.f$. Should I be considered it as composition? $\endgroup$ – Sriti Mallick Jul 17 '13 at 12:36
  • $\begingroup$ Then what's wrong with my argument? $\endgroup$ – Sriti Mallick Jul 17 '13 at 12:40
  • 1
    $\begingroup$ @SritiMallick you've deduced that $\int_0^1 f^2$ is a number between $0$ and $1$, but why does that mean $f$ actually attains it? $\endgroup$ – citedcorpse Jul 17 '13 at 12:43
8
$\begingroup$

Read the task carefully. You proved that $\int_0^1 f^2(t)dt\in[0,1]$. The task however asks, whether $f$ assumes the value of the integral, that is: Is there a $x\in[0,1]$, such that $f(x)=\int_0^1 f^2(t)dt$? This is indeed wrong, you can think of a counterexample using the following hint:

Hint: Let $f$ be a constant function, $f(x)=c$ for some $c\in[0,1]$. Then $f$ assumes the value $\int_0^1 f^2(t)dt$, iff $\int_0^1 f^2(t)dt=c$. Think about how to choose $c$ to get a counterexample.

$\endgroup$

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.