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

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.

• 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 – citedcorpse Jul 17 '13 at 12:32
• I thought it as multiplication i.e. $f.f$. Should I be considered it as composition? – Sriti Mallick Jul 17 '13 at 12:36
• @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? – citedcorpse Jul 17 '13 at 12:43
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.