Suppose $f,g$ are functions from $[0,1]$ to $[0,1]$. If $f$ and $g$ are bijective, is $\frac{f^2 + g^2}{2}$ bijective? I would guess no but cannot think of a counterexample.
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Consider $f = \sin{\frac{\pi x}{2}}$ and $g = \cos{\frac{\pi x}{2}}$. Then they are both bijective over $[0,1]$ but $\frac{1}{2}(f^2+g^2) = \frac{1}{2}$.
$\begingroup$
$\endgroup$
As an example, the functions $$ f(x) = x\\ g(x) = \sqrt{1 - x^2} $$ are both bijective, but $(f^2 + g^2)/2 = 1/2$.
answered Oct 23, 2016 at 22:49
$\begingroup$
$\endgroup$
0
The other two answers are extreme cases where $\frac{f^2 + g^2}2$ is constant, but they lack the simplicity of $f(x) = x, g(x) = 1-x$. We get $\frac{f^2 + g^2}{2} = x^2 - x + \frac12$, which doesn't go below $\frac14$, and thus is not surjective / onto. Also, for any function value $c>\frac14$, there are two $x$-values that make $\frac{f^2(x) + g^2(x)}{2} = c$, so it's not injective / one-to-one either.
