1. $f:\mathbb{R}\to\mathbb{R}$ is continuous and injective, then it is strictly monotone.


  1. If $f\in C[0,2]$ with $f(0)=f(2)$, then $\exists x_1,x_2\in [0,2]\ni x_1-x_2=1$ and $f(x_1)=f(x_2)$

False, as $|f(x_1)-f(x_2)|<\epsilon$ but $|x_1-x_2|=1$

3.Let $f$ and $g$ be continuous real valued function on $\mathbb R$ such that for all $x\in \mathbb R$ wehave $f(g(x))=g(f(x.))$ If there exists $x_0\in \mathbb R $ such that $f(f(x_0))=g(g(x_0))$ then there exists $x_1\in\mathbb R$ such that $f(x_1)=g(x_1).$

I have no idea how to do prove or disprove this one.

Thanks for helping and correcting me.

  • 2
    $\begingroup$ (2) is true, consider the function $g: [0, 1] \rightarrow \mathbb{R}$ by $g(x) = f(x) - f(x + 1)$ and use IVP. $\endgroup$ – Krish Nov 28 '14 at 14:01

Regarding part 2, see Krish's comment.

Regarding part 3: Let $a:=f(x_0)$, $b:=g(x_0)$ and consider the continuous function $x\mapsto h(x)=f(x)-g(x)$. We have $$f(a)=f(f(x_0))=g(g(x_0))=g(b)$$ and $$g(a)=g(f(x_0))=f(g(x_0))=f(b), $$ hence $$ h(a)=-h(b).$$ If $h(a)=0$, we can let $x_1=a$ (or $b$). And if $h(a)\ne 0$, the IVT gives us $x_1$ between $a$ and $b$ with $h(x_1)=0$.


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