Let $f(x)$ be a continuous function defined on $[0,1]$ with range $[0,1]$. Show that there is some $c$ in $[0,1]$ such that $f(c)=1-c$
My attempt is as follows:-
$$f(c)=1-c$$
$$f(c)-1+c=0$$
$$g(c)=f(c)-1+c$$
$g(c)$ will be continuous function in $[0,1]$ as sum of two continuous functions will also be continuous.
If we can prove that there $g(c)$ will have atleast one root in $[0,1]$, then we are done.
$$g(0)=f(0)-1$$
$$g(0)\in [0,1]-1$$
$$g(0)\in[-1,0]\tag{1}$$
$$g(1)=f(1)-1+1$$
$$g(1)=f(1)$$
$$g(1)\in[0,1]\tag{2}$$
Case $1$: If $g(0)$ or $g(1)$ goes $0$, then we are done because then $g(c)$ will be having at least one root in $[0,1]$
Case $2$: If $g(0)$ is negative and $g(1)$ is positive, then what we can say?
By intermediate value theorem, we can say that as $g(0)<0<g(1)$, so there will exists some c in $(0,1)$ for which $g(c)$ is zero.
But I have a counter, what if $g(c)$ is like following:-
Here we are getting no $c$ in $(0,1)$ for which $g(c)$ is zero.
Best Answer
Let $g(x)=1-f(x)-x$ Then $g$ is continuous and $g(0) \geq 0$, $g(1) \leq 0$. Hence there exists $c$ such that $g(c)=0$ which gives $f(c)=1-c$.