So suppose that we have an operator $T:C[0,\infty)\to C[0,\infty)$ such that for all $M\in\mathbb{R}^+$ the restriction of $T|_{C[0,M]}$ maps into $C[0,M]$ and has a unique fixed point, then is that enough to deduce that $T$ itself has a fixed point.
I want to say yes, but I am not sure. Here's my idea, so we want to build the fixed point $f$ from the restrictions of $T.$ So if $x\in\mathbb{R}_{\geq 0},$ then we look at $x+1$ and let $g$ be the fixed point of $T|_{C[0,x+1]}.$ We then define $f(x)=g(x).$ This clearly defines a function from $[0,\infty)\to[0,\infty)$. We need verify that it is a fixed point of $T,$ and it is continuous.
Note, that if $h$ is the fixed point of $T_{C[0,y]}$ and $l$ the fixed point of $T_{C[0,z]}$ with $y<z,$ then $Tl(x)=l(x)$ for all $x\in[0,z],$ hence for all $x\in[0,y],$ so $T(l|_{[0,y]})=l|_{[0,y]}.$ Then $l|_{[0,y]}$ is a fixed point of $T|_{C[0,y]}$ so by uniqueness $l|_{[0,y]}=h.$ Well then $f|_{[0,y]}$ is a fixed point of $T|_{C[0,y]},$ hence $f$ is continuous on $[0,y],$ and $Tf(x)=f(x)$ for all $x\in[0,y].$ So since $y>0$ was arbitrary $f$ is continuous on $[0,\infty),$ and for all $x\geq 0$ we have $Tf(x)=f(x),$ so $f$ is a fixed point of $T.$
Could anyone verify if what I'm doing is correct, or maybe tell me why it's wrong? Thank you very much.
I will try to guess what you are trying to do and give a solution for the specific $T$ you are considering. You already know that for each $M$ there is a unique continuous function $f_M$ on $[0,M]$ such that $f_M(x)=1+\int_0^{\sqrt x} f_M(t)dt$. In this case there surely exists a continuous function $f$ on $[0,\infty)$ such that $f(x)=1+\int_0^{\sqrt x} f(t)dt$. All you have to do is observe that $f_{N+1}(x)=f_N(x)$ for $ x \leq N$ so $f(x)=f_n(x)$ where $n$ is any integer greater than $x$ gives a well defined continuous function. This function is a fixed point.