2

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.

Melody
  • 2,793
  • How do you restrict $T$ to $C[0,M]$? Shouldn't it be the case that $C[0,\infty)\subset C[0,M]$? – lEm Apr 24 '19 at 05:24
  • Well we can restrict in the sense that every $C[0,M]$ function can be extended to a $C[0,\infty)$ function. I suppose I should have stated more about that.

    Anyways, the example I am working with is $Tf(x)=1+\int_0^{\sqrt{x}}f(t).$ In which case what I mean should be clear. In that case by restrict it, I mean by limiting ourselves to functions with smaller domains. I guess restrict probably isn't the best word. Do you have a suggestion for a better way to phrase it?

     – Melody Apr 24 '19 at 05:38
  • What do you mean by an operator on $C[0,\infty)$? What is the norm on $C[0,\infty)$? – Kavi Rama Murthy Apr 24 '19 at 05:54
  • @Kavi Rama Murthy I simply mean a map sending $C[0,\infty)$ to itself. I don't understand why the norm is relevant in this case? Does not the map fixes points regardless of norm we define? In the problem I am considering the norm is equivalent to the supremum norm. But the question I am asking here should work regardless of norm, correct? – Melody Apr 24 '19 at 05:58

1 Answers1

1

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.

Kavi Rama Murthy
  • 311,013
  • Thank you. This was what I was trying to do. I tried to ask it in a more general way, but you're answer to the specific problem gives the exact idea of how to do it in other case. Thank you very much. As always, you give great advice. – Melody Apr 24 '19 at 06:11