Show that the given map F is a contraction map & find a solution of F, if any.

Question

Given a space of all real-valued continuous function $C[0,1]$ with $sup$ norm, define $$F:C[0,1]\rightarrow C[0,1]\text{ by } F(x)(t)=x(0)+\lambda\int_{0}^{t}x(s)ds$$ $\lambda\in\mathbb{R}$ with $|\lambda|<1$ and $t\in [0,1]$.

I cannot understand whether $x(0)$ will be treated as a constant or it will change during the proof.

Thanks in advance.

Answer 1

In my opinion $x(0)$ has to be a given constant. Imagine you have found a fix point $x$ of $F$, i.e. $F(x)=x$. Then for every $\alpha\in\mathbb{R}$ the function $\alpha\cdot x$ satisfies $F(\alpha x)=\alpha x$. This would contradict the uniqueness part of the contraction mapping principle.

EDIT: Let's sketch a proof: Let $y,z\in C^0([0,1])$, then $$|F(z)(t)-F(y)(t)|=|x(0)-x(0) + \lambda \int_0^tz(s)-y(s)ds|\leq\lambda\int_0^t|z(s)-y(s)|ds.$$ Hence $$\sup_{t\in[0,1]}|F(z)(t)-F(y)(t)|\leq \lambda \sup_{t\in[0,1]}|z(t)-y(t)|.$$ Since $0<\lambda<1$ this gives you a contraction. This yields a $y\in C^0([0,1])$ such that $$y(t)=F(y)(t)=x(0)+\lambda\int_0^ty(s)\, ds.$$ If you like to know how $y$ looks like, you either derive the whole equation by $t$ and obtain an ordinary differential equation or you can try a fix-point iteration, i.e. $z_n:=F(z_{n-1})$, $z_0=0$. The result will be something along the lines of an exponential function.

Answer 2

$|F(x)(t)-f(y)(t)| = |\lambda\int_{0}^{t}(x(s)-y(s))ds| \le | \lambda| \int_{0}^{t}|x(s)-y(s)|ds \le | \lambda| \int_{0}^{1}|x(s)-y(s)|ds \le | \lambda| \int_{0}^{1}||x-y||ds =|\lambda| \cdot ||x-y||$

for all $t \in [0,1]$.

Hence $||F(x)-F(y)|| \le |\lambda| \cdot ||x-y||$ .

Answer 3

According to the definition

$$F:C[0,1]\rightarrow C[0,1]\text{ by } F(x)(t)=x(0)+\lambda\int_{0}^{t}x(s)ds$$

$$ F(x)(0) = x(0)+\lambda\int_{0}^{0}x(s)ds= x(0)$$ Thus, $x(0)$ stays constant.

Answer 4

This is just a discussion, not an answer.

Notice that $(C[0,1],\|\cdot\|_{\infty})$ is a Banach space. Besides, the given map $F$ is linear (thus $F(0) = 0$ and therefore $0\in C[0,1]$ is already a fixed point for $F$).

If $F$ were a contraction it would have a unique fixed point, by Banach fixed point theorem.

However, if there exists a $\bar{x}\in C[0,1]$ such that $\bar{x}\neq 0$ and $F(\bar{x}) = \bar{x}$, then $F(\alpha\bar{x}) = \alpha F(\bar{x}) = \alpha\bar{x}$ for every $\alpha\in\mathbb R$ ($\alpha\bar{x}$ would be a fixed point for $F$, for all $\alpha\in\mathbb R$), which means that $F$ would've infinitely many fixed points.

Therefore, we have two possibilities:

(a) $F$ is not a contraction map; or

(b) $0$ is the only fixed point for $F$.