Fixed point theorem with $\frac12 tx(t^2)=x(t)+2t^2$

Question

Let $V=C[0,1]$ with the norm $\| \|_\infty$ and let $M=${$x\in V |\int_0^1|x(t)|dt\le 1$}.

a. Show that there exists a unique $x_\infty \in M$ that satisfies the equation $\frac12 tx(t^2)=x(t)+2t^2$.

b. If $x_0(t)=-1$ calculate successive approximations of $x_1$ and $x_2$.

c. Calculate $\| x_2-x_1\| _\infty$.

d. Using the previous results, give the upper bound of the value $x_\infty(\frac12)$.

I'm very confused about the fixed point theorem overall. Judging from the premise, I suppose that I need to show that the application is contractant. But how do I set the function in the first place? Should I isolate x(t)? And for the approximations we saw that $\|x_\infty -x_1\| \le \frac{c^n}{1-c} \|x_1 -x_0 \|$ and $\|x_\infty -x_n\| \le \frac{c}{1-c} \|x_n -x_{n-1} \|$. I replaced $x_0$ in the first inequality but that doesn't really give us anything, does it. For part c, same problem. I would use the second inequality, but how do you get $x_n$? And for the last part, I would use the second inequality as well, but would you then plug $\frac12$ in the equation at the start?

Any explanations of the fixed point theorem are welcome, I am very confused about how to apply it and what it means in general.

Answer 1

Hint: The Banach Fixed Point Theorem says that if $(V, \| \cdot \|)$ is a Banach space (i.e a complete normed space) and $F :V \to V$ satisfies $\| F(x) - F(y) \| \leqslant C \| x - y\| $ with $0<C<1$, that is $F$ is a contraction, then $F$ has a unique fixed point.

For your question, $V = C[0,1]$ which is a Banach space with the norm $\| \cdot \|_\infty$. Your job is to find a specific $F : C[0,1] \to C[0,1]$ that is

1. a contraction; and
2. if $F(x)(t)=x(t)$ then \begin{equation} \tag{1}\frac 1 2 t x(t^2) = x(t) + 2t^2. \label{eq:1}\end{equation}

In general, this could be very difficult, but in this case there is an obvious guess. Rearranging \eqref{eq:1} gives $$ x(t) = \frac 1 2 t x(t^2) - 2t^2,$$ so the obvious guess would be $F(x)(t) := \frac 1 2 t x(t^2) - 2t^2$. Now you need to show that:

1. If $x \in C[0,1]$ then $F(x) \in C[0,1]$. (Hence, $F$ is well-defined).
2. $F$ is a contraction.
3. If $x \in M$ then $F(x) \in M$. This implies that the fixed point will be in $M$.

This gets you to the end of (a). Are you able to do the rest?