Is $T:C[0,1]\to C[0,1]: Tf(x)=\frac12 x f(x^2)+1$ a contraction?

Question

I am a non-mathematician (a physicist) who wants to find a continuous function on $[0,1]$ that satisfies the relation $f(x)=\frac12 x f(x^2)+1$ with $0\le x\le 1$. I need it for one of my models.

This is what I've got sofar.

I want to use Banach's contraction theorem.

Define $T:C[0,1]\to C[0,1]: (Tf)(x)=\frac12 x f(x^2)+1$. If $T$ is a contraction, the Banach's contraction theorem yields that $Tf=f$ and that solves my problem. I know that C[0,1] is a complete normed space with regard to $\lVert\cdot\rVert_{\infty}$

I do not know how to show that $T$ is a contraction. Can someone show me how do do this?

Answer 1

Define $\|f\|_\infty := \max_{x\in [0,1]} |f(x)|$, for all $f\in C[0,1]$. Given $f, g\in C[0,1]$ and $x\in [0,1]$,

$$|Tf(x) - Tg(x)| = \left|\left(\frac{1}{2}xf(x^2) + 1\right) -\left(\frac{1}{2}xg(x^2) + 1\right)\right| = \frac{|x|}{2}|f(x^2) - g(x^2)| \le \frac{1}{2}\|f - g\|_\infty.$$

Thus $$\|Tf - Tg\|_\infty \le \frac{1}{2}\|f - g\|.$$ So $T$ is a contraction (with contraction constant $\frac{1}{2}$).

Answer 2

Now that @kobe has demonstrated that the operator is a contraction, this justifies the search for the unique fixed point of this operator. This can be achieved by assuming that $f$ can be represented by a power series $f(x) = \sum_{n=0}^\infty a_n x^n$. It is clear that $f(0)=1$, so $a_0=1$. Now if you plug in this series into your equation, and then compare coefficients, you will arrive at the unique fixed point: $$f(x) = 1 + \sum_{n=1}^\infty \frac{x^{2^n-1}}{2^n}.$$ You can verify that this works by plugging it into $$f(x) = \frac12 x f(x^2) + 1.$$