Define $T:C[0,1] \rightarrow C[0,1]$ by $$T(f)(x) = \int_0^x f(t^3)dt$$ I think I've shown that it is not a contraction mapping under uniform norm. The question is - does $T$ have a unique fixed point?
I thought that if it does, it could be possible to find another norm on $C[0,1]$, under which it is a contraction.. but I do not really know what norms on this space exist, and which one would be suitable. Or is it the case that there is not a unique fixed point?
Thank you!
$T(f)$ has an unique fixed point. Namely, the trivial one where $f(x)$ is identically zero.
Let $f$ be any fixed point of $T(f)$. Since $f$ is continuous on $[0,1]$, it is bounded there. Define $M$ as
$$M = \| f \|_{\infty} =\sup \big\{\; \big|f(x)\big| : x \in [0,1]\; \big\} < \infty $$
For any $x \in [0,1]$, we have $$|f(x)| = \left| \int_0^x f(t^3) dt \right| \le \int_0^x |f(t^3)| dt = \int_0^x M dt = M x$$ Repeat this estimation one more time, we get
$$|f(x)| \le \int_0^x M t^3 dt = \frac{M}{4} x^4 \le \frac{M}{4}$$
This leads to $\displaystyle\;M \le \frac{M}{4}\;$ and this is possible only when $M = 0$, i.e $f(x)$ is identically zero.
In terms of language of contraction mapping, $T^2 = T\circ T$ is a contraction mapping with respect to the supremium norm.