Show that $T : C([-1,1],\mathbb{R}) \to C([-1,1],\mathbb{R})$ is a contraction

Question

I am trying to show that $T : C([-1,1],\mathbb{R}) \to C([-1,1],\mathbb{R})$ is a contraction, where $T$ is defined by $$T(f)(x) = \pi x + \int\limits_{-1}^1\frac{f(t)dt}{\pi+(x-t)^4}.$$ I am not having much luck in my attempts so far.

Answer 1

Note that $f$ is affine; we can write $T(f)$ as the sum of a fixed function (independent of $f$) $\pi x$, plus a linear map: $$S : C([0, 1], \Bbb{R}) \to C([0, 1], \Bbb{R}) : f \mapsto \int_{-1}^1 \frac{f(t) \, \mathrm{d}t}{\pi + (x - t)^4}.$$ It's easy to see that $T$ is a contraction if and only if $S$ is a contraction. Using the linearity of $S$, we simply need to show that $\|S(f)\|_\infty < 1$ whenever $\|f\|_\infty \le 1$.

Suppose $\|f\|_\infty \le 1$. Then $|f(x)| \le 1$ for all $x \in [-1, 1]$. Consequently, if $x \in [-1, 1]$, \begin{align*} |S(f)(x)| &= \left|\int_{-1}^1 \frac{f(t) \, \mathrm{d}t}{\pi + (x - t)^4}\right| \\ &\le \int_{-1}^1 \frac{|f(t)| \, \mathrm{d}t}{\pi + (x - t)^4} \\ &\le \int_{-1}^1 \frac{\mathrm{d}t}{\pi} = \frac{2}{\pi}. \end{align*} Hence $\|S(f)\|_\infty \le 2/\pi < 1$. That is, $S$ (and hence $T$) is a contraction. Not only that, $S$ and $T$ are Banach contractions, meaning that there is some $k \in [0, 1)$ such that $$\|T(f_1) - T(f_2)\|_\infty \le k\|f_1 - f_2\|_\infty.$$ This is useful if you wish to apply the Banach Fixed Point Theorem to $T$ (applying it to $S$ would be boring, as $0$ is a fixed point of any linear map!).