Problem 1

Exercise 4.1 (Non-contractive mapping)   Define $T : \mathbb{R} \to \mathbb{R}$ by

$$T(x) := \frac{\pi}{2} + x - \arctan(x)$$ (4.1)

Show that $\vert T(x)-T(y)\vert \leq \vert x-y\vert$ for all $x,y\in\mathbb{R}$ and that $T$ has no fixed points in $\mathbb{R}$. State the contraction mapping theorem and explain why this example does not contradict the theorem.

Proof. For the Lipschitz estimate, we estimate the first derivative of $T$ by formal differentiation rules:

$$T'(w) = 1 - \frac{1}{1+w^2} \leq 1$$ (4.2)

The fundamental theorem reveals

$$T(x) = \int_0^x T'(w)dw$$ (4.3)

$$T(y) = \int_0^y T'(w)dw$$ (4.4)

Therefore the difference can be estimated

$$\vert T(x)-T(y)\vert = \left\vert \int_y^x T'(w)dw \right\vert \leq \vert x-y\vert$$ (4.5)

The contraction mapping theorem states that if $\vert T(x)-T(y)\vert\leq c\vert x-y\vert$ for some $c<1$, then the map $T$ has a fixed point. In this example we did not select such a $c<1$, so we are comfortable now proving that actually is no fixed point. Suppose $T(x)=x$. Then $\pi/2=\arctan(x)$, which is never true. $\qedsymbol$