Problem:

1.Show that none of the following mappings $\displaystyle f:X \rightarrow X $ have a fixed point and explain why the contraction mapping principle is not contradicted.

a. $\displaystyle (0,1) \subseteq R$ and $\displaystyle f(x)=\frac{x}{2}$

b. X = R and $\displaystyle f(x)=x+1 $ for x in X.

c. $\displaystyle {(x,y) \in R^2 | x^2+y^2=1}$ and $\displaystyle f(x,y)=(-y,x)$ for (x,y) in X

2. Define the function $\displaystyle f:[1, \infty) \rightarrow R$ by

$\displaystyle f(x)=1+\sqrt{x}$ for $\displaystyle x \geq 1$

Show that this function has exactly one fixed point.

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Attempt:

Problem 1:

By a definition a point x in X is a fixed point for the mapping $\displaystyle f:X \rightarrow X $ provided that $\displaystyle f(x) = x $

For (a) and (b) is it sufficient to show that $\displaystyle x \neq \frac{x}{2} $ and $\displaystyle x \neq x+1 $?

For (c), the only fixed point would be $\displaystyle x= \frac{\sqrt{2}}{2},y= \frac{\sqrt{2}}{2}$, but $\displaystyle f \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) \neq f \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

Problem 2:

If this function has a fixed point, then $\displaystyle f(x)=x= 1 +\sqrt{x} $

$\displaystyle \implies x - \sqrt{x} = 1 \implies -x^{\frac{1}{2}} + x - 1 =0$

I should slap myself, because I can't figure a way to solve for x. First glance, I thought of using the quadratic formula, but I can't do that.

Thank you for your time.