# Thread: Real analysis questions - thanks alot!

1. ## Real analysis questions - thanks alot!

Hey guys,
i'm working on real analysis and i'm stuck on these 2 questions:
1) Prove that the annulus A = {z belongs to R2| r <=|z|<=R} (where R>r>0) is connected.

2) f: R -> R is differentiable
Suppose there exists L < 1 such that f'(x) < L for all R
(a) Prove that f has a unique fixed point
(b) Show by example that (a) fails if L = 1
Thanks alot guys!

2. Originally Posted by pc31
2) f: R -> R is differentiable
Suppose there exists L < 1 such that f'(x) < L for all R
(a) Prove that f has a unique fixed point
(b) Show by example that (a) fails if L = 1
2(a)
Let $\displaystyle g(x) = x-f(x)$. Then $\displaystyle g'(x) = 1-f'(x)>0$. Hence g is a strictly increasing function. A strictly increasing function can only cross the x-axis at most once.

Suppose $\displaystyle g(x)$ doesn’t cross the x-axis at all. The only way this is possible for a strictly increasing differentiable (and thus continuous) function is if either (i) $\displaystyle g(x)$ is negative for all x and $\displaystyle g'(x)\to0$ as $\displaystyle x\to\infty$ or (ii) $\displaystyle g(x)$ is positive for all x and $\displaystyle g'(x)\to0$ as $\displaystyle x\to-\infty$. In either case, we have that $\displaystyle f'(x)\to1$ as $\displaystyle x\to\infty$. This contradicts the fact that $\displaystyle f'(x)$ is bounded above by $\displaystyle L<1$. (Get it? $\displaystyle \lim_{x\to\infty}{f'(x)}=1$ means that $\displaystyle f'(x)$ gets arbitrarily close to 1 for sufficiently large values of x; in particular, it can get within $\displaystyle 1-L$ of 1, in which case, we would have the contradiction that $\displaystyle f'(x)>L$ for these sufficiently large values of x.)

Hence g must cross the horizontal axis exactly once; in other words, f has a unique fixed point.

2(b)
For this, just choose any strictly increasing function $\displaystyle g(x)<0$ such that $\displaystyle \lim_{x\to\infty}{g'(x)}=0$ (or $\displaystyle g(x)>0$ such that $\displaystyle \lim_{x\to-\infty}{g'(x)}=0$). For example, $\displaystyle g(x)=-\mathrm{e}^{-x}$, i.e. $\displaystyle f(x)=x+\mathrm{e}^{-x}$.

3. Originally Posted by pc31
Hey guys,
i'm working on real analysis and i'm stuck on these 2 questions:
1) Prove that the annulus A = {z belongs to R2| r <=|z|<=R} (where R>r>0) is connected.
Hint: It is sufficient to sohw that the annulus is pathwise connected.

4. Originally Posted by ThePerfectHacker
Hint: It is sufficient to sohw that the annulus is pathwise connected.

Can you please specify? Thank you!

5. Originally Posted by pc31
Can you please specify? Thank you!
A set $\displaystyle S$ is pathwise connected iff between any two distinct points there exists a continous function (path) joining those two point and lying wholly within the set. If $\displaystyle S$ is pathwise connected then it is connected. (The converse is not true).

6. Originally Posted by pc31
1) Prove that the annulus A = {z belongs to R2| r <=|z|<=R} (where R>r>0) is connected.
1
To prove path-connectedness …

Let $\displaystyle (r_1,\theta_1)$ and $\displaystyle (r_2,\theta_2)$, where $\displaystyle r\leq r_1,r_2\leq R$, be any two points in A, expressed in polar co-ordinates. Define $\displaystyle f:[0,1]\to \mathbb{R}^2$ by

$\displaystyle f(t)=\left((1-t)r_1+tr_2,(1-t)\theta_1+t\theta_2\right)$

Note that $\displaystyle f(0)=(r_1,\theta_1)$ and $\displaystyle f(1)=(r_2,\theta_2)$. Now all you need to do is to show that for any $\displaystyle t\in[0,1]$, $\displaystyle r\leq(1-t)r_1+tr_2\leq R$. Then f would be a path in A connecting $\displaystyle (r_1,\theta_1)$ and $\displaystyle (r_2,\theta_2)$; hence any two points in A can be connected by a path in A and so A is path-connected.