Solutions to Homework 7

Rudin Chapter 4.

Problem 1.
The condition

$\displaystyle \lim_{h\to0}[f(x+h)-f(x-h)]=0$    

does not imply continuity ofr $ f$ at $ x.$ It certainly holds if $ f$ is continuous, so consider a function which is not continuous at one point, such as

$\displaystyle f(x)=\begin{cases}0&x\not=0\\ 1&x=0.\end{cases}$    

Certainly the condition holds for $ x\not=0$ by continuity. For $ x=0$ it also holds, since the definition of limit excludes the value $ h=0,$ thus for $ h\not=0,$ $ f(h)-f(-h)=0.$

Problem 4.
If $ f:X\longrightarrow Y$ is continuous and $ E\subset X$ is dense, then for every $ x\in X$ there is a sequence $ z_n\in E$ such that $ z_n\to x$ in $ X$ as $ n\to\infty.$ Then, by the continuity of $ f,$ $ f(z_n)\to f(x)$ in $ Y,$ which shows that $ f(E)$ is dense in $ f(X)\subset
Y.$ Now, if $ f$ and $ g:X\longrightarrow Y$ are both continuoues and $ f(z)=g(z)$ for all $ z\in E,$ where $ E\subset X$ is dense it follows that $ f=g,$ that is $ f(x)=g(x)$ for all $ x\in X.$ Indeed, by the density of $ E$ in $ X$ there exists a sequence $ z_n\to x$ in $ X$ with $ z_n\in E$ for all $ n.$ Then

$\displaystyle f(x)=\lim_{n\to\infty}f(z_n)=\lim_{n\to\infty}g(z_n)=g(x).$    

Problem 15.
Let $ f:\mathbb{R}\longrightarrow \mathbb{R}$ be continuous and open. Thus if $ V\subset\mathbb{R}$ is open then both $ f^{-1}(V)$ and $ f(V)$ are open. If $ a<b$ in $ \mathbb{R}$ then $ f$ assumes its maximum, and minimum, on $ [a,b],$ by the continuity of $ f.$ If there is an interior point at which $ f$ assumes its maximum or minumum then $ f((a,b))$ cannot be open, since the maximum, or minimum is not an interior point of it. Since $ [a,b]$ is connected, $ f([a,b])=[c,d]$ is an interval and $ f((a,b))=(c,d)=[c,d]\setminus\{f(a),f(b)\}.$ In particular $ f(a)\not=f(b)$ since otherwise these cannot be the endpoints. Thus $ f$ is continuous and injective. Such a map is necessarily monotonic. To see this explicitly, suppose $ f(0)<f(1)$ (otherwise replace $ f$ by $ -f).$ Then if $ x>0,$ $ f(x)>f(0)$ since if not, $ f([0,x+1])$ contains point on both sides of $ f(0),$ $ f(1)>f(0)$ and $ f(x)<f(0)$ and so would contain $ f(0),$ violating the injetivity. Similarly, if $ x<0$ then $ f(x)<0.$ The same argument now shows that given $ r>0,$ $ x>r$ implies $ f(x)>f(r)$ and similarly for $ r<0.$ Thus $ f$ is monotonic.

Richard B. Melrose 2004-05-18