Solutions to Homework 7

Rudin Chapter 4.

Problem 1.

The condition

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

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

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

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

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

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.