proof of Tietze extension theorem

To prove the Tietze Extension Theorem, we first need a lemma.

Lemma 1.

If $X$ is a normal topological space and $A$ is closed in $X$ , then for any continuous function $f\colon A\to\mathbb{R}$ such that $|f(x)|\leq 1$ , there is a continuous function $g\colon X\to\mathbb{R}$ such that $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$ , and $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ for $x\in A$ .

Proof.

The sets $f^{-1}\bigl{(}(-\infty,-\frac{1}{3}]\bigr{)}$ and $f^{-1}\bigl{(}[\frac{1}{3},\infty)\bigr{)}$ are disjoint and closed in $A$ . Since $A$ is closed, they are also closed in $X$ . Since $X$ is normal, then by Urysohn’s lemma and the fact that $[0,1]$ is homeomorphic to $[-\frac{1}{3},\frac{1}{3}]$ , there is a continuous function $g\colon X\to[-\frac{1}{3},\frac{1}{3}]$ such that $g\Bigl{(}f^{-1}\bigl{(}(-\infty,-\frac{1}{3}c]\bigr{)}\Bigr{)}=-\frac{1}{3}$ and $g\Bigl{(}f^{-1}\bigl{(}[\frac{1}{3},\infty)\bigr{)}\Bigr{)}=\frac{1}{3}$ . Thus $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$ . Now if $-\leq f(x)\leq-\frac{1}{3}$ , then $g(x)=-\frac{1}{3}$ and thus $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ . Similarly if $\frac{1}{3}\leq f(x)\leq 1$ , then $g(x)=\frac{1}{3}$ and thus $|f(x)-g(x)|\leq\frac{2}{3}$ . Finally, for $\left\lvert f(x)\right\rvert\leq\frac{1}{3}$ we have that $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ , and so $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ . Hence $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ holds for all $x\in A$ . ∎

This puts us in a position to prove the main theorem.

Proof of the Tietze extension theorem.

First suppose that for any continuous function on a closed subset there is a continuous extension. Let $C$ and $D$ be disjoint and closed in $X$ . Define $f\colon C\cup D\to\mathbb{R}$ by $f(x)=0$ for $x\in C$ and $f(x)=1$ for $x\in D$ . Now $f$ is continuous and we can extend it to a continuous function $F\colon X\to\mathbb{R}$ . By Urysohn’s lemma, $X$ is normal because $F$ is a continuous function such that $F(x)=0$ for $x\in C$ and $F(x)=1$ for $x\in D$ .

Conversely, let $X$ be normal and $A$ be closed in $X$ . By the lemma, there is a continuous function $g_{0}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{0}(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)\right\rvert\leq\frac{2}{3}$ for $x\in A$ . Since $(f-g_{0})\colon A\rightarrow\mathbb{R}$ is continuous, the lemma tells us there is a continuous function $g_{1}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{1}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)\right\rvert\leq\frac{2}{3}(\frac{2}{3})$ for $x\in A$ . By repeated application of the lemma we can construct a sequence of continuous functions $g_{0},g_{1},g_{2},\ldots$ such that $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ for all $x\in X$ , and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)-g_{2}(x)-\cdots\right\rvert\leq(\frac{2}{3}% )^{n}$ for $x\in A$ .

Define $F(x)=\sum_{n=0}^{\infty}g_{n}(x)$ . Since $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ and $\sum_{n=0}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}$ converges as a geometric series, then $\sum_{n=0}^{\infty}g_{n}(x)$ converges absolutely and uniformly, so $F$ is a continuous function defined everywhere. Moreover $\sum_{n=0}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}=1$ implies that $\left\lvert F(x)\right\rvert\leq 1$ .

Now for $x\in A$ , we have that $\left\lvert f(x)-\sum_{n=0}^{k}g_{n}(x)\right\rvert\leq(\frac{2}{3})^{k+1}$ and as $k$ goes to infinity, the right side goes to zero and so the sum goes to $F(x)$ . Thus $\left\lvert f(x)-F(x)\right\rvert=0$ Therefore $F$ extends $f$ .∎

Remarks: If $f$ was a function satisfying $\left\lvert f(x)\right\rvert<1$ , then the theorem can be strengthened as follows. Find an extension $F$ of $f$ as above. The set $B=F^{-1}(\{-1\}\cup\{1\})$ is closed and disjoint from $A$ because $\left\lvert F(x)\right\rvert=\left\lvert f(x)\right\rvert<1$ for $x\in A$ . By Urysohn’s lemma there is a continuous function $\phi$ such that $\phi(A)=\{1\}$ and $\phi(B)=\{0\}$ . Hence $F(x)\phi(x)$ is a continuous extension of $f(x)$ , and has the property that $\left\lvert F(x)\phi(x)\right\rvert<1$ .

If $f$ is unbounded, then Tietze extension theorem holds as well. To see that consider $t(x)=\tan^{-1}(x)/(\pi/2)$ . The function $t\circ f$ has the property that $(t\circ f)(x)<1$ for $x\in A$ , and so it can be extended to a continuous function $h\colon X\to\mathbb{R}$ which has the property $\left\lvert h(x)\right\rvert<1$ . Hence $t^{-1}\circ h$ is a continuous extension of $f$ .

Title	proof of Tietze extension theorem
Canonical name	ProofOfTietzeExtensionTheorem
Date of creation	2013-03-22 14:08:58
Last modified on	2013-03-22 14:08:58
Owner	bbukh (348)
Last modified by	bbukh (348)
Numerical id	10
Author	bbukh (348)
Entry type	Proof
Classification	msc 54C20