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 $f\mathrm{:}A\mathrm{\to }\mathrm{R}$ such that $\mathrm{|}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{|}\mathrm{\le }\mathrm{1}$, there is a continuous function $g\mathrm{:}X\mathrm{\to }\mathrm{R}$ such that $\mathrm{\left|}g\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\right|}\mathrm{\le }\frac{\mathrm{1}}{\mathrm{3}}$ for $x\mathrm{\in }X$, and $\mathrm{\left|}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{-}g\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\right|}\mathrm{\le }\frac{\mathrm{2}}{\mathrm{3}}$ for $x\mathrm{\in }A$.

Proof.

The sets $f^{-1}\left((-\mathrm{\infty },-\frac{1}{3}]\right)$ and $f^{-1}\left([\frac{1}{3},\mathrm{\infty })\right)$ 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 $g:X\to [-\frac{1}{3},\frac{1}{3}]$ such that $g\left(f^{-1}\left((-\mathrm{\infty },-\frac{1}{3}c]\right)\right)=-\frac{1}{3}$ and $g\left(f^{-1}\left([\frac{1}{3},\mathrm{\infty })\right)\right)=\frac{1}{3}$. Thus $\left|g(x)\right|\le \frac{1}{3}$ for $x\in X$. Now if $-\le f(x)\le -\frac{1}{3}$, then $g(x)=-\frac{1}{3}$ and thus $\left|f(x)-g(x)\right|\le \frac{2}{3}$. Similarly if $\frac{1}{3}\le f(x)\le 1$, then $g(x)=\frac{1}{3}$ and thus $|f(x)-g(x)|\le \frac{2}{3}$. Finally, for $\left|f(x)\right|\le \frac{1}{3}$ we have that $\left|g(x)\right|\le \frac{1}{3}$, and so $\left|f(x)-g(x)\right|\le \frac{2}{3}$. Hence $\left|f(x)-g(x)\right|\le \frac{2}{3}$ 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 $f:C\cup D\to ℝ$ 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 $F:X\to ℝ$. 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 $g_0:X\to ℝ$ such that $\left|g_0(x)\right|\le \frac{1}{3}$ for $x\in X$ and $\left|f(x)-g_0(x)\right|\le \frac{2}{3}$ for $x\in A$. Since $(f-g_0):A\to ℝ$ is continuous, the lemma tells us there is a continuous function $g_1:X\to ℝ$ such that $\left|g_1(x)\right|\le \frac{1}{3}(\frac{2}{3})$ for $x\in X$ and $\left|f(x)-g_0(x)-g_1(x)\right|\le \frac{2}{3}(\frac{2}{3})$ for $x\in A$. By repeated application of the lemma we can construct a sequence of continuous functions $g_0,g_1,g_2,\mathrm{\dots }$ such that $\left|g_n(x)\right|\le \frac{1}{3}{(\frac{2}{3})}^n$ for all $x\in X$, and $\left|f(x)-g_0(x)-g_1(x)-g_2(x)-\mathrm{\cdots }\right|\le {(\frac{2}{3})}^n$ for $x\in A$.

Define $F(x)={\sum }_{n=0}^{\mathrm{\infty }}{g}_n(x)$. Since $\left|g_n(x)\right|\le \frac{1}{3}{(\frac{2}{3})}^n$ and ${\sum }_{n=0}^{\mathrm{\infty }}\frac{1}{3}{(\frac{2}{3})}^n$ converges as a geometric series, then ${\sum }_{n=0}^{\mathrm{\infty }}{g}_n(x)$ converges absolutely and uniformly, so $F$ is a continuous function defined everywhere. Moreover ${\sum }_{n=0}^{\mathrm{\infty }}\frac{1}{3}{(\frac{2}{3})}^n=1$ implies that $\left|F(x)\right|\le 1$.

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

Remarks: If $f$ was a function satisfying , 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 for $x\in A$. By Urysohn’s lemma there is a continuous function $\varphi$ such that $\varphi (A)=\{1\}$ and $\varphi (B)=\{0\}$. Hence $F(x)\varphi (x)$ is a continuous extension of $f(x)$, and has the property that .

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 for $x\in A$, and so it can be extended to a continuous function $h:X\to ℝ$ which has the property . 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