Tietze extension theorem

Proof

The function ${\displaystyle F}$ is constructed iteratively. Firstly, we define $${\displaystyle \begin{array}{cc}{c}_0& =sup\{|f(a)|:a\in A\}\\ E_0& =\{a\in A:f(a)\ge c_0/3\}\\ F_0& =\{a\in A:f(a)\le -c_0/3\}.\end{array}}$$ Observe that ${\displaystyle E_0}$ and ${\displaystyle F_0}$ are closed and disjoint subsets of ${\displaystyle A}$. By taking a linear combination of the function obtained from the proof of Urysohn's lemma, there exists a continuous function ${\displaystyle g_0:X\to \mathbb{R}}$ such that $${\displaystyle \begin{array}{cc}{g}_0& =\frac{c_0}{3}\text{ on }{E}_0\\ g_0& =-\frac{c_0}{3}\text{ on }{F}_0\end{array}}$$ and furthermore $${\displaystyle -\frac{c_0}{3}\le g_0\le \frac{c_0}{3}}$$ on ${\displaystyle X}$. In particular, it follows that $${\displaystyle \begin{array}{cc}|g_0|& \le \frac{c_0}{3}\\ |f-g_0|& \le \frac{2c_0}{3}\end{array}}$$ on ${\displaystyle A}$. We now use induction to construct a sequence of continuous functions ${\displaystyle (g_n{)}_{n=0}^{\mathrm{\infty }}}$ such that $${\displaystyle \begin{array}{cc}|g_n|& \le \frac{2^n{c}_0}{3^{n+1}}\\ |f-g_0-...-g_n|& \le \frac{2^{n+1}{c}_0}{3^{n+1}}.\end{array}}$$ We've shown that this holds for ${\displaystyle n=0}$ and assume that ${\displaystyle g_0,...,g_{n-1}}$ have been constructed. Define $${\displaystyle c_{n-1}=sup\{|f(a)-g_0(a)-...-g_{n-1}(a)|:a\in A\}}$$ and repeat the above argument replacing ${\displaystyle c_0}$ with ${\displaystyle c_{n-1}}$ and replacing ${\displaystyle f}$ with ${\displaystyle f-g_0-...-g_{n-1}}$. Then we find that there exists a continuous function ${\displaystyle g_n:X\to \mathbb{R}}$ such that $${\displaystyle \begin{array}{cc}|g_n|& \le \frac{c_{n-1}}{3}\\ |f-g_0-...-g_n|& \le \frac{2c_{n-1}}{3}.\end{array}}$$ By the inductive hypothesis, ${\displaystyle c_{n-1}\le 2^n{c}_0/3^n}$ hence we obtain the required identities and the induction is complete. Now, we define a continuous function ${\displaystyle F_n:X\to \mathbb{R}}$ as $${\displaystyle F_n=g_0+...+g_n.}$$ Given ${\displaystyle n\ge m}$, $${\displaystyle \begin{array}{cc}|F_n-F_m|& =|g_{m+1}+...+g_n|\\ & \le \left({\left(\frac{2}{3}\right)}^{m+1}+...+{\left(\frac{2}{3}\right)}^n\right)\frac{c_0}{3}\\ & \le {\left(\frac{2}{3}\right)}^{m+1}{c}_0.\end{array}}$$ Therefore, the sequence ${\displaystyle (F_n{)}_{n=0}^{\mathrm{\infty }}}$ is Cauchy. Since the space of continuous functions on ${\displaystyle X}$ together with the sup norm is a complete metric space, it follows that there exists a continuous function ${\displaystyle F:X\to \mathbb{R}}$ such that ${\displaystyle F_n}$ converges uniformly to ${\displaystyle F}$. Since $${\displaystyle |f-F_n|\le \frac{2^n{c}_0}{3^{n+1}}}$$ on ${\displaystyle A}$, it follows that ${\displaystyle F=f}$ on ${\displaystyle A}$. Finally, we observe that $${\displaystyle |F_n|\le \sum _{n=0}^{\mathrm{\infty }}|g_n|\le c_0}$$ hence ${\displaystyle F}$ is bounded and has the same bound as ${\displaystyle f}$. ${\displaystyle ◻}$

History

L. E. J. Brouwer and Henri Lebesgue proved a special case of the theorem, when ${\displaystyle X}$ is a finite-dimensional real vector space. Heinrich Tietze extended it to all metric spaces, and Pavel Urysohn proved the theorem as stated here, for normal topological spaces.

Equivalent statements

This theorem is equivalent to Urysohn's lemma (which is also equivalent to the normality of the space) and is widely applicable, since all metric spaces and all compact Hausdorff spaces are normal. It can be generalized by replacing ${\displaystyle \mathbb{R}}$ with ${\displaystyle {\mathbb{R}}^J}$ for some indexing set ${\displaystyle J,}$ any retract of ${\displaystyle {\mathbb{R}}^J,}$ or any normal absolute retract whatsoever.

Variations

If ${\displaystyle X}$ is a metric space, ${\displaystyle A}$ a non-empty subset of ${\displaystyle X}$ and ${\displaystyle f:A\to \mathbb{R}}$ is a Lipschitz continuous function with Lipschitz constant ${\displaystyle K,}$ then ${\displaystyle f}$ can be extended to a Lipschitz continuous function ${\displaystyle F:X\to \mathbb{R}}$ with same constant ${\displaystyle K.}$ This theorem is also valid for Hölder continuous functions, that is, if ${\displaystyle f:A\to \mathbb{R}}$ is Hölder continuous function with constant less than or equal to ${\displaystyle 1,}$ then ${\displaystyle f}$ can be extended to a Hölder continuous function ${\displaystyle F:X\to \mathbb{R}}$ with the same constant.

Another variant (in fact, generalization) of Tietze's theorem is due to H.Tong and Z. Ercan: Let ${\displaystyle A}$ be a closed subset of a normal topological space ${\displaystyle X.}$ If ${\displaystyle f:X\to \mathbb{R}}$ is an upper semicontinuous function, ${\displaystyle g:X\to \mathbb{R}}$ a lower semicontinuous function, and ${\displaystyle h:A\to \mathbb{R}}$ a continuous function such that ${\displaystyle f(x)\le g(x)}$ for each ${\displaystyle x\in X}$ and ${\displaystyle f(a)\le h(a)\le g(a)}$ for each ${\displaystyle a\in A}$, then there is a continuous extension ${\displaystyle H:X\to \mathbb{R}}$ of ${\displaystyle h}$ such that ${\displaystyle f(x)\le H(x)\le g(x)}$ for each ${\displaystyle x\in X.}$ This theorem is also valid with some additional hypothesis if ${\displaystyle \mathbb{R}}$ is replaced by a general locally solid Riesz space.

Dugundji (1951) extends the theorem as follows: If ${\displaystyle X}$ is a metric space, ${\displaystyle Y}$ is a locally convex topological vector space, ${\displaystyle A}$ is a closed subset of ${\displaystyle X}$ and ${\displaystyle f:A\to Y}$ is continuous, then it could be extended to a continuous function ${\displaystyle \tilde{f}}$ defined on all of ${\displaystyle X}$. Moreover, the extension could be chosen such that ${\displaystyle \tilde{f}(X)\subseteq \text{conv}f(A)}$