# Metrizable Space and continuous extension function

Let A be a closed subspace of a metrizable space $X$, then for every continuous function $f:A\rightarrow I$ and every metric $\rho$ on the space $X$ , I want to show that the formula

$$F(x) = \begin{cases} \inf \left\{f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1: a\in A \right\}, & \textrm{ if x\in X\backslash A  ,} \\\\ f(x), & \textrm{ if x\in A}, \\ \end{cases}$$

define a continuous extension $F$ of $f$ over $X$.

Remark:- What I know I have to use TIETZE-URYSOHN THEOREM,which states:-

Every continuous function from a closed subspace $M$ of a normal space $X$ to $I$ or $\mathbb{R}$ is continuously extendable over $X$. .

From the comment I figure out this remark is not useful.

• You don't have to use Tietze. You have to show that $F$ is continuous, then you have an explicit extension of $f$, while Tietze/Urysohn only gives existence [and the proof a way to construct one, but very cumbersome]. This shows that metric spaces are particularly nice. – Daniel Fischer Nov 30 '14 at 20:32
• Now, I understand we don't need to use the Tietze/Urysohn theorem. thanks – Birkary Nov 30 '14 at 20:51
• Historical aside: this extension formula first appeared in 1919 paper by Hausdorff, following the 1915 paper by Tietze from which the extension result originates (although Tietze proved it for metric spaces only). References here – user147263 Dec 1 '14 at 0:27
• More importantly, I think you have to assume $f$ is bounded from below. For example, suppose $A$ is the closed unit ball of a Hilbert space: then there is a continuous real-valued $f$ with $\inf_A f= -\infty$, hence $F(x)\equiv -\infty$ according to this definition. – user147263 Dec 1 '14 at 0:32

In order for this extension to work, $f$ must be assumed bounded from below. Indeed, if $A$ is the $x$-axis of $\mathbb R^2$ and $f(x,0)=-x^2$ , the extension gives $F(x)=-\infty$ for every $x\in \mathbb R^2\setminus A$. Such examples exist even if $A$ is a bounded closed set.

The idea of this formula is that even though the infimum allows $a$ to be any point of $A$, the "far-away" points have large ratio $\rho(x,a)/\rho (x,A)$ which effectively drops them from the competition for the infimum. In contrast, when $a$ is an almost-nearest point of $A$ to $x$, we have $\rho(x,a)\approx \rho(x,A)$ and so the "penalty" $\frac{\rho(x,a)}{\rho(x,A)}-1$ drops down to nearly zero.

### Continuity at $x_0\in X\setminus A$

Focus on the ball of radius $\frac12 \rho(x_0,A)$ centered at $x_0$. Within this ball, $\rho(x,A)\ge \frac12 \rho(x_0,A)$. The ratio of two Lipschitz continuous functions is Lipschitz continuous, provided that the denominator is bounded away from $0$. More precisely, the function $$g_a(x) = f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1$$ is Lipschitz on this ball with constant $4/\rho(x_0,A)$, independent of $a$. Taking the infimum preserves this property.

### Continuity at $x_0\in A$

Given $\epsilon>0$, take $R>0$ such that $|f(a)-f(x_0)|<\epsilon$ when $\rho(a,x_0)<R$. If $x\notin A$ and $\rho(x,x_0)<r\ll R$ ($r$ much smaller than $R$), then every point $a\in A$ with $\rho(a,x_0)\ge R$ has $$f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1 \ge \inf_A f+\frac{R}{r}-1$$ So, choose $r$ small enough so that $$\inf_A f+\frac{R}{r}-1> f(x_0)$$ Use this to conclude that $F(x)\ge f(x_0)-\epsilon$ when $\rho(x,x_0)<r$.

The inequality $F(x)\le f(x_0)+\epsilon$ follows by considering $a\in A$ such that $\rho(x,a)/\rho(x,A)$ is very close to $1$; note that such $a$ will be within $B(x_0,R)$.