# Extension of continuous function on a closed set of a metric space

Let $$(S,d)$$ be a metric space, $$F$$ a closed set in $$S$$, and $$f$$ a continuous, bounded, nonnegative function on $$F$$. For $$x\in S$$, let $$g(x):=\text{sup}_{t\in F}f(t)/(1+d(t,x)^2)^{1/d(x,F)},x\notin F;g(x):=f(x),x\in F$$. Prove $$g$$ is bounded and continuous on $$S$$. $$Hint$$: Prove $$g$$ is both upper and lower semicontinuous, $$\text{limsup}_{y\rightarrow x}g(y)\leq g(x)\leq\text{liminf}_{y\rightarrow x}g(y)$$: (a) for $$x\notin F$$, and (b) for $$x\in F,y\notin F$$. In case (b), as $$y=y_n\rightarrow x$$ consider $$t_n\in F$$ with $$d(y_n,t_n)/d(y_n,F)\rightarrow 1$$ as $$n\rightarrow\infty$$.

My efforts:

First we show the boundedness in two cases: (1) $$x\in F$$ and (2) $$x\notin F$$.

(1) $$x\in F$$. $$g(x)=f(x)$$ is bounded since $$f$$ is bounded.

(2) $$x\notin F$$. $$d(x,F)>0$$ since $$F$$ is closed. Thus $$1/d(x,F)>0$$. For any $$t\in F$$ and $$x\notin F$$, $$d(t,x)>0$$ since $$F$$ is closed. Thus $$d(t,x)^2>0$$ and $$1+d(t,x)^2>1$$. Then $$(1+d(t,x)^2)^{1/d(x,F)}>1$$. Since $$f$$ is bounded, it is still bounded when divided by a number greater than 1. Taking sup of a bounded function over a closed set, the resulting function is still bounded.

Then we show $$g$$ is continuous by showing it is both upper and lower semicontinuous, i.e., $$\text{limsup}_{y\rightarrow x}g(y)\leq g(x)\leq\text{liminf}_{y\rightarrow x}g(y)$$. The case of $$x\in F,y\in F$$ is trivial since $$g(x)=f(x)$$ on $$F$$ and $$f$$ is continuous on $$F$$. Thus we only need to consider two cases: (a) $$x\notin F$$ and (b) $$x\in F,y\notin F$$.

The general definition of limsup is $$\text{limsup}_{y\rightarrow x}g(y):=$$ inf{sup{$$g(y):y\in U,y\neq x$$}: $$x\in U$$ open} where sup $$\emptyset:=-\infty$$. In a metric space $$(S,d)$$, the definition of limsup becomes $$\text{limsup}_{y\rightarrow x}g(y):=$$ inf{sup{$$g(y):d(x,y)}: $$r>0$$} where sup $$\emptyset:=-\infty$$. If $$r_1, then $$B(x,r_1)\subset B(x,r_2)$$ and sup{$$g(y):d(x,y)} $$\leq$$ sup{$$g(y):d(x,y)}. Thus $$\text{limsup}_{y\rightarrow x}g(y)=\text{lim}_{r\rightarrow 0}$$sup{$$g(y):d(x,y)}. Similarly we have $$\text{liminf}_{y\rightarrow x}g(y)=\text{lim}_{r\rightarrow 0}$$inf{$$g(y):d(x,y)}.

(a) $$x\notin F$$. Since $$F$$ is closed, there exists a $$r>0$$ such that $$B(x,r)\cap F=\emptyset$$. For any $$s, $$B(x,s)\cap F=\emptyset$$ since $$B(x,s)\subset B(x,r)$$. Thus $$\text{limsup}_{y\rightarrow x}g(y)=\text{lim}_{r\rightarrow 0}$$sup{$$\text{sup}_{t\in F}f(t)/(1+d(t,y)^2)^{1/d(y,F)}:d(x,y)}. Given any $$y\notin F$$, $$f(t)/(1+d(t,y)^2)^{1/d(y,F)}$$ is a continuous function of $$t$$ on $$F$$ and $$f(t)>f(t)/(1+d(t,y)^2)^{1/d(y,F)}$$.

(b) $$x\in F,y\notin F$$. In this case, $$x$$ must be the boundary point of $$F$$. Since $$d(y,F)$$ is a continuous function of $$y$$, $$\text{lim}_{y\rightarrow x}d(y,F)=0$$. As $$y=y_n\rightarrow x$$ consider $$t_n\in F$$ with $$d(y_n,t_n)/d(y_n,F)\rightarrow 1$$ as $$n\rightarrow\infty$$. Then $$\text{lim}_{n\rightarrow\infty}(1+d(t_n,y_n)^2)^{1/d(y_n,F)}=\text{lim}_{n\rightarrow\infty}(1+d(y_n,F)d(t_n,y_n)^2/d(y_n,F))^{1/d(y_n,F)}=\text{lim}_{n\rightarrow\infty}\text{exp}(d(t_n,y_n)^2/d(y_n,F))$$ = $$\text{lim}_{n\rightarrow\infty}\text{exp}(d(t_n,y_n))=\text{lim}_{n\rightarrow\infty}\text{exp}(d(y_n,F))=\text{exp}(d(x,F))=e^0=1$$. Thus $$\text{lim}_{n\rightarrow\infty}g(y_n)\geq\text{lim}_{n\rightarrow\infty}f(t_n)=f(x)$$

No matter (a) or (b), I don't know how to proceed.

• So what is your question? Aug 24 '20 at 8:32
• @PaulFrost No matter (a) or (b), I don't know how to proceed. Aug 24 '20 at 15:27