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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{$g(y):d(x,y)<r,y\neq x$}: $r>0$} where sup $\emptyset:=-\infty$. If $r_1<r_2$, then $B(x,r_1)\subset B(x,r_2)$ and sup{$g(y):d(x,y)<r_1,y\neq x$} $\leq$ sup{$g(y):d(x,y)<r_2,y\neq x$}. Thus $\text{limsup}_{y\rightarrow x}g(y)=\text{lim}_{r\rightarrow 0}$sup{$g(y):d(x,y)<r,y\neq x$}. Similarly we have $\text{liminf}_{y\rightarrow x}g(y)=\text{lim}_{r\rightarrow 0}$inf{$g(y):d(x,y)<r,y\neq x$}.

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

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

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