Real Analysis Real Analysis Math Forum

 August 31st, 2015, 05:52 PM #11 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Yes, zylo indeed it can. This statement is enough: Consider $\displaystyle U\subset\mathbb R^m$ and $\displaystyle f:U\rightarrow\mathbb R^n$ uniformly continuous. Then there exist $\displaystyle \overline{f}:\overline{U}\rightarrow\mathbb R^n$ continuous such that $\displaystyle \overline{f}(x)=f(x)$ for all $\displaystyle x\in U$. With this result in hand you have that if $\displaystyle U$ is bounded then $\displaystyle \overline{U}$ is compact and therefore $\displaystyle \overline{f}$ is bounded which gives us that $\displaystyle f$ is bounded. As soon as my wife let go of my laptop I will post the proof. It's a little bit boring writing it all in the cell phone. Thanks from zylo Last edited by parasio; August 31st, 2015 at 05:58 PM. September 1st, 2015, 05:53 PM #12 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 As promissed here is my solution.. Afirmation: Consider $\displaystyle U\subset\mathbb R^m$ and consider $\displaystyle f:U\rightarrow\mathbb R^n$ a uniformly continuous function. Then for each $\displaystyle a\in U'$, $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)$ exists. Proof: Take $\displaystyle a\in U'$ then there exist a sequence $\displaystyle (a_n)$ in $\displaystyle U\backslash\{a\}$ such that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}a_n=a$ which gives us that $\displaystyle (a_n)$ is a Cauchy sequence. Since $\displaystyle f$ is uniformly continuous we have that the sequence $\displaystyle (f(a_n))$ is a Cauchy sequence. Since $\displaystyle \mathbb R^n$ is complete we have that there exist $\displaystyle L=\displaystyle\lim_{n\rightarrow\infty}f(a_n)$. Take another sequence $\displaystyle (b_n)$ in $\displaystyle U\backslash\{a\}$ with $\displaystyle \displaystyle\lim_{n\rightarrow\infty}b_n=a$. Consider the sequence $\displaystyle (c_n)$ where $\displaystyle c_{2n-1}=a_n$ and $\displaystyle c_{2n}=b_n$. We have that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}c_n=a$ and therefore it's a Cauchy sequence which gives us that there exist $\displaystyle M=\displaystyle\lim_{n\rightarrow\infty}f(c_n)$. Since $\displaystyle (f(a_n))$ is a subsequence of $\displaystyle (f(c_n))$ we have that $\displaystyle M=L$. Since $\displaystyle \displaystyle\lim_{n\rightarrow\infty}b_n=a$ we have that $\displaystyle (b_n)$ is a Cauchy sequence which gives us that there exist $\displaystyle N=\displaystyle\lim_{n\rightarrow\infty}f(b_n)$. Since $\displaystyle (f(b_n))$ is a subsequence of $\displaystyle (f(c_n))$ we have that $\displaystyle N=M=L$. Hence for any sequence $\displaystyle (a_n)$ in $\displaystyle U\backslash\{a\}$ converging to $\displaystyle a$ we have that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}f(a_n)=L$ and therefore there exist $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)$ and $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)=L$. Consider now the function $\displaystyle \overline{f}:\overline{U}\rightarrow\mathbb R^n$ (remember $\displaystyle \overline{U}=U'\cup U$) defined as $\displaystyle \overline{f}(a)=f(a)$ if $\displaystyle a\in U$ and $\displaystyle \overline{f}(a)=\displaystyle\lim_{x\rightarrow a}f(x)$ if $\displaystyle a\in U'\backslash U$. $\displaystyle \overline{f}$ is obviously continuous. Tags bounded, continuity, interval, uniform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post neseka Real Analysis 0 April 22nd, 2012 10:38 AM baz Real Analysis 3 March 6th, 2012 11:00 AM watson Real Analysis 7 January 9th, 2012 12:59 PM seams192 Real Analysis 6 January 29th, 2011 02:27 AM outsos Real Analysis 8 April 28th, 2010 04:51 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      