A uniformly Cauchy sequence of functions is uniformly convergent proof












0














Let $f_n:Asubseteqmathbb{R}tomathbb{R}$



$f_n$ is uniformly Cauchy $implies$ $exists f:Atomathbb{R}$ : $f_nxrightarrow{u}f$ in A



proof.



$forallvarepsilon>0$ $existsnu$ : $forall n,m>nu$ $sup_{xin A}|f_n(x)-f_m(x)|<varepsilon$



I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n,m>nu$ $forall xin A$ $|f_n(x)-f_m(x)|<varepsilon$



Where is sup now?



So if $mto+infty$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $forall xin A$ $|f_n(x)-f(x)|leqvarepsilon$



Ok it's clear why m disappears but I don't understand why $<varepsilon$ becomes $leqvarepsilon$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $sup_{xin A}|f_n(x)-f(x)|leqvarepsilon$



This is not clear, where does sup come from?



$implieslim_{ntoinfty}sup_{xin A}|f_n(x)-f(x)|=0$



This is directly from the definition of limit, and this means that



$implies f_nxrightarrow{u}f$ in $A$






real-analysis







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  • Please consider accepting my answer if it has helped :)
    – user667
    Nov 19 at 14:11
















0














Let $f_n:Asubseteqmathbb{R}tomathbb{R}$



$f_n$ is uniformly Cauchy $implies$ $exists f:Atomathbb{R}$ : $f_nxrightarrow{u}f$ in A



proof.



$forallvarepsilon>0$ $existsnu$ : $forall n,m>nu$ $sup_{xin A}|f_n(x)-f_m(x)|<varepsilon$



I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n,m>nu$ $forall xin A$ $|f_n(x)-f_m(x)|<varepsilon$



Where is sup now?



So if $mto+infty$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $forall xin A$ $|f_n(x)-f(x)|leqvarepsilon$



Ok it's clear why m disappears but I don't understand why $<varepsilon$ becomes $leqvarepsilon$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $sup_{xin A}|f_n(x)-f(x)|leqvarepsilon$



This is not clear, where does sup come from?



$implieslim_{ntoinfty}sup_{xin A}|f_n(x)-f(x)|=0$



This is directly from the definition of limit, and this means that



$implies f_nxrightarrow{u}f$ in $A$






real-analysis







share|cite|improve this question






















  • Please consider accepting my answer if it has helped :)
    – user667
    Nov 19 at 14:11














0












0








0







Let $f_n:Asubseteqmathbb{R}tomathbb{R}$



$f_n$ is uniformly Cauchy $implies$ $exists f:Atomathbb{R}$ : $f_nxrightarrow{u}f$ in A



proof.



$forallvarepsilon>0$ $existsnu$ : $forall n,m>nu$ $sup_{xin A}|f_n(x)-f_m(x)|<varepsilon$



I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n,m>nu$ $forall xin A$ $|f_n(x)-f_m(x)|<varepsilon$



Where is sup now?



So if $mto+infty$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $forall xin A$ $|f_n(x)-f(x)|leqvarepsilon$



Ok it's clear why m disappears but I don't understand why $<varepsilon$ becomes $leqvarepsilon$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $sup_{xin A}|f_n(x)-f(x)|leqvarepsilon$



This is not clear, where does sup come from?



$implieslim_{ntoinfty}sup_{xin A}|f_n(x)-f(x)|=0$



This is directly from the definition of limit, and this means that



$implies f_nxrightarrow{u}f$ in $A$






real-analysis







share|cite|improve this question













Let $f_n:Asubseteqmathbb{R}tomathbb{R}$



$f_n$ is uniformly Cauchy $implies$ $exists f:Atomathbb{R}$ : $f_nxrightarrow{u}f$ in A



proof.



$forallvarepsilon>0$ $existsnu$ : $forall n,m>nu$ $sup_{xin A}|f_n(x)-f_m(x)|<varepsilon$



I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n,m>nu$ $forall xin A$ $|f_n(x)-f_m(x)|<varepsilon$



Where is sup now?



So if $mto+infty$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $forall xin A$ $|f_n(x)-f(x)|leqvarepsilon$



Ok it's clear why m disappears but I don't understand why $<varepsilon$ becomes $leqvarepsilon$



$implies forallvarepsilon>0$ $existsnuinmathbb{N}$ : $forall n>nu$ $sup_{xin A}|f_n(x)-f(x)|leqvarepsilon$



This is not clear, where does sup come from?



$implieslim_{ntoinfty}sup_{xin A}|f_n(x)-f(x)|=0$



This is directly from the definition of limit, and this means that



$implies f_nxrightarrow{u}f$ in $A$






real-analysis




real-analysis






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share|cite|improve this question











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asked Nov 18 at 15:16









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  • Please consider accepting my answer if it has helped :)
    – user667
    Nov 19 at 14:11


















  • Please consider accepting my answer if it has helped :)
    – user667
    Nov 19 at 14:11
















Please consider accepting my answer if it has helped :)
– user667
Nov 19 at 14:11




Please consider accepting my answer if it has helped :)
– user667
Nov 19 at 14:11










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Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $sup_{xin A}|f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$ then we are guaranteed that $forall xin A, |f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<epsilon$ for all $xin A$, we have that $epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $sup_{x in A}|f_n(x)-f|leqepsilon$ since equality could hold.



I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x in A$. We have, $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence ${f_n(x)}$. We know that there is a pointwise limit because $mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<frac{epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<frac{epsilon}{2}$. Now set $m>max{N_1,N}$ (this step is a rigorous way of saying $m rightarrow infty$). Putting it all together $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $xin A$) such that, $$forall xin A, n>N, |f_n(x)-f(x)|<epsilon$$






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    Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $sup_{xin A}|f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$ then we are guaranteed that $forall xin A, |f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<epsilon$ for all $xin A$, we have that $epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $sup_{x in A}|f_n(x)-f|leqepsilon$ since equality could hold.



    I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x in A$. We have, $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence ${f_n(x)}$. We know that there is a pointwise limit because $mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<frac{epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<frac{epsilon}{2}$. Now set $m>max{N_1,N}$ (this step is a rigorous way of saying $m rightarrow infty$). Putting it all together $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $xin A$) such that, $$forall xin A, n>N, |f_n(x)-f(x)|<epsilon$$






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      Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $sup_{xin A}|f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$ then we are guaranteed that $forall xin A, |f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<epsilon$ for all $xin A$, we have that $epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $sup_{x in A}|f_n(x)-f|leqepsilon$ since equality could hold.



      I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x in A$. We have, $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence ${f_n(x)}$. We know that there is a pointwise limit because $mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<frac{epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<frac{epsilon}{2}$. Now set $m>max{N_1,N}$ (this step is a rigorous way of saying $m rightarrow infty$). Putting it all together $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $xin A$) such that, $$forall xin A, n>N, |f_n(x)-f(x)|<epsilon$$






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        Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $sup_{xin A}|f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$ then we are guaranteed that $forall xin A, |f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<epsilon$ for all $xin A$, we have that $epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $sup_{x in A}|f_n(x)-f|leqepsilon$ since equality could hold.



        I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x in A$. We have, $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence ${f_n(x)}$. We know that there is a pointwise limit because $mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<frac{epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<frac{epsilon}{2}$. Now set $m>max{N_1,N}$ (this step is a rigorous way of saying $m rightarrow infty$). Putting it all together $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $xin A$) such that, $$forall xin A, n>N, |f_n(x)-f(x)|<epsilon$$






        share|cite|improve this answer














        Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $sup_{xin A}|f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$ then we are guaranteed that $forall xin A, |f_n(x)-f_m(x)|<epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<epsilon$ for all $xin A$, we have that $epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $sup_{x in A}|f_n(x)-f|leqepsilon$ since equality could hold.



        I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x in A$. We have, $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence ${f_n(x)}$. We know that there is a pointwise limit because $mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<frac{epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<frac{epsilon}{2}$. Now set $m>max{N_1,N}$ (this step is a rigorous way of saying $m rightarrow infty$). Putting it all together $$|f_n(x)-f(x)|leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $xin A$) such that, $$forall xin A, n>N, |f_n(x)-f(x)|<epsilon$$







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        edited Nov 18 at 18:21

























        answered Nov 18 at 18:15









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