Proving that continuity and the lim sup of a given set being 0 are equivalent











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Suppose we have a function defined as follows:



$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$



with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.



It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.



However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!






real-analysis continuity epsilon-delta limsup-and-liminf







share|cite|improve this question
























  • Write out the definition of lim sup. It will become very obvious.
    – Don Thousand
    Nov 21 at 2:59










  • Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
    – notadoctor
    Nov 21 at 3:07










  • That's not the definition of lim sup...
    – Don Thousand
    Nov 21 at 3:08






  • 1




    @notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
    – user25959
    Nov 21 at 3:10










  • I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
    – notadoctor
    Nov 21 at 4:02















up vote
0
down vote

favorite












Suppose we have a function defined as follows:



$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$



with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.



It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.



However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!






real-analysis continuity epsilon-delta limsup-and-liminf







share|cite|improve this question
























  • Write out the definition of lim sup. It will become very obvious.
    – Don Thousand
    Nov 21 at 2:59










  • Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
    – notadoctor
    Nov 21 at 3:07










  • That's not the definition of lim sup...
    – Don Thousand
    Nov 21 at 3:08






  • 1




    @notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
    – user25959
    Nov 21 at 3:10










  • I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
    – notadoctor
    Nov 21 at 4:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose we have a function defined as follows:



$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$



with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.



It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.



However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!






real-analysis continuity epsilon-delta limsup-and-liminf







share|cite|improve this question















Suppose we have a function defined as follows:



$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$



with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.



It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.



However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!






real-analysis continuity epsilon-delta limsup-and-liminf




real-analysis continuity epsilon-delta limsup-and-liminf






share|cite|improve this question















share|cite|improve this question













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edited Nov 21 at 3:35









Chinnapparaj R

4,9951825




4,9951825










asked Nov 21 at 2:52









notadoctor

947




947












  • Write out the definition of lim sup. It will become very obvious.
    – Don Thousand
    Nov 21 at 2:59










  • Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
    – notadoctor
    Nov 21 at 3:07










  • That's not the definition of lim sup...
    – Don Thousand
    Nov 21 at 3:08






  • 1




    @notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
    – user25959
    Nov 21 at 3:10










  • I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
    – notadoctor
    Nov 21 at 4:02


















  • Write out the definition of lim sup. It will become very obvious.
    – Don Thousand
    Nov 21 at 2:59










  • Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
    – notadoctor
    Nov 21 at 3:07










  • That's not the definition of lim sup...
    – Don Thousand
    Nov 21 at 3:08






  • 1




    @notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
    – user25959
    Nov 21 at 3:10










  • I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
    – notadoctor
    Nov 21 at 4:02
















Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 at 2:59




Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 at 2:59












Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 at 3:07




Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 at 3:07












That's not the definition of lim sup...
– Don Thousand
Nov 21 at 3:08




That's not the definition of lim sup...
– Don Thousand
Nov 21 at 3:08




1




1




@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 at 3:10




@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 at 3:10












I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 at 4:02




I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 at 4:02










1 Answer
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Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.






share|cite|improve this answer





















  • Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
    – notadoctor
    Nov 21 at 6:44










  • $|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
    – Kavi Rama Murthy
    Nov 21 at 7:42











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Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.






share|cite|improve this answer





















  • Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
    – notadoctor
    Nov 21 at 6:44










  • $|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
    – Kavi Rama Murthy
    Nov 21 at 7:42















up vote
1
down vote



accepted










Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.






share|cite|improve this answer





















  • Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
    – notadoctor
    Nov 21 at 6:44










  • $|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
    – Kavi Rama Murthy
    Nov 21 at 7:42













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.






share|cite|improve this answer












Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 5:12









Kavi Rama Murthy

47.4k31854




47.4k31854












  • Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
    – notadoctor
    Nov 21 at 6:44










  • $|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
    – Kavi Rama Murthy
    Nov 21 at 7:42


















  • Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
    – notadoctor
    Nov 21 at 6:44










  • $|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
    – Kavi Rama Murthy
    Nov 21 at 7:42
















Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 at 6:44




Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 at 6:44












$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 at 7:42




$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 at 7:42


















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