Let $f:mathbbR rightarrowmathbbR$ such that $f$ is continuous and limit at $infty$ and $-infty $ exist then function is uniformly continuous?
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Let $f:mathbbR rightarrowmathbbR$ such that $f$ is continuous and limit at $infty$ and $-infty $ exist then function is uniformly continuous?
I do not think this is true as limit may exist at end points but what if function wiggle in between. May be I am wrong. Can I get some hint?
real-analysis general-topology
asked Sep 5 at 7:46
StammeringMathematician
56616
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Let $f:mathbbR rightarrowmathbbR$ such that $f$ is continuous and limit at $infty$ and $-infty $ exist then function is uniformly continuous?
I do not think this is true as limit may exist at end points but what if function wiggle in between. May be I am wrong. Can I get some hint?
real-analysis general-topology
asked Sep 5 at 7:46
StammeringMathematician
56616
1
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbbR rightarrowmathbbR$ such that $f$ is continuous and limit at $infty$ and $-infty $ exist then function is uniformly continuous?
I do not think this is true as limit may exist at end points but what if function wiggle in between. May be I am wrong. Can I get some hint?
real-analysis general-topology
asked Sep 5 at 7:46
StammeringMathematician
56616
Let $f:mathbbR rightarrowmathbbR$ such that $f$ is continuous and limit at $infty$ and $-infty $ exist then function is uniformly continuous?
I do not think this is true as limit may exist at end points but what if function wiggle in between. May be I am wrong. Can I get some hint?
real-analysis general-topology
real-analysis general-topology
asked Sep 5 at 7:46
StammeringMathematician
56616
asked Sep 5 at 7:46
StammeringMathematician
56616
asked Sep 5 at 7:46
StammeringMathematician
56616
asked Sep 5 at 7:46
StammeringMathematician
56616
asked Sep 5 at 7:46
StammeringMathematician
56616
56616
1
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49
add a comment |Â
1
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49
1
1
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49
add a comment |Â
1 Answer
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Let $L=lim_xrightarrow +inftyf(x), L'=lim_xrightarrow -inftyf(x)$, for every $c>0$, there exists $M>0$ such that $x>M$ implies that $|f(x)-L|<c/4, x<-M$ implies that $|f(x)-L'|<c/4$.
The function is uniformly continuous on $[-3M,3M]$ since $[-3M,3M]$ is compact, there exists $e$ such that for every $x,yin [-3M,3M], |x-y|<e$ implies that $|f(x)-f(y)|<c$.
Let $d=inf(e,M)$. Consider $x,yin mathbbR$, such that $|x-y|<d$, if $x,yin [-M,M]subset [-3M,3M], |x-y|<e$ and $|f(x)-f(y)|<c$. If $x,y>M$ or $x,y<-M$, $|f(x)-f(y)|<c$.
Suppose that $xin [-M,M]$ since $|x-y|<M, |y|<|x|+|M|=2M$, and $|x-y|<3M$, we deduce that $|f(x)-f(y)|<e$.
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $L=lim_xrightarrow +inftyf(x), L'=lim_xrightarrow -inftyf(x)$, for every $c>0$, there exists $M>0$ such that $x>M$ implies that $|f(x)-L|<c/4, x<-M$ implies that $|f(x)-L'|<c/4$.
The function is uniformly continuous on $[-3M,3M]$ since $[-3M,3M]$ is compact, there exists $e$ such that for every $x,yin [-3M,3M], |x-y|<e$ implies that $|f(x)-f(y)|<c$.
Let $d=inf(e,M)$. Consider $x,yin mathbbR$, such that $|x-y|<d$, if $x,yin [-M,M]subset [-3M,3M], |x-y|<e$ and $|f(x)-f(y)|<c$. If $x,y>M$ or $x,y<-M$, $|f(x)-f(y)|<c$.
Suppose that $xin [-M,M]$ since $|x-y|<M, |y|<|x|+|M|=2M$, and $|x-y|<3M$, we deduce that $|f(x)-f(y)|<e$.
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
add a comment |Â
up vote
1
down vote
accepted
Let $L=lim_xrightarrow +inftyf(x), L'=lim_xrightarrow -inftyf(x)$, for every $c>0$, there exists $M>0$ such that $x>M$ implies that $|f(x)-L|<c/4, x<-M$ implies that $|f(x)-L'|<c/4$.
The function is uniformly continuous on $[-3M,3M]$ since $[-3M,3M]$ is compact, there exists $e$ such that for every $x,yin [-3M,3M], |x-y|<e$ implies that $|f(x)-f(y)|<c$.
Let $d=inf(e,M)$. Consider $x,yin mathbbR$, such that $|x-y|<d$, if $x,yin [-M,M]subset [-3M,3M], |x-y|<e$ and $|f(x)-f(y)|<c$. If $x,y>M$ or $x,y<-M$, $|f(x)-f(y)|<c$.
Suppose that $xin [-M,M]$ since $|x-y|<M, |y|<|x|+|M|=2M$, and $|x-y|<3M$, we deduce that $|f(x)-f(y)|<e$.
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $L=lim_xrightarrow +inftyf(x), L'=lim_xrightarrow -inftyf(x)$, for every $c>0$, there exists $M>0$ such that $x>M$ implies that $|f(x)-L|<c/4, x<-M$ implies that $|f(x)-L'|<c/4$.
The function is uniformly continuous on $[-3M,3M]$ since $[-3M,3M]$ is compact, there exists $e$ such that for every $x,yin [-3M,3M], |x-y|<e$ implies that $|f(x)-f(y)|<c$.
Let $d=inf(e,M)$. Consider $x,yin mathbbR$, such that $|x-y|<d$, if $x,yin [-M,M]subset [-3M,3M], |x-y|<e$ and $|f(x)-f(y)|<c$. If $x,y>M$ or $x,y<-M$, $|f(x)-f(y)|<c$.
Suppose that $xin [-M,M]$ since $|x-y|<M, |y|<|x|+|M|=2M$, and $|x-y|<3M$, we deduce that $|f(x)-f(y)|<e$.
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
Let $L=lim_xrightarrow +inftyf(x), L'=lim_xrightarrow -inftyf(x)$, for every $c>0$, there exists $M>0$ such that $x>M$ implies that $|f(x)-L|<c/4, x<-M$ implies that $|f(x)-L'|<c/4$.
The function is uniformly continuous on $[-3M,3M]$ since $[-3M,3M]$ is compact, there exists $e$ such that for every $x,yin [-3M,3M], |x-y|<e$ implies that $|f(x)-f(y)|<c$.
Let $d=inf(e,M)$. Consider $x,yin mathbbR$, such that $|x-y|<d$, if $x,yin [-M,M]subset [-3M,3M], |x-y|<e$ and $|f(x)-f(y)|<c$. If $x,y>M$ or $x,y<-M$, $|f(x)-f(y)|<c$.
Suppose that $xin [-M,M]$ since $|x-y|<M, |y|<|x|+|M|=2M$, and $|x-y|<3M$, we deduce that $|f(x)-f(y)|<e$.
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
answered Sep 5 at 9:05
Tsemo Aristide
52.5k11244
52.5k11244
add a comment |Â
add a comment |Â
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1
It cannot wiggle in between because any continuous function on a compact set is uniformly continuous. Your question has been answered before on MSE.
– Kavi Rama Murthy
Sep 5 at 7:49