Analyse Equicontinuity (uniform continuity) of a function $ frac{sin(x)}{x},; x > 0 $
$begingroup$
How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$
When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?
calculus continuity equicontinuity
edited Dec 9 '18 at 23:37
user376343
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
$endgroup$
add a comment |
$begingroup$
How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$
When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?
calculus continuity equicontinuity
edited Dec 9 '18 at 23:37
user376343
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
$endgroup$
1
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55
add a comment |
$begingroup$
How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$
When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?
calculus continuity equicontinuity
edited Dec 9 '18 at 23:37
user376343
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
$endgroup$
How can I analyse equicontinuity of the function $$
frac{sin(x)}{x},; x > 0?
$$
When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically.
Tried to prove using definition. I am stuck.
Any hints or better ways to do this kind of problems?
calculus continuity equicontinuity
calculus continuity equicontinuity
edited Dec 9 '18 at 23:37
user376343
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
edited Dec 9 '18 at 23:37
user376343
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
edited Dec 9 '18 at 23:37
user376343
3,9584829
edited Dec 9 '18 at 23:37
user376343
3,9584829
edited Dec 9 '18 at 23:37
user376343
3,9584829
3,9584829
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
asked Dec 9 '18 at 17:31
R0xx0rZzzR0xx0rZzz
747
747
1
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55
add a comment |
1
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55
1
1
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.
Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
$endgroup$
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
add a comment |
$begingroup$
If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
$endgroup$
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.
Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
$endgroup$
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
add a comment |
$begingroup$
This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.
Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
$endgroup$
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
add a comment |
$begingroup$
This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.
Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
$endgroup$
This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.
Here we are given the function
$${rm sinc}(x):=left{eqalign{{sin xover x}quad&(xne0)cr 1qquad&(x=0) .cr}right.$$
Note that we can write this function in the form
$${rm sinc}(x)=int_0^1cos(t x)>dtqquad(-infty<x<infty) ,$$
so that
$$bigl|{rm sinc}(y)-{rm sinc}(x)bigr|leqint_0^1bigl|cos(ty)-cos (tx)bigr|>dtleqint_0^1bigl|ty-txbigr|dt={1over2}|y-x| .$$
This shows that the ${rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${mathbb R}$.
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
answered Dec 9 '18 at 18:32
Christian BlatterChristian Blatter
175k8115327
175k8115327
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
add a comment |
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
1
1
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
$begingroup$
To the proposer: Another method is to show that any continuous $f:[0,infty)to Bbb R,$ such that $lim_{xto infty}f(x)$ exists, is uniformly continuous.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 19:07
add a comment |
$begingroup$
If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
$endgroup$
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
add a comment |
$begingroup$
If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
$endgroup$
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
add a comment |
$begingroup$
If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
$endgroup$
If $f$ is continuous on $mathbb R$ and $lim_{|x|to infty}f(x)=0,$ then $f$ is uniformly continuous on $mathbb R.$ Clearly $f(x)=(sin x)/x$ satisfies these criteria.
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
edited Dec 24 '18 at 17:05
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
answered Dec 10 '18 at 0:10
zhw.zhw.
74.5k43175
74.5k43175
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
add a comment |
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
$begingroup$
typo corrected$,$
$endgroup$
– zhw.
Dec 24 '18 at 17:06
add a comment |
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1
$begingroup$
By "equicontinuous", do you mean "uniformly continuous"? If that is the case, you can try proving that the function is actually Lipschitz continuous, because its derivative is bounded.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:39
$begingroup$
I am preparing for an exam and we didn't cover Lipschitz continuity yet. I am not allowed to use
$endgroup$
– R0xx0rZzz
Dec 9 '18 at 17:41
$begingroup$
Well, whatever, you can try showing that your function $f=f(x)$ is uniformly continuous because it has a bounded derivative. It is not difficult. Let $f(0):=1$ and write $f(x)=1+int_0^x f'(y), dy$.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:55