if $f^2(x)$ is uniformly continuous then f is uniformly continuous
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this is the real question.1I saw this question like if f is from R to infinity and continuous if g(x)=$f^2(x)$ if g(x) is uniformly continuous so is f(x)
$f^2(x)=f(x)*f(x)$
My Thought:
1)for sequence $x_n$ and $y_n$ are like $d(x_n,y_n)$ tends to 0 then $d(g(x_n),g(y_n))$ tends to 0 as g is uniformly continuous .considering usual metric of R(real line) $|g(x)-g(y)|=|f^2(x) -f^2(y)|=|f(x)-f(y)|*|f(x)+f(y)|$ from here how can I show that $|f(x)-f(y)|$ tends to 0
2)first I thought about $sqrt(x)$ , as it is uniformly continuous and so is g(x) so $sqrt(g(x))$ will also be uniformly continuous.But as $sqrt(x)$ is not defined on whole R so I cannot conclude f is uniformly continuous on whole R.
or there is any counter example.
pleas
real-analysis continuity
asked Dec 5 '18 at 7:56
onlymathonlymath
749
$endgroup$
add a comment |
$begingroup$
this is the real question.1I saw this question like if f is from R to infinity and continuous if g(x)=$f^2(x)$ if g(x) is uniformly continuous so is f(x)
$f^2(x)=f(x)*f(x)$
My Thought:
1)for sequence $x_n$ and $y_n$ are like $d(x_n,y_n)$ tends to 0 then $d(g(x_n),g(y_n))$ tends to 0 as g is uniformly continuous .considering usual metric of R(real line) $|g(x)-g(y)|=|f^2(x) -f^2(y)|=|f(x)-f(y)|*|f(x)+f(y)|$ from here how can I show that $|f(x)-f(y)|$ tends to 0
2)first I thought about $sqrt(x)$ , as it is uniformly continuous and so is g(x) so $sqrt(g(x))$ will also be uniformly continuous.But as $sqrt(x)$ is not defined on whole R so I cannot conclude f is uniformly continuous on whole R.
or there is any counter example.
pleas
real-analysis continuity
asked Dec 5 '18 at 7:56
onlymathonlymath
749
$endgroup$
$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
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Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
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– Kavi Rama Murthy
Dec 5 '18 at 8:14
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@freakish ...so it cant be proved for the negative part ?
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– onlymath
Dec 5 '18 at 8:16
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@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
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@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24
add a comment |
$begingroup$
this is the real question.1I saw this question like if f is from R to infinity and continuous if g(x)=$f^2(x)$ if g(x) is uniformly continuous so is f(x)
$f^2(x)=f(x)*f(x)$
My Thought:
1)for sequence $x_n$ and $y_n$ are like $d(x_n,y_n)$ tends to 0 then $d(g(x_n),g(y_n))$ tends to 0 as g is uniformly continuous .considering usual metric of R(real line) $|g(x)-g(y)|=|f^2(x) -f^2(y)|=|f(x)-f(y)|*|f(x)+f(y)|$ from here how can I show that $|f(x)-f(y)|$ tends to 0
2)first I thought about $sqrt(x)$ , as it is uniformly continuous and so is g(x) so $sqrt(g(x))$ will also be uniformly continuous.But as $sqrt(x)$ is not defined on whole R so I cannot conclude f is uniformly continuous on whole R.
or there is any counter example.
pleas
real-analysis continuity
asked Dec 5 '18 at 7:56
onlymathonlymath
749
$endgroup$
this is the real question.1I saw this question like if f is from R to infinity and continuous if g(x)=$f^2(x)$ if g(x) is uniformly continuous so is f(x)
$f^2(x)=f(x)*f(x)$
My Thought:
1)for sequence $x_n$ and $y_n$ are like $d(x_n,y_n)$ tends to 0 then $d(g(x_n),g(y_n))$ tends to 0 as g is uniformly continuous .considering usual metric of R(real line) $|g(x)-g(y)|=|f^2(x) -f^2(y)|=|f(x)-f(y)|*|f(x)+f(y)|$ from here how can I show that $|f(x)-f(y)|$ tends to 0
2)first I thought about $sqrt(x)$ , as it is uniformly continuous and so is g(x) so $sqrt(g(x))$ will also be uniformly continuous.But as $sqrt(x)$ is not defined on whole R so I cannot conclude f is uniformly continuous on whole R.
or there is any counter example.
pleas
real-analysis continuity
real-analysis continuity
asked Dec 5 '18 at 7:56
onlymathonlymath
749
asked Dec 5 '18 at 7:56
onlymathonlymath
749
edited Dec 5 '18 at 8:09
onlymath
asked Dec 5 '18 at 7:56
onlymathonlymath
749
asked Dec 5 '18 at 7:56
onlymathonlymath
749
asked Dec 5 '18 at 7:56
onlymathonlymath
749
749
$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
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Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 8:14
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@freakish ...so it cant be proved for the negative part ?
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– onlymath
Dec 5 '18 at 8:16
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@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
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@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24
add a comment |
$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
$begingroup$
Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 8:14
$begingroup$
@freakish ...so it cant be proved for the negative part ?
$endgroup$
– onlymath
Dec 5 '18 at 8:16
$begingroup$
@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
$begingroup$
@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24
$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
$begingroup$
Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 8:14
$begingroup$
Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 8:14
$begingroup$
@freakish ...so it cant be proved for the negative part ?
$endgroup$
– onlymath
Dec 5 '18 at 8:16
$begingroup$
@freakish ...so it cant be proved for the negative part ?
$endgroup$
– onlymath
Dec 5 '18 at 8:16
$begingroup$
@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
$begingroup$
@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
$begingroup$
@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24
$begingroup$
@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
With your point 2 you have proven that $|f|$ is uniformly continuous.
Let $epsilon>0$. Let $delta>0$ such that for all $x,yinmathbb{R}$, $|x-y|<delta$, $||f(x)|-|f(y)||<frac{epsilon}{2}$.
Let $x,yinmathbb{R}$, $|x-y|<delta$.
If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<frac{epsilon}{2} < epsilon$.
If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
$endgroup$
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
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Good catch, editing
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– RcnSc
Dec 5 '18 at 8:41
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Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
add a comment |
$begingroup$
Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???
answered Dec 5 '18 at 8:08
G C RG C R
215
$endgroup$
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
But $h:[0,infty) to [0,infty)$
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– G C R
Dec 5 '18 at 8:15
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@onlymath I'm not sure about this answer either. Ask @G C R
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– Anvit
Dec 5 '18 at 8:16
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h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
|
show 3 more comments
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Following your second idea, you need to show that if $f$ is continuous and $sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: ${|f(x)| ge varepsilon}$ and ${|f(x)| < varepsilon}$.
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
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I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
add a comment |
$begingroup$
$f^2$ is uniformly cont $implies (|x-y|<deltaimplies|f^2(x)-f^2(y)|<epsilon) implies
|f^2(x)-f^2(y)|<{epsilonoverdelta}|x-y|implies {|f^2(x)-f^2(y)|over|x-y|}<{epsilonoverdelta}implies|2f(x)f'(x)|<{epsilonoverdelta}implies |f'(x)|<{kover |f(x)|}<frac k {inf ,f}$
If $inf fne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part.
Let $K={x||f(x)|<1}$. Since $f$ is cont, $f'(x)<M,,forall xin K$ for some $Min mathbb N,$ also $f'(x)<k,,forall xinmathbb Rcap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
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$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
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– onlymath
Dec 5 '18 at 9:37
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Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
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– Anvit
Dec 5 '18 at 10:17
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With your point 2 you have proven that $|f|$ is uniformly continuous.
Let $epsilon>0$. Let $delta>0$ such that for all $x,yinmathbb{R}$, $|x-y|<delta$, $||f(x)|-|f(y)||<frac{epsilon}{2}$.
Let $x,yinmathbb{R}$, $|x-y|<delta$.
If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<frac{epsilon}{2} < epsilon$.
If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
$endgroup$
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
add a comment |
$begingroup$
With your point 2 you have proven that $|f|$ is uniformly continuous.
Let $epsilon>0$. Let $delta>0$ such that for all $x,yinmathbb{R}$, $|x-y|<delta$, $||f(x)|-|f(y)||<frac{epsilon}{2}$.
Let $x,yinmathbb{R}$, $|x-y|<delta$.
If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<frac{epsilon}{2} < epsilon$.
If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
$endgroup$
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
add a comment |
$begingroup$
With your point 2 you have proven that $|f|$ is uniformly continuous.
Let $epsilon>0$. Let $delta>0$ such that for all $x,yinmathbb{R}$, $|x-y|<delta$, $||f(x)|-|f(y)||<frac{epsilon}{2}$.
Let $x,yinmathbb{R}$, $|x-y|<delta$.
If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<frac{epsilon}{2} < epsilon$.
If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
$endgroup$
With your point 2 you have proven that $|f|$ is uniformly continuous.
Let $epsilon>0$. Let $delta>0$ such that for all $x,yinmathbb{R}$, $|x-y|<delta$, $||f(x)|-|f(y)||<frac{epsilon}{2}$.
Let $x,yinmathbb{R}$, $|x-y|<delta$.
If $f(x)$ and $f(y)$ have the same sign, then $|f(x)-f(y)| = ||f(x)|-|f(y)||<frac{epsilon}{2} < epsilon$.
If $f(x)$ and $f(y)$ have different signs, then by continuity of $f$ there is $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
edited Dec 5 '18 at 8:41
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
answered Dec 5 '18 at 8:39
RcnScRcnSc
1014
1014
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
add a comment |
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
You mean $f(x)$ and $f(y)$ have same/different signs, right?
$endgroup$
– Anvit
Dec 5 '18 at 8:40
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Good catch, editing
$endgroup$
– RcnSc
Dec 5 '18 at 8:41
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
$begingroup$
Cute answer by RcnSc
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 6:11
add a comment |
$begingroup$
Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???
answered Dec 5 '18 at 8:08
G C RG C R
215
$endgroup$
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
But $h:[0,infty) to [0,infty)$
$endgroup$
– G C R
Dec 5 '18 at 8:15
$begingroup$
@onlymath I'm not sure about this answer either. Ask @G C R
$endgroup$
– Anvit
Dec 5 '18 at 8:16
$begingroup$
h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
|
show 3 more comments
$begingroup$
Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???
answered Dec 5 '18 at 8:08
G C RG C R
215
$endgroup$
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
But $h:[0,infty) to [0,infty)$
$endgroup$
– G C R
Dec 5 '18 at 8:15
$begingroup$
@onlymath I'm not sure about this answer either. Ask @G C R
$endgroup$
– Anvit
Dec 5 '18 at 8:16
$begingroup$
h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
|
show 3 more comments
$begingroup$
Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???
answered Dec 5 '18 at 8:08
G C RG C R
215
$endgroup$
Composition of two uniformly continuous is uniformly continuous . Now take $g(x)=f^2(x)$ and $h(x)=sqrt(x)$ it is also uniformly continuous .Now think $h(g(x))$???
answered Dec 5 '18 at 8:08
G C RG C R
215
answered Dec 5 '18 at 8:08
G C RG C R
215
answered Dec 5 '18 at 8:08
G C RG C R
215
answered Dec 5 '18 at 8:08
G C RG C R
215
215
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
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– onlymath
Dec 5 '18 at 8:14
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But $h:[0,infty) to [0,infty)$
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– G C R
Dec 5 '18 at 8:15
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@onlymath I'm not sure about this answer either. Ask @G C R
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– Anvit
Dec 5 '18 at 8:16
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h=f for the positive part ...so how can I show f is uniform for nagetive part
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– onlymath
Dec 5 '18 at 8:19
|
show 3 more comments
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I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
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– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
But $h:[0,infty) to [0,infty)$
$endgroup$
– G C R
Dec 5 '18 at 8:15
$begingroup$
@onlymath I'm not sure about this answer either. Ask @G C R
$endgroup$
– Anvit
Dec 5 '18 at 8:16
$begingroup$
h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
I thought like that but square root function isn't defined for all R..so I am having a confusion about uniform continuity on whole R
$endgroup$
– onlymath
Dec 5 '18 at 8:10
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
@Anvit I didnt get that ...we are composing two functions and $sqrt(g(x))=sqrt(f(x)*f(x))$ uniformly continuous for the positive part ..but what if f(x) for x less than 0 is confusing me
$endgroup$
– onlymath
Dec 5 '18 at 8:14
$begingroup$
But $h:[0,infty) to [0,infty)$
$endgroup$
– G C R
Dec 5 '18 at 8:15
$begingroup$
But $h:[0,infty) to [0,infty)$
$endgroup$
– G C R
Dec 5 '18 at 8:15
$begingroup$
@onlymath I'm not sure about this answer either. Ask @G C R
$endgroup$
– Anvit
Dec 5 '18 at 8:16
$begingroup$
@onlymath I'm not sure about this answer either. Ask @G C R
$endgroup$
– Anvit
Dec 5 '18 at 8:16
$begingroup$
h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
$begingroup$
h=f for the positive part ...so how can I show f is uniform for nagetive part
$endgroup$
– onlymath
Dec 5 '18 at 8:19
|
show 3 more comments
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Following your second idea, you need to show that if $f$ is continuous and $sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: ${|f(x)| ge varepsilon}$ and ${|f(x)| < varepsilon}$.
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
$endgroup$
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
add a comment |
$begingroup$
Following your second idea, you need to show that if $f$ is continuous and $sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: ${|f(x)| ge varepsilon}$ and ${|f(x)| < varepsilon}$.
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
$endgroup$
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
add a comment |
$begingroup$
Following your second idea, you need to show that if $f$ is continuous and $sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: ${|f(x)| ge varepsilon}$ and ${|f(x)| < varepsilon}$.
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
$endgroup$
Following your second idea, you need to show that if $f$ is continuous and $sqrt{g}=|f|$ is uniformly continuous, then $f$ is also uniformly continuous. To get some intuition about the situation, come up with an example of a function $f$ such that $|f|$ is continuous, but $f$ is not. From this, you should see why it might be natural to divide the domain into two parts: ${|f(x)| ge varepsilon}$ and ${|f(x)| < varepsilon}$.
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
edited Dec 5 '18 at 8:22
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
answered Dec 5 '18 at 8:16
maridiamaridia
1,065113
1,065113
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
add a comment |
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
$begingroup$
I have thought that f(x)=x+1 for x greater than equal to 0 and x-1 otherwise....but didnt get the point you are making..please help
$endgroup$
– onlymath
Dec 5 '18 at 9:45
add a comment |
$begingroup$
$f^2$ is uniformly cont $implies (|x-y|<deltaimplies|f^2(x)-f^2(y)|<epsilon) implies
|f^2(x)-f^2(y)|<{epsilonoverdelta}|x-y|implies {|f^2(x)-f^2(y)|over|x-y|}<{epsilonoverdelta}implies|2f(x)f'(x)|<{epsilonoverdelta}implies |f'(x)|<{kover |f(x)|}<frac k {inf ,f}$
If $inf fne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part.
Let $K={x||f(x)|<1}$. Since $f$ is cont, $f'(x)<M,,forall xin K$ for some $Min mathbb N,$ also $f'(x)<k,,forall xinmathbb Rcap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
$endgroup$
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
add a comment |
$begingroup$
$f^2$ is uniformly cont $implies (|x-y|<deltaimplies|f^2(x)-f^2(y)|<epsilon) implies
|f^2(x)-f^2(y)|<{epsilonoverdelta}|x-y|implies {|f^2(x)-f^2(y)|over|x-y|}<{epsilonoverdelta}implies|2f(x)f'(x)|<{epsilonoverdelta}implies |f'(x)|<{kover |f(x)|}<frac k {inf ,f}$
If $inf fne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part.
Let $K={x||f(x)|<1}$. Since $f$ is cont, $f'(x)<M,,forall xin K$ for some $Min mathbb N,$ also $f'(x)<k,,forall xinmathbb Rcap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
$endgroup$
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
add a comment |
$begingroup$
$f^2$ is uniformly cont $implies (|x-y|<deltaimplies|f^2(x)-f^2(y)|<epsilon) implies
|f^2(x)-f^2(y)|<{epsilonoverdelta}|x-y|implies {|f^2(x)-f^2(y)|over|x-y|}<{epsilonoverdelta}implies|2f(x)f'(x)|<{epsilonoverdelta}implies |f'(x)|<{kover |f(x)|}<frac k {inf ,f}$
If $inf fne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part.
Let $K={x||f(x)|<1}$. Since $f$ is cont, $f'(x)<M,,forall xin K$ for some $Min mathbb N,$ also $f'(x)<k,,forall xinmathbb Rcap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
$endgroup$
$f^2$ is uniformly cont $implies (|x-y|<deltaimplies|f^2(x)-f^2(y)|<epsilon) implies
|f^2(x)-f^2(y)|<{epsilonoverdelta}|x-y|implies {|f^2(x)-f^2(y)|over|x-y|}<{epsilonoverdelta}implies|2f(x)f'(x)|<{epsilonoverdelta}implies |f'(x)|<{kover |f(x)|}<frac k {inf ,f}$
If $inf fne 0$, we are done since Derivative is bounded implies function is uniformly continous.
Now comes the really non rigourous part.
Let $K={x||f(x)|<1}$. Since $f$ is cont, $f'(x)<M,,forall xin K$ for some $Min mathbb N,$ also $f'(x)<k,,forall xinmathbb Rcap K^c$
Thus, $f'$ is bounded and hence $f$ in uniformly continuous
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
edited Dec 5 '18 at 10:15
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
answered Dec 5 '18 at 9:04
AnvitAnvit
1,607419
1,607419
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
add a comment |
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
I may be wrong but g(x) or f(x) might not be differentiable...and one more question you have written R K what it means?
$endgroup$
– onlymath
Dec 5 '18 at 9:37
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
$begingroup$
Was Supposed to be RK meaning $Rcap K^c$ but you cant directly type '' in mathjax. Edited in. and Yes you are correct. They may not be differentiable. My solution assumes that though. The one you have selected is great though
$endgroup$
– Anvit
Dec 5 '18 at 10:17
add a comment |
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$begingroup$
$sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
$endgroup$
– freakish
Dec 5 '18 at 8:11
$begingroup$
Just a small observation: as a multiple choice question this is trivial since all other possibilities can be ruled out easily. (Of course, the fact that $f$ must be uniformly continuous is not obvious and the question is interesting too. I have given an upvote).
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– Kavi Rama Murthy
Dec 5 '18 at 8:14
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@freakish ...so it cant be proved for the negative part ?
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– onlymath
Dec 5 '18 at 8:16
$begingroup$
@onlymath Yes, right. Well you can extend it to negatives by taking $sqrt{-x}$.
$endgroup$
– freakish
Dec 5 '18 at 8:20
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@Anvit possibly the above might solve the problem
$endgroup$
– onlymath
Dec 5 '18 at 8:24