if $f^2(x)$ is uniformly continuous then f is uniformly continuous












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
    $endgroup$
    – freakish
    Dec 5 '18 at 8:11












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{x}$ is not defined on whole $mathbb{R}$ but you are only interested in $[0,infty)$.
    $endgroup$
    – freakish
    Dec 5 '18 at 8:11












  • $begingroup$
    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














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 8:09







onlymath

















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












  • $begingroup$
    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












  • $begingroup$
    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










4 Answers
4






active

oldest

votes


















1












$begingroup$

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 $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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 $h(g(x))$???






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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 ${|f(x)| < varepsilon}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

$f^2$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.






share|cite|improve this answer











$endgroup$



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 $z in[x,y], f(z)=0$. $|f(x)-f(y)| = |f(x)-f(z)| + |f(z)-f(y)| = ||f(x)|-|f(z)|| + ||f(z)|-|f(y)|| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 8:41

























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


















  • $begingroup$
    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











0












$begingroup$

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 $h(g(x))$???






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 $h(g(x))$???






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 $h(g(x))$???






share|cite|improve this answer









$endgroup$



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 $h(g(x))$???







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
    $endgroup$
    – 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










  • $begingroup$
    @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











0












$begingroup$

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 ${|f(x)| < varepsilon}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 ${|f(x)| < varepsilon}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 ${|f(x)| < varepsilon}$.






share|cite|improve this answer











$endgroup$



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 ${|f(x)| < varepsilon}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 8:22

























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


















  • $begingroup$
    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











0












$begingroup$

$f^2$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















0












$begingroup$

$f^2$ 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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

$f^2$ 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






share|cite|improve this answer











$endgroup$



$f^2$ 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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 10:15

























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




















  • $begingroup$
    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




















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