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Quotient of uniformly continuous bounded functions is uniformly continuous
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Let $f,g:mathbbR^n rightarrow mathbbR$ be uniformly continuous and bounded functions. Suppose there exists $k gt 0$ such that $|g(x)| geq k$ for any $xin mathbbR^n$. Prove that $fracfg$ is uniformly continuous.
I'm trying to come up with an epsilon delta proof.
Let $epsilon gt 0$.
By the uniform continuity of $f$ and $g$ we have that
$exists delta_1$ such that if $| x - y | lt delta_1$ then $|f(x) - f(y)| lt epsilon$
$exists delta_2$ such that if $| x - y | lt delta_2$ then $|g(x) - g(y)| lt epsilon$
Now I've tried to bound $big|fracf(x)g(x) - fracf(y)g(y)big|$ with no success:
$$bigg|fracf(x)g(x) - fracf(y)g(y)bigg|=bigg|fracf(x)g(y)-f(y)g(x)g(x)g(y)bigg|=frac leq frack^2$$
but I don't know how to continue from here.
real-analysis multivariable-calculus uniform-continuity
asked Jul 29 at 23:07
Yagger
5271315
add a comment |Â
up vote
0
down vote
favorite
Let $f,g:mathbbR^n rightarrow mathbbR$ be uniformly continuous and bounded functions. Suppose there exists $k gt 0$ such that $|g(x)| geq k$ for any $xin mathbbR^n$. Prove that $fracfg$ is uniformly continuous.
I'm trying to come up with an epsilon delta proof.
Let $epsilon gt 0$.
By the uniform continuity of $f$ and $g$ we have that
$exists delta_1$ such that if $| x - y | lt delta_1$ then $|f(x) - f(y)| lt epsilon$
$exists delta_2$ such that if $| x - y | lt delta_2$ then $|g(x) - g(y)| lt epsilon$
Now I've tried to bound $big|fracf(x)g(x) - fracf(y)g(y)big|$ with no success:
$$bigg|fracf(x)g(x) - fracf(y)g(y)bigg|=bigg|fracf(x)g(y)-f(y)g(x)g(x)g(y)bigg|=frac leq frack^2$$
but I don't know how to continue from here.
real-analysis multivariable-calculus uniform-continuity
asked Jul 29 at 23:07
Yagger
5271315
2
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
I'm writing it right now :)
– Suzet
Jul 30 at 0:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f,g:mathbbR^n rightarrow mathbbR$ be uniformly continuous and bounded functions. Suppose there exists $k gt 0$ such that $|g(x)| geq k$ for any $xin mathbbR^n$. Prove that $fracfg$ is uniformly continuous.
I'm trying to come up with an epsilon delta proof.
Let $epsilon gt 0$.
By the uniform continuity of $f$ and $g$ we have that
$exists delta_1$ such that if $| x - y | lt delta_1$ then $|f(x) - f(y)| lt epsilon$
$exists delta_2$ such that if $| x - y | lt delta_2$ then $|g(x) - g(y)| lt epsilon$
Now I've tried to bound $big|fracf(x)g(x) - fracf(y)g(y)big|$ with no success:
$$bigg|fracf(x)g(x) - fracf(y)g(y)bigg|=bigg|fracf(x)g(y)-f(y)g(x)g(x)g(y)bigg|=frac leq frack^2$$
but I don't know how to continue from here.
real-analysis multivariable-calculus uniform-continuity
asked Jul 29 at 23:07
Yagger
5271315
Let $f,g:mathbbR^n rightarrow mathbbR$ be uniformly continuous and bounded functions. Suppose there exists $k gt 0$ such that $|g(x)| geq k$ for any $xin mathbbR^n$. Prove that $fracfg$ is uniformly continuous.
I'm trying to come up with an epsilon delta proof.
Let $epsilon gt 0$.
By the uniform continuity of $f$ and $g$ we have that
$exists delta_1$ such that if $| x - y | lt delta_1$ then $|f(x) - f(y)| lt epsilon$
$exists delta_2$ such that if $| x - y | lt delta_2$ then $|g(x) - g(y)| lt epsilon$
Now I've tried to bound $big|fracf(x)g(x) - fracf(y)g(y)big|$ with no success:
$$bigg|fracf(x)g(x) - fracf(y)g(y)bigg|=bigg|fracf(x)g(y)-f(y)g(x)g(x)g(y)bigg|=frac leq frack^2$$
but I don't know how to continue from here.
real-analysis multivariable-calculus uniform-continuity
asked Jul 29 at 23:07
Yagger
5271315
asked Jul 29 at 23:07
Yagger
5271315
asked Jul 29 at 23:07
Yagger
5271315
asked Jul 29 at 23:07
Yagger
5271315
5271315
2
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
I'm writing it right now :)
– Suzet
Jul 30 at 0:09
add a comment |Â
2
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
I'm writing it right now :)
– Suzet
Jul 30 at 0:09
2
2
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
I'm writing it right now :)
– Suzet
Jul 30 at 0:09
I'm writing it right now :)
– Suzet
Jul 30 at 0:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Using your notations and continuing what you already did, we have
$$frack^2 = frack^2$$
We then use the triangular inequality to get $$frack^2leq frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|leq fracepsilonk^2operatornamesup|f|+fracepsilonk^2operatornamesup|g|$$
This finishes the proof. If one wants to conclude it with a single $epsilon$ at the end, then just switch $epsilon$ with $$epsilon':=frack^2fepsilon$$ when using the hypothesis of uniform continuity.
answered Jul 30 at 0:20
Suzet
2,203427
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using your notations and continuing what you already did, we have
$$frack^2 = frack^2$$
We then use the triangular inequality to get $$frack^2leq frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|leq fracepsilonk^2operatornamesup|f|+fracepsilonk^2operatornamesup|g|$$
This finishes the proof. If one wants to conclude it with a single $epsilon$ at the end, then just switch $epsilon$ with $$epsilon':=frack^2fepsilon$$ when using the hypothesis of uniform continuity.
answered Jul 30 at 0:20
Suzet
2,203427
add a comment |Â
up vote
2
down vote
accepted
Using your notations and continuing what you already did, we have
$$frack^2 = frack^2$$
We then use the triangular inequality to get $$frack^2leq frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|leq fracepsilonk^2operatornamesup|f|+fracepsilonk^2operatornamesup|g|$$
This finishes the proof. If one wants to conclude it with a single $epsilon$ at the end, then just switch $epsilon$ with $$epsilon':=frack^2fepsilon$$ when using the hypothesis of uniform continuity.
answered Jul 30 at 0:20
Suzet
2,203427
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using your notations and continuing what you already did, we have
$$frack^2 = frack^2$$
We then use the triangular inequality to get $$frack^2leq frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|leq fracepsilonk^2operatornamesup|f|+fracepsilonk^2operatornamesup|g|$$
This finishes the proof. If one wants to conclude it with a single $epsilon$ at the end, then just switch $epsilon$ with $$epsilon':=frack^2fepsilon$$ when using the hypothesis of uniform continuity.
answered Jul 30 at 0:20
Suzet
2,203427
Using your notations and continuing what you already did, we have
$$frack^2 = frack^2$$
We then use the triangular inequality to get $$frack^2leq frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$frack^2|g(y)|+fracg(x)-g(y)k^2|f(y)|leq fracepsilonk^2operatornamesup|f|+fracepsilonk^2operatornamesup|g|$$
This finishes the proof. If one wants to conclude it with a single $epsilon$ at the end, then just switch $epsilon$ with $$epsilon':=frack^2fepsilon$$ when using the hypothesis of uniform continuity.
answered Jul 30 at 0:20
Suzet
2,203427
answered Jul 30 at 0:20
Suzet
2,203427
answered Jul 30 at 0:20
Suzet
2,203427
answered Jul 30 at 0:20
Suzet
2,203427
2,203427
add a comment |Â
add a comment |Â
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2
You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous.
– Suzet
Jul 29 at 23:08
@Suzet great I think I got it. Write it as an answer if you want so I can accept it.
– Yagger
Jul 29 at 23:33
I'm writing it right now :)
– Suzet
Jul 30 at 0:09