Mixing
Proving $1/ x^2$ is not uniformly continuous on (0,2]
$begingroup$
I am trying to prove the function in the title is not uniformly continuous on (0,2].
Different from the proof given in proof, I proceed to show the proof following lecture as follows:
begin{equation}
|1/x^2 - 1/y^2|=frac{|y-x||y+x|}{x^2y^2}
end{equation}
(after simplification).
Since both $x,y in (0,2]$ above eq. reduces to begin{equation}
|1/x^2 - 1/y^2|= frac{|y-x|}{4}<frac{delta}{4} (text{by the definition of continuity}).
end{equation}
Now if I pick $delta=4epsilon$, I don't arrive at the contradiction. Any hint where I am wrong will help.
real-analysis uniform-continuity
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to prove the function in the title is not uniformly continuous on (0,2].
Different from the proof given in proof, I proceed to show the proof following lecture as follows:
begin{equation}
|1/x^2 - 1/y^2|=frac{|y-x||y+x|}{x^2y^2}
end{equation}
(after simplification).
Since both $x,y in (0,2]$ above eq. reduces to begin{equation}
|1/x^2 - 1/y^2|= frac{|y-x|}{4}<frac{delta}{4} (text{by the definition of continuity}).
end{equation}
Now if I pick $delta=4epsilon$, I don't arrive at the contradiction. Any hint where I am wrong will help.
real-analysis uniform-continuity
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
$endgroup$
$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02
|
show 1 more comment
$begingroup$
I am trying to prove the function in the title is not uniformly continuous on (0,2].
Different from the proof given in proof, I proceed to show the proof following lecture as follows:
begin{equation}
|1/x^2 - 1/y^2|=frac{|y-x||y+x|}{x^2y^2}
end{equation}
(after simplification).
Since both $x,y in (0,2]$ above eq. reduces to begin{equation}
|1/x^2 - 1/y^2|= frac{|y-x|}{4}<frac{delta}{4} (text{by the definition of continuity}).
end{equation}
Now if I pick $delta=4epsilon$, I don't arrive at the contradiction. Any hint where I am wrong will help.
real-analysis uniform-continuity
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
$endgroup$
I am trying to prove the function in the title is not uniformly continuous on (0,2].
Different from the proof given in proof, I proceed to show the proof following lecture as follows:
begin{equation}
|1/x^2 - 1/y^2|=frac{|y-x||y+x|}{x^2y^2}
end{equation}
(after simplification).
Since both $x,y in (0,2]$ above eq. reduces to begin{equation}
|1/x^2 - 1/y^2|= frac{|y-x|}{4}<frac{delta}{4} (text{by the definition of continuity}).
end{equation}
Now if I pick $delta=4epsilon$, I don't arrive at the contradiction. Any hint where I am wrong will help.
real-analysis uniform-continuity
real-analysis uniform-continuity
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
edited Feb 3 at 12:53
J. W. Tanner
4,8471420
4,8471420
asked Feb 3 at 9:44
chandreshchandresh
1,000816
asked Feb 3 at 9:44
chandreshchandresh
1,000816
asked Feb 3 at 9:44
chandreshchandresh
1,000816
1,000816
$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02
|
show 1 more comment
$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02
$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $delta>0$, choose $xin (0,2]$ such that $x<frac{delta}2$ and choose $y=frac{x}2$. Then
$$
|y-x| = frac x2 < frac delta{2}<delta
$$
and since $xleq 2$
$$
left|frac1 {x^2}-frac 1{y^2}right| = left|frac1 {x^2}-frac 4{x^2}right|=frac{3}{x^2} geq frac 34.
$$
This contradicts the definition of uniform continuity on $(0,2]$.
If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:Arightarrow mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $delta>0$, choose $xin (0,2]$ such that $x<frac{delta}2$ and choose $y=frac{x}2$. Then
$$
|y-x| = frac x2 < frac delta{2}<delta
$$
and since $xleq 2$
$$
left|frac1 {x^2}-frac 1{y^2}right| = left|frac1 {x^2}-frac 4{x^2}right|=frac{3}{x^2} geq frac 34.
$$
This contradicts the definition of uniform continuity on $(0,2]$.
If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:Arightarrow mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
$begingroup$
Let $delta>0$, choose $xin (0,2]$ such that $x<frac{delta}2$ and choose $y=frac{x}2$. Then
$$
|y-x| = frac x2 < frac delta{2}<delta
$$
and since $xleq 2$
$$
left|frac1 {x^2}-frac 1{y^2}right| = left|frac1 {x^2}-frac 4{x^2}right|=frac{3}{x^2} geq frac 34.
$$
This contradicts the definition of uniform continuity on $(0,2]$.
If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:Arightarrow mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
$begingroup$
Let $delta>0$, choose $xin (0,2]$ such that $x<frac{delta}2$ and choose $y=frac{x}2$. Then
$$
|y-x| = frac x2 < frac delta{2}<delta
$$
and since $xleq 2$
$$
left|frac1 {x^2}-frac 1{y^2}right| = left|frac1 {x^2}-frac 4{x^2}right|=frac{3}{x^2} geq frac 34.
$$
This contradicts the definition of uniform continuity on $(0,2]$.
If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:Arightarrow mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
$endgroup$
Let $delta>0$, choose $xin (0,2]$ such that $x<frac{delta}2$ and choose $y=frac{x}2$. Then
$$
|y-x| = frac x2 < frac delta{2}<delta
$$
and since $xleq 2$
$$
left|frac1 {x^2}-frac 1{y^2}right| = left|frac1 {x^2}-frac 4{x^2}right|=frac{3}{x^2} geq frac 34.
$$
This contradicts the definition of uniform continuity on $(0,2]$.
If you are allowed to use some results on uniformly continuous functions there are several faster ways to solve the exercise. For instance, the extension theorem states that if $f:Arightarrow mathbb{R}$ is uniformly continuous then it is possible to extend $f$ to the closure of $A$ in such a way that the extension is uniformly continuous (and, in particular, continuous). Since there is no way to extend $f(x)=frac 1{x^2}$ to $[0,2]$ to a continuous function, then $f$ is not uniformly continuous on $(0,2]$.
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
answered Feb 3 at 12:33
AlessioDVAlessioDV
1,208114
1,208114
add a comment |
add a comment |
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$begingroup$
How did you get $|x^{-2}-y^{-2}|<|y-x|/4$? Somehow I doubt this is true.
$endgroup$
– Blazej
Feb 3 at 9:50
$begingroup$
max value of $|x+y|$ could be 4 since both $x, y in (0,2]$. Does it make sense now?
$endgroup$
– chandresh
Feb 3 at 9:52
$begingroup$
But denominator $x^{-2} y^{-2}$ may be arbitrarily large.
$endgroup$
– Blazej
Feb 3 at 9:53
$begingroup$
How come? if Both $x$ and $y$ are bounded.
$endgroup$
– chandresh
Feb 3 at 9:54
$begingroup$
They may be as close to zero as you want. In the denominator it is always the inverse inequality-the smaller the denominator is, the bigger the fraction is.
$endgroup$
– Mark
Feb 3 at 10:02