Prove that $f=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
$begingroup$
Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$
real-analysis continuity
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
$endgroup$
add a comment |
$begingroup$
Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$
real-analysis continuity
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
$endgroup$
2
$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
1
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08
add a comment |
$begingroup$
Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$
real-analysis continuity
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
$endgroup$
Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.
I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$
real-analysis continuity
real-analysis continuity
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
asked Feb 29 '16 at 1:23
e2DAeyePie2DAeyePi
124110
124110
2
$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
1
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08
add a comment |
2
$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
1
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08
2
2
$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
1
1
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
$endgroup$
add a comment |
$begingroup$
Here we can use the following sequential criterion:
If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$
Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
$endgroup$
add a comment |
$begingroup$
Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
$endgroup$
add a comment |
$begingroup$
Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
$endgroup$
Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,
$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
answered Feb 29 '16 at 2:10
Math1000Math1000
19.4k31746
19.4k31746
add a comment |
add a comment |
$begingroup$
Here we can use the following sequential criterion:
If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$
Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
add a comment |
$begingroup$
Here we can use the following sequential criterion:
If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$
Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
$endgroup$
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
add a comment |
$begingroup$
Here we can use the following sequential criterion:
If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$
Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
$endgroup$
Here we can use the following sequential criterion:
If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$
Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
edited Apr 27 '16 at 10:52
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
answered Apr 27 '16 at 7:52
BumblebeeBumblebee
9,74912551
9,74912551
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
add a comment |
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
$begingroup$
One would need the reverse implication.
$endgroup$
– Did
Apr 27 '16 at 8:19
add a comment |
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$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30
1
$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39
$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08