Let $f: [0,infty] to mathbb{R}$ be continuous such that its limit tends to $0$ as $x to infty$. Prove that...
$begingroup$
This question has been answered in the past but I am confused about a point in the proof.
Here is the problem statement:
Let $f$ be a continuous function from $[0, infty)$ to $mathbb{R}$ such that $$lim_{xtoinfty} f(x) = 0$$. Prove that $f$ is uniformly continuous on $[0, infty)$.
Proof:
Since $$lim_{xtoinfty} f(x) = 0$$ given $epsilon >0$ there exists an $N>0$ such that for all $x,y > N$, $|f(x)-f(y)| < epsilon$.
Now, since $[0,N]$ is a compact set, and $f$ is continuous, $f$ is uniformly continuous on $[0,N]$. That is, given the same $epsilon >0$, there exists a $delta > 0$ such that $|f(x)-f(y)| < epsilon$ for all $x,y in [0,N]$ with $|x-y| < delta$.
Now we just need to show that $f$ is uniformly continuous on $(N,infty)$.
My question is, can't we just say that for all $x,y > N$ with $|x-y| < delta$, $|f(x) - f(y)| < epsilon$ and hence, we have $f$ is uniformly continuous on all of $[0, infty)$?
A previous prove says to let $delta_j = min{1,delta}$ before proving uniform continuity of $f$ on $(N, infty)$ and I don't see the need for this step.
real-analysis
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
$endgroup$
add a comment |
$begingroup$
This question has been answered in the past but I am confused about a point in the proof.
Here is the problem statement:
Let $f$ be a continuous function from $[0, infty)$ to $mathbb{R}$ such that $$lim_{xtoinfty} f(x) = 0$$. Prove that $f$ is uniformly continuous on $[0, infty)$.
Proof:
Since $$lim_{xtoinfty} f(x) = 0$$ given $epsilon >0$ there exists an $N>0$ such that for all $x,y > N$, $|f(x)-f(y)| < epsilon$.
Now, since $[0,N]$ is a compact set, and $f$ is continuous, $f$ is uniformly continuous on $[0,N]$. That is, given the same $epsilon >0$, there exists a $delta > 0$ such that $|f(x)-f(y)| < epsilon$ for all $x,y in [0,N]$ with $|x-y| < delta$.
Now we just need to show that $f$ is uniformly continuous on $(N,infty)$.
My question is, can't we just say that for all $x,y > N$ with $|x-y| < delta$, $|f(x) - f(y)| < epsilon$ and hence, we have $f$ is uniformly continuous on all of $[0, infty)$?
A previous prove says to let $delta_j = min{1,delta}$ before proving uniform continuity of $f$ on $(N, infty)$ and I don't see the need for this step.
real-analysis
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
$endgroup$
$begingroup$
I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44
add a comment |
$begingroup$
This question has been answered in the past but I am confused about a point in the proof.
Here is the problem statement:
Let $f$ be a continuous function from $[0, infty)$ to $mathbb{R}$ such that $$lim_{xtoinfty} f(x) = 0$$. Prove that $f$ is uniformly continuous on $[0, infty)$.
Proof:
Since $$lim_{xtoinfty} f(x) = 0$$ given $epsilon >0$ there exists an $N>0$ such that for all $x,y > N$, $|f(x)-f(y)| < epsilon$.
Now, since $[0,N]$ is a compact set, and $f$ is continuous, $f$ is uniformly continuous on $[0,N]$. That is, given the same $epsilon >0$, there exists a $delta > 0$ such that $|f(x)-f(y)| < epsilon$ for all $x,y in [0,N]$ with $|x-y| < delta$.
Now we just need to show that $f$ is uniformly continuous on $(N,infty)$.
My question is, can't we just say that for all $x,y > N$ with $|x-y| < delta$, $|f(x) - f(y)| < epsilon$ and hence, we have $f$ is uniformly continuous on all of $[0, infty)$?
A previous prove says to let $delta_j = min{1,delta}$ before proving uniform continuity of $f$ on $(N, infty)$ and I don't see the need for this step.
real-analysis
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
$endgroup$
This question has been answered in the past but I am confused about a point in the proof.
Here is the problem statement:
Let $f$ be a continuous function from $[0, infty)$ to $mathbb{R}$ such that $$lim_{xtoinfty} f(x) = 0$$. Prove that $f$ is uniformly continuous on $[0, infty)$.
Proof:
Since $$lim_{xtoinfty} f(x) = 0$$ given $epsilon >0$ there exists an $N>0$ such that for all $x,y > N$, $|f(x)-f(y)| < epsilon$.
Now, since $[0,N]$ is a compact set, and $f$ is continuous, $f$ is uniformly continuous on $[0,N]$. That is, given the same $epsilon >0$, there exists a $delta > 0$ such that $|f(x)-f(y)| < epsilon$ for all $x,y in [0,N]$ with $|x-y| < delta$.
Now we just need to show that $f$ is uniformly continuous on $(N,infty)$.
My question is, can't we just say that for all $x,y > N$ with $|x-y| < delta$, $|f(x) - f(y)| < epsilon$ and hence, we have $f$ is uniformly continuous on all of $[0, infty)$?
A previous prove says to let $delta_j = min{1,delta}$ before proving uniform continuity of $f$ on $(N, infty)$ and I don't see the need for this step.
real-analysis
real-analysis
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
asked Apr 8 '15 at 21:56
JohnverJohnver
892610
892610
$begingroup$
I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44
add a comment |
$begingroup$
I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44
$begingroup$
I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44
add a comment |
1 Answer
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A possible leak in the proof follows if we ask: Is $f$ continuous on $N$.
A simple way to solve this:
In stead of taking $[0,N]$, take $[0,N+1]$ as the compact set.
Where $N$ is chosen such that: $$forall x>N: |f(x)-0|<frac{epsilon}{2}$$
Than we have:
$$forall x,y>N: |f(x)-f(y)| leq |f(x)-0|+|f(y)-0|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon : : (1)$$
As in the OP we have a $delta_1>0$ such that: $$forall x,y: x,y in [0,N+1]: |x-y|<delta_1 Rightarrow |f(x)-f(y)| <epsilon : : (2)$$
Now take $delta$ such that: $0<delta <min {frac{1}{2},delta_1 }$.
Final step: Take $x,y in [0,+infty[$.
There can be three cases:
Case 1: $:$ $x,y in [0,N+frac{1}{2}]$.
Then it follows directly from $(2)$ that: $|f(x)-f(y)|<epsilon$.
Case 2: $:$ $x,y in [N+frac{1}{2},+infty]$. Then: it follows directly from $(1)$.
Case 3: By the choice of $delta$ we can see that $y<N+1$ so both $x,y$ are smaller than $N+1$. And thus case 1 applies.
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
A possible leak in the proof follows if we ask: Is $f$ continuous on $N$.
A simple way to solve this:
In stead of taking $[0,N]$, take $[0,N+1]$ as the compact set.
Where $N$ is chosen such that: $$forall x>N: |f(x)-0|<frac{epsilon}{2}$$
Than we have:
$$forall x,y>N: |f(x)-f(y)| leq |f(x)-0|+|f(y)-0|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon : : (1)$$
As in the OP we have a $delta_1>0$ such that: $$forall x,y: x,y in [0,N+1]: |x-y|<delta_1 Rightarrow |f(x)-f(y)| <epsilon : : (2)$$
Now take $delta$ such that: $0<delta <min {frac{1}{2},delta_1 }$.
Final step: Take $x,y in [0,+infty[$.
There can be three cases:
Case 1: $:$ $x,y in [0,N+frac{1}{2}]$.
Then it follows directly from $(2)$ that: $|f(x)-f(y)|<epsilon$.
Case 2: $:$ $x,y in [N+frac{1}{2},+infty]$. Then: it follows directly from $(1)$.
Case 3: By the choice of $delta$ we can see that $y<N+1$ so both $x,y$ are smaller than $N+1$. And thus case 1 applies.
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
$endgroup$
add a comment |
$begingroup$
A possible leak in the proof follows if we ask: Is $f$ continuous on $N$.
A simple way to solve this:
In stead of taking $[0,N]$, take $[0,N+1]$ as the compact set.
Where $N$ is chosen such that: $$forall x>N: |f(x)-0|<frac{epsilon}{2}$$
Than we have:
$$forall x,y>N: |f(x)-f(y)| leq |f(x)-0|+|f(y)-0|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon : : (1)$$
As in the OP we have a $delta_1>0$ such that: $$forall x,y: x,y in [0,N+1]: |x-y|<delta_1 Rightarrow |f(x)-f(y)| <epsilon : : (2)$$
Now take $delta$ such that: $0<delta <min {frac{1}{2},delta_1 }$.
Final step: Take $x,y in [0,+infty[$.
There can be three cases:
Case 1: $:$ $x,y in [0,N+frac{1}{2}]$.
Then it follows directly from $(2)$ that: $|f(x)-f(y)|<epsilon$.
Case 2: $:$ $x,y in [N+frac{1}{2},+infty]$. Then: it follows directly from $(1)$.
Case 3: By the choice of $delta$ we can see that $y<N+1$ so both $x,y$ are smaller than $N+1$. And thus case 1 applies.
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
$endgroup$
add a comment |
$begingroup$
A possible leak in the proof follows if we ask: Is $f$ continuous on $N$.
A simple way to solve this:
In stead of taking $[0,N]$, take $[0,N+1]$ as the compact set.
Where $N$ is chosen such that: $$forall x>N: |f(x)-0|<frac{epsilon}{2}$$
Than we have:
$$forall x,y>N: |f(x)-f(y)| leq |f(x)-0|+|f(y)-0|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon : : (1)$$
As in the OP we have a $delta_1>0$ such that: $$forall x,y: x,y in [0,N+1]: |x-y|<delta_1 Rightarrow |f(x)-f(y)| <epsilon : : (2)$$
Now take $delta$ such that: $0<delta <min {frac{1}{2},delta_1 }$.
Final step: Take $x,y in [0,+infty[$.
There can be three cases:
Case 1: $:$ $x,y in [0,N+frac{1}{2}]$.
Then it follows directly from $(2)$ that: $|f(x)-f(y)|<epsilon$.
Case 2: $:$ $x,y in [N+frac{1}{2},+infty]$. Then: it follows directly from $(1)$.
Case 3: By the choice of $delta$ we can see that $y<N+1$ so both $x,y$ are smaller than $N+1$. And thus case 1 applies.
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
$endgroup$
A possible leak in the proof follows if we ask: Is $f$ continuous on $N$.
A simple way to solve this:
In stead of taking $[0,N]$, take $[0,N+1]$ as the compact set.
Where $N$ is chosen such that: $$forall x>N: |f(x)-0|<frac{epsilon}{2}$$
Than we have:
$$forall x,y>N: |f(x)-f(y)| leq |f(x)-0|+|f(y)-0|<frac{epsilon}{2}+frac{epsilon}{2}=epsilon : : (1)$$
As in the OP we have a $delta_1>0$ such that: $$forall x,y: x,y in [0,N+1]: |x-y|<delta_1 Rightarrow |f(x)-f(y)| <epsilon : : (2)$$
Now take $delta$ such that: $0<delta <min {frac{1}{2},delta_1 }$.
Final step: Take $x,y in [0,+infty[$.
There can be three cases:
Case 1: $:$ $x,y in [0,N+frac{1}{2}]$.
Then it follows directly from $(2)$ that: $|f(x)-f(y)|<epsilon$.
Case 2: $:$ $x,y in [N+frac{1}{2},+infty]$. Then: it follows directly from $(1)$.
Case 3: By the choice of $delta$ we can see that $y<N+1$ so both $x,y$ are smaller than $N+1$. And thus case 1 applies.
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
answered Apr 8 '15 at 22:26
abcdefabcdef
912416
912416
add a comment |
add a comment |
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I would argue differently. For $varepsilon>0$ choose $N$ such that $|f(x)-f(y)|<varepsilon/2$ for $x,y >N$. Then for $[0,N]$ choose the $delta$ corresponding to $varepsilon/2$ from the uniform continuity property. This $delta$, then works for all $x,y$ since if $x,y>N$ or $x,y<N$ the property is obviously satisfied. If $x<N<y$, then apply the uniform continuity bound on $|f(x)-f(N)| <varepsilon/2$ and $|f(N)-f(y)|<varepsilon/2$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:14
$begingroup$
Another way to do it: Since $f$ has a limit at $+infty$, you can say that $f(x) = f(tan y)$ with $y in [0,pi/2]$. If you denote $g: [0,pi/2], g(y) = f(tan y)$, then $g$ is continuous. Furthermore, $f(x) = g(arctan x)$. Now use the fact that $g$ is uniformly continuous and that $|arctan x-arctan y| leq |x-y|$.
$endgroup$
– Beni Bogosel
Apr 8 '15 at 22:19
$begingroup$
Possible Duplicate
$endgroup$
– Empty
Apr 9 '15 at 11:44