Proving a sequence of functions does not converge uniformly. Can it be assumed that $Ngeq 1$ an epsilon N...
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
$endgroup$
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
$endgroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
edited Dec 5 '18 at 3:54
Trigginometric
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
113
113
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
$endgroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
edited Dec 5 '18 at 4:16
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
answered Dec 5 '18 at 3:55
obscuransobscurans
1,027311
1,027311
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
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Required, but never shown
Required, but never shown
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We have not covered simple convergence as of yet, only uniform and pointwise.
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– Trigginometric
Dec 5 '18 at 3:53
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Simple = "pointwise."
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– Will M.
Dec 5 '18 at 3:53
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I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
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– Trigginometric
Dec 5 '18 at 3:55
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Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
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– Trigginometric
Dec 5 '18 at 4:03
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But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
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– Trigginometric
Dec 5 '18 at 4:07