Prove that $(f_n)_{ninmathbb{N}},f_n:mathbb{R}rightarrowmathbb{R}$ with $f_n(x)=frac{n+1}{n}x$ is uniformly...
$begingroup$
Maybe the Statement of the Question is also false, I am trying to come up with an easy example for uniform convergence.
It is clear that $lim_{nrightarrowinfty}(f_n)_{ninmathbb{N}}=$id,id$(x)=x$
To Show that the functionsequence is uniformly convergent I have to show
$forallepsilon>0forall xin mathbb{R} exists n_0inmathbb{N}forall n geq n_0:|frac{n}{n+1}x-x|<epsiloniff |-frac{1}{n+1}x|<epsilon$
Now if I I Claim that I have found such a $n_0in mathbb{N}$ If I would pick an $epsilon>1$ and $x=n_0+1$ I would get $1<1$
What am I doing wrong here?
real-analysis
asked Jan 9 at 19:20
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
Maybe the Statement of the Question is also false, I am trying to come up with an easy example for uniform convergence.
It is clear that $lim_{nrightarrowinfty}(f_n)_{ninmathbb{N}}=$id,id$(x)=x$
To Show that the functionsequence is uniformly convergent I have to show
$forallepsilon>0forall xin mathbb{R} exists n_0inmathbb{N}forall n geq n_0:|frac{n}{n+1}x-x|<epsiloniff |-frac{1}{n+1}x|<epsilon$
Now if I I Claim that I have found such a $n_0in mathbb{N}$ If I would pick an $epsilon>1$ and $x=n_0+1$ I would get $1<1$
What am I doing wrong here?
real-analysis
asked Jan 9 at 19:20
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
Maybe the Statement of the Question is also false, I am trying to come up with an easy example for uniform convergence.
It is clear that $lim_{nrightarrowinfty}(f_n)_{ninmathbb{N}}=$id,id$(x)=x$
To Show that the functionsequence is uniformly convergent I have to show
$forallepsilon>0forall xin mathbb{R} exists n_0inmathbb{N}forall n geq n_0:|frac{n}{n+1}x-x|<epsiloniff |-frac{1}{n+1}x|<epsilon$
Now if I I Claim that I have found such a $n_0in mathbb{N}$ If I would pick an $epsilon>1$ and $x=n_0+1$ I would get $1<1$
What am I doing wrong here?
real-analysis
asked Jan 9 at 19:20
RM777RM777
38312
$endgroup$
Maybe the Statement of the Question is also false, I am trying to come up with an easy example for uniform convergence.
It is clear that $lim_{nrightarrowinfty}(f_n)_{ninmathbb{N}}=$id,id$(x)=x$
To Show that the functionsequence is uniformly convergent I have to show
$forallepsilon>0forall xin mathbb{R} exists n_0inmathbb{N}forall n geq n_0:|frac{n}{n+1}x-x|<epsiloniff |-frac{1}{n+1}x|<epsilon$
Now if I I Claim that I have found such a $n_0in mathbb{N}$ If I would pick an $epsilon>1$ and $x=n_0+1$ I would get $1<1$
What am I doing wrong here?
real-analysis
real-analysis
asked Jan 9 at 19:20
RM777RM777
38312
asked Jan 9 at 19:20
RM777RM777
38312
asked Jan 9 at 19:20
RM777RM777
38312
asked Jan 9 at 19:20
RM777RM777
38312
asked Jan 9 at 19:20
RM777RM777
38312
38312
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I am trying to come up with an easy example for uniform convergence.
It is very hard for a sequence of polynomials to converge uniformly on all of $mathbb{R}$. Because all non-constant polynomials go to $pminfty$ at $pminfty$, a sequence of polynomials $p_n(x)$ can converge uniformly to $p(x)$ if and only if there is some $N$ such that $p_n(x)-p(x)$ is a constant for each $n>N$, and these constants go to zero. It isn't good enough to have the polynomial's coefficients converge - we need everything but the constant term to be equal after some point.
So, two options. Either we run the whole thing on some smaller interval, or we use examples that aren't polynomials. Here's one: $f_n(x)=frac1{(x+frac1n)^2+1}$.
answered Jan 9 at 19:49
jmerryjmerry
17k11633
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Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
$endgroup$
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
add a comment |
$begingroup$
Your second last line;
$|x|(frac{n_0+1}{n_0}-1)= |x|frac{1}{n_0} < epsilon$.
Let $epsilon =1$, and choose $x =n_0+1$.
$1>1$; Contradiction .
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
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$begingroup$
I am trying to come up with an easy example for uniform convergence.
It is very hard for a sequence of polynomials to converge uniformly on all of $mathbb{R}$. Because all non-constant polynomials go to $pminfty$ at $pminfty$, a sequence of polynomials $p_n(x)$ can converge uniformly to $p(x)$ if and only if there is some $N$ such that $p_n(x)-p(x)$ is a constant for each $n>N$, and these constants go to zero. It isn't good enough to have the polynomial's coefficients converge - we need everything but the constant term to be equal after some point.
So, two options. Either we run the whole thing on some smaller interval, or we use examples that aren't polynomials. Here's one: $f_n(x)=frac1{(x+frac1n)^2+1}$.
answered Jan 9 at 19:49
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
I am trying to come up with an easy example for uniform convergence.
It is very hard for a sequence of polynomials to converge uniformly on all of $mathbb{R}$. Because all non-constant polynomials go to $pminfty$ at $pminfty$, a sequence of polynomials $p_n(x)$ can converge uniformly to $p(x)$ if and only if there is some $N$ such that $p_n(x)-p(x)$ is a constant for each $n>N$, and these constants go to zero. It isn't good enough to have the polynomial's coefficients converge - we need everything but the constant term to be equal after some point.
So, two options. Either we run the whole thing on some smaller interval, or we use examples that aren't polynomials. Here's one: $f_n(x)=frac1{(x+frac1n)^2+1}$.
answered Jan 9 at 19:49
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
I am trying to come up with an easy example for uniform convergence.
It is very hard for a sequence of polynomials to converge uniformly on all of $mathbb{R}$. Because all non-constant polynomials go to $pminfty$ at $pminfty$, a sequence of polynomials $p_n(x)$ can converge uniformly to $p(x)$ if and only if there is some $N$ such that $p_n(x)-p(x)$ is a constant for each $n>N$, and these constants go to zero. It isn't good enough to have the polynomial's coefficients converge - we need everything but the constant term to be equal after some point.
So, two options. Either we run the whole thing on some smaller interval, or we use examples that aren't polynomials. Here's one: $f_n(x)=frac1{(x+frac1n)^2+1}$.
answered Jan 9 at 19:49
jmerryjmerry
17k11633
$endgroup$
I am trying to come up with an easy example for uniform convergence.
It is very hard for a sequence of polynomials to converge uniformly on all of $mathbb{R}$. Because all non-constant polynomials go to $pminfty$ at $pminfty$, a sequence of polynomials $p_n(x)$ can converge uniformly to $p(x)$ if and only if there is some $N$ such that $p_n(x)-p(x)$ is a constant for each $n>N$, and these constants go to zero. It isn't good enough to have the polynomial's coefficients converge - we need everything but the constant term to be equal after some point.
So, two options. Either we run the whole thing on some smaller interval, or we use examples that aren't polynomials. Here's one: $f_n(x)=frac1{(x+frac1n)^2+1}$.
answered Jan 9 at 19:49
jmerryjmerry
17k11633
answered Jan 9 at 19:49
jmerryjmerry
17k11633
answered Jan 9 at 19:49
jmerryjmerry
17k11633
answered Jan 9 at 19:49
jmerryjmerry
17k11633
17k11633
add a comment |
add a comment |
$begingroup$
Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
$endgroup$
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
add a comment |
$begingroup$
Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
$endgroup$
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
add a comment |
$begingroup$
Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
$endgroup$
Note that $|f_n(n)-f(n)| = 1$ for all $n$. Hence the convergence is not uniform.
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
answered Jan 9 at 19:23
copper.hatcopper.hat
128k561161
128k561161
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
add a comment |
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
Okey so the uniform convergence is a Statement About the sequence of functions but not About the limesfunction itself.
$endgroup$
– RM777
Jan 9 at 19:31
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
$begingroup$
@RM777: I don't understand what you mean by limesfunction...
$endgroup$
– copper.hat
Jan 9 at 19:56
add a comment |
$begingroup$
Your second last line;
$|x|(frac{n_0+1}{n_0}-1)= |x|frac{1}{n_0} < epsilon$.
Let $epsilon =1$, and choose $x =n_0+1$.
$1>1$; Contradiction .
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
$begingroup$
Your second last line;
$|x|(frac{n_0+1}{n_0}-1)= |x|frac{1}{n_0} < epsilon$.
Let $epsilon =1$, and choose $x =n_0+1$.
$1>1$; Contradiction .
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
$begingroup$
Your second last line;
$|x|(frac{n_0+1}{n_0}-1)= |x|frac{1}{n_0} < epsilon$.
Let $epsilon =1$, and choose $x =n_0+1$.
$1>1$; Contradiction .
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
Your second last line;
$|x|(frac{n_0+1}{n_0}-1)= |x|frac{1}{n_0} < epsilon$.
Let $epsilon =1$, and choose $x =n_0+1$.
$1>1$; Contradiction .
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
answered Jan 9 at 19:42
Peter SzilasPeter Szilas
11.7k2822
11.7k2822
add a comment |
add a comment |
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