Mixing
Is this sequence of functions uniformly convergent?
$begingroup$
Let $a,b in mathbb{R}$ with $0<a<b$. For each $nin mathbb{N}$, let $f_n:[a,b]longrightarrow mathbb{R}$ be defined by $f_n(x):=left(1+frac{ln x}{n}right)^n$.
Then it is easy to see that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ pointwise, where $f(x)=x$.
Is it true that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ uniformly on $[a,b]$ ?
analysis
edited Jan 18 at 20:32
MPW
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
Let $a,b in mathbb{R}$ with $0<a<b$. For each $nin mathbb{N}$, let $f_n:[a,b]longrightarrow mathbb{R}$ be defined by $f_n(x):=left(1+frac{ln x}{n}right)^n$.
Then it is easy to see that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ pointwise, where $f(x)=x$.
Is it true that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ uniformly on $[a,b]$ ?
analysis
edited Jan 18 at 20:32
MPW
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
$endgroup$
1
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38
add a comment |
$begingroup$
Let $a,b in mathbb{R}$ with $0<a<b$. For each $nin mathbb{N}$, let $f_n:[a,b]longrightarrow mathbb{R}$ be defined by $f_n(x):=left(1+frac{ln x}{n}right)^n$.
Then it is easy to see that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ pointwise, where $f(x)=x$.
Is it true that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ uniformly on $[a,b]$ ?
analysis
edited Jan 18 at 20:32
MPW
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
$endgroup$
Let $a,b in mathbb{R}$ with $0<a<b$. For each $nin mathbb{N}$, let $f_n:[a,b]longrightarrow mathbb{R}$ be defined by $f_n(x):=left(1+frac{ln x}{n}right)^n$.
Then it is easy to see that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ pointwise, where $f(x)=x$.
Is it true that $limlimits_{nlongrightarrow infty}f_n(x)=f(x)$ uniformly on $[a,b]$ ?
analysis
analysis
edited Jan 18 at 20:32
MPW
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
edited Jan 18 at 20:32
MPW
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
edited Jan 18 at 20:32
MPW
30.4k12157
edited Jan 18 at 20:32
MPW
30.4k12157
edited Jan 18 at 20:32
MPW
30.4k12157
30.4k12157
asked Jan 18 at 20:20
serenusserenus
24316
asked Jan 18 at 20:20
serenusserenus
24316
asked Jan 18 at 20:20
serenusserenus
24316
24316
1
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38
add a comment |
1
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38
1
1
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $$f_n(x)le f(x)$$ therefore we need to show that there exists $N$ such that for $0<ale xle b$ and $n>N$ $$0le f(x)-f_n(x)<epsilon$$we need to show that $$max_{ale xle b} f(x)-f_n(x)$$can be bounded arbitrarily small for $N$ being sufficiently large. Now note that$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}$$where $x_0$ is a critical point of $x-f_n(x)$ since $x-f_n(x)$ is differentiable over $[a,b]$ i.e. for $x_0$:$$1=f'_n(x_0)to left(1+{ln xover n}right)^{n-1}=x_0$$by substituting we obtain $$x_0-f_n(x_0)=-{x_0ln x_0over n}$$therefore$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}\=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),{-x_0ln x_0over n}Big}$$but both $a-f_n(a)$ and $b-f_n(b)$ tend to $0$ and ${-x_0ln x_0over n}$ is bounded whenever $ale x_0le b$. Therefore the proof is complete.
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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$begingroup$
Note that $$f_n(x)le f(x)$$ therefore we need to show that there exists $N$ such that for $0<ale xle b$ and $n>N$ $$0le f(x)-f_n(x)<epsilon$$we need to show that $$max_{ale xle b} f(x)-f_n(x)$$can be bounded arbitrarily small for $N$ being sufficiently large. Now note that$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}$$where $x_0$ is a critical point of $x-f_n(x)$ since $x-f_n(x)$ is differentiable over $[a,b]$ i.e. for $x_0$:$$1=f'_n(x_0)to left(1+{ln xover n}right)^{n-1}=x_0$$by substituting we obtain $$x_0-f_n(x_0)=-{x_0ln x_0over n}$$therefore$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}\=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),{-x_0ln x_0over n}Big}$$but both $a-f_n(a)$ and $b-f_n(b)$ tend to $0$ and ${-x_0ln x_0over n}$ is bounded whenever $ale x_0le b$. Therefore the proof is complete.
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Note that $$f_n(x)le f(x)$$ therefore we need to show that there exists $N$ such that for $0<ale xle b$ and $n>N$ $$0le f(x)-f_n(x)<epsilon$$we need to show that $$max_{ale xle b} f(x)-f_n(x)$$can be bounded arbitrarily small for $N$ being sufficiently large. Now note that$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}$$where $x_0$ is a critical point of $x-f_n(x)$ since $x-f_n(x)$ is differentiable over $[a,b]$ i.e. for $x_0$:$$1=f'_n(x_0)to left(1+{ln xover n}right)^{n-1}=x_0$$by substituting we obtain $$x_0-f_n(x_0)=-{x_0ln x_0over n}$$therefore$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}\=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),{-x_0ln x_0over n}Big}$$but both $a-f_n(a)$ and $b-f_n(b)$ tend to $0$ and ${-x_0ln x_0over n}$ is bounded whenever $ale x_0le b$. Therefore the proof is complete.
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Note that $$f_n(x)le f(x)$$ therefore we need to show that there exists $N$ such that for $0<ale xle b$ and $n>N$ $$0le f(x)-f_n(x)<epsilon$$we need to show that $$max_{ale xle b} f(x)-f_n(x)$$can be bounded arbitrarily small for $N$ being sufficiently large. Now note that$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}$$where $x_0$ is a critical point of $x-f_n(x)$ since $x-f_n(x)$ is differentiable over $[a,b]$ i.e. for $x_0$:$$1=f'_n(x_0)to left(1+{ln xover n}right)^{n-1}=x_0$$by substituting we obtain $$x_0-f_n(x_0)=-{x_0ln x_0over n}$$therefore$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}\=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),{-x_0ln x_0over n}Big}$$but both $a-f_n(a)$ and $b-f_n(b)$ tend to $0$ and ${-x_0ln x_0over n}$ is bounded whenever $ale x_0le b$. Therefore the proof is complete.
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Note that $$f_n(x)le f(x)$$ therefore we need to show that there exists $N$ such that for $0<ale xle b$ and $n>N$ $$0le f(x)-f_n(x)<epsilon$$we need to show that $$max_{ale xle b} f(x)-f_n(x)$$can be bounded arbitrarily small for $N$ being sufficiently large. Now note that$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}$$where $x_0$ is a critical point of $x-f_n(x)$ since $x-f_n(x)$ is differentiable over $[a,b]$ i.e. for $x_0$:$$1=f'_n(x_0)to left(1+{ln xover n}right)^{n-1}=x_0$$by substituting we obtain $$x_0-f_n(x_0)=-{x_0ln x_0over n}$$therefore$$max_{ale xle b} f(x)-f_n(x)=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),x_0-f_n(x_0)Big}\=max_{x_0text{ is a critical point in }[a,b]}Big{a-f_n(a),b-f_n(b),{-x_0ln x_0over n}Big}$$but both $a-f_n(a)$ and $b-f_n(b)$ tend to $0$ and ${-x_0ln x_0over n}$ is bounded whenever $ale x_0le b$. Therefore the proof is complete.
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 21:26
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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1
$begingroup$
For a>1, you can use Dini's theorem.
$endgroup$
– J.F
Jan 18 at 20:30
$begingroup$
@G.F: If $a<1$, then it seems possible to handle the intervals $[a,1]$ and $[1,b]$ separately?
$endgroup$
– W-t-P
Jan 18 at 20:38