Construct a sequence of continuous functions $f_n$ on $[0,infty)$ such that $f_n leq f_{n+1}$ and $f_n to 1$.












1












$begingroup$


I need to construct a sequence of continuous functions $f_n$ on $[0,infty)$ such that $f_n leq f_{n+1}$ for each $n$ and $f_n to 1$ pointwisely not uniformly.



I have been trying for 2 hours, the condition that $f_n to 1$ is easy and being continuous is also easy, but the condition that $f_n leq f_{n+1}$ is rather difficult for me. I can only construct decreasing ones.



Can anyone help with this?






calculus real-analysis analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    I'd recommend a variant of $arctan nx$.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:39






  • 2




    $begingroup$
    @GerryMyerson well $tanh(nx)$ requires even less effort :D
    $endgroup$
    – მამუკა ჯიბლაძე
    Nov 30 '18 at 12:01






  • 2




    $begingroup$
    if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 12:29










  • $begingroup$
    What about $f_n=1$ for all $n$?
    $endgroup$
    – dan_fulea
    Nov 30 '18 at 20:04










  • $begingroup$
    @dan_fulea, your example converges uniformly
    $endgroup$
    – bbw
    Nov 30 '18 at 21:18
















1












$begingroup$


I need to construct a sequence of continuous functions $f_n$ on $[0,infty)$ such that $f_n leq f_{n+1}$ for each $n$ and $f_n to 1$ pointwisely not uniformly.



I have been trying for 2 hours, the condition that $f_n to 1$ is easy and being continuous is also easy, but the condition that $f_n leq f_{n+1}$ is rather difficult for me. I can only construct decreasing ones.



Can anyone help with this?






calculus real-analysis analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    I'd recommend a variant of $arctan nx$.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:39






  • 2




    $begingroup$
    @GerryMyerson well $tanh(nx)$ requires even less effort :D
    $endgroup$
    – მამუკა ჯიბლაძე
    Nov 30 '18 at 12:01






  • 2




    $begingroup$
    if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 12:29










  • $begingroup$
    What about $f_n=1$ for all $n$?
    $endgroup$
    – dan_fulea
    Nov 30 '18 at 20:04










  • $begingroup$
    @dan_fulea, your example converges uniformly
    $endgroup$
    – bbw
    Nov 30 '18 at 21:18














1












1








1





$begingroup$


I need to construct a sequence of continuous functions $f_n$ on $[0,infty)$ such that $f_n leq f_{n+1}$ for each $n$ and $f_n to 1$ pointwisely not uniformly.



I have been trying for 2 hours, the condition that $f_n to 1$ is easy and being continuous is also easy, but the condition that $f_n leq f_{n+1}$ is rather difficult for me. I can only construct decreasing ones.



Can anyone help with this?






calculus real-analysis analysis







share|cite|improve this question









$endgroup$




I need to construct a sequence of continuous functions $f_n$ on $[0,infty)$ such that $f_n leq f_{n+1}$ for each $n$ and $f_n to 1$ pointwisely not uniformly.



I have been trying for 2 hours, the condition that $f_n to 1$ is easy and being continuous is also easy, but the condition that $f_n leq f_{n+1}$ is rather difficult for me. I can only construct decreasing ones.



Can anyone help with this?






calculus real-analysis analysis




calculus real-analysis analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 11:33









bbwbbw

50038




50038








  • 3




    $begingroup$
    I'd recommend a variant of $arctan nx$.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:39






  • 2




    $begingroup$
    @GerryMyerson well $tanh(nx)$ requires even less effort :D
    $endgroup$
    – მამუკა ჯიბლაძე
    Nov 30 '18 at 12:01






  • 2




    $begingroup$
    if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 12:29










  • $begingroup$
    What about $f_n=1$ for all $n$?
    $endgroup$
    – dan_fulea
    Nov 30 '18 at 20:04










  • $begingroup$
    @dan_fulea, your example converges uniformly
    $endgroup$
    – bbw
    Nov 30 '18 at 21:18














  • 3




    $begingroup$
    I'd recommend a variant of $arctan nx$.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:39






  • 2




    $begingroup$
    @GerryMyerson well $tanh(nx)$ requires even less effort :D
    $endgroup$
    – მამუკა ჯიბლაძე
    Nov 30 '18 at 12:01






  • 2




    $begingroup$
    if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 12:29










  • $begingroup$
    What about $f_n=1$ for all $n$?
    $endgroup$
    – dan_fulea
    Nov 30 '18 at 20:04










  • $begingroup$
    @dan_fulea, your example converges uniformly
    $endgroup$
    – bbw
    Nov 30 '18 at 21:18








3




3




$begingroup$
I'd recommend a variant of $arctan nx$.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:39




$begingroup$
I'd recommend a variant of $arctan nx$.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:39




2




2




$begingroup$
@GerryMyerson well $tanh(nx)$ requires even less effort :D
$endgroup$
– მამუკა ჯიბლაძე
Nov 30 '18 at 12:01




$begingroup$
@GerryMyerson well $tanh(nx)$ requires even less effort :D
$endgroup$
– მამუკა ჯიბლაძე
Nov 30 '18 at 12:01




2




2




$begingroup$
if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
$endgroup$
– Calvin Khor
Nov 30 '18 at 12:29




$begingroup$
if $f_nto 1$ point wise and not uniformly but $f_n$ are decreasing, you can use $2-f_n$ instead.
$endgroup$
– Calvin Khor
Nov 30 '18 at 12:29












$begingroup$
What about $f_n=1$ for all $n$?
$endgroup$
– dan_fulea
Nov 30 '18 at 20:04




$begingroup$
What about $f_n=1$ for all $n$?
$endgroup$
– dan_fulea
Nov 30 '18 at 20:04












$begingroup$
@dan_fulea, your example converges uniformly
$endgroup$
– bbw
Nov 30 '18 at 21:18




$begingroup$
@dan_fulea, your example converges uniformly
$endgroup$
– bbw
Nov 30 '18 at 21:18










2 Answers
2






active

oldest

votes


















3












$begingroup$

You may take $f_n$ to be $1$ on $[0,n]$, decrease linearly from $1$ to $0$ on $[n,n-1]$, and be $0$ on $[n+1, +infty[$. The formula is



$$f_n(x) = begin{cases}
1 & text{if } x le n\
1 - (x-n) & text{if } n < x le n+1 \
0 & text{if } n+1 < x\
end{cases}$$



The sequence $(f_n)_{n ge 0}$ is increasing, and converges point wise $f_n to 1$, but not uniformly (what is $|f_{n}-1|_{infty}$ ?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh right ! Thanks for spotting my mistake !
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:10



















1












$begingroup$

Let $$f_n(x)=begin{cases}cos {pi xover 2n}&,quad 0le xle n\0& ,quad x>nend{cases}$$and you're done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:19













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

You may take $f_n$ to be $1$ on $[0,n]$, decrease linearly from $1$ to $0$ on $[n,n-1]$, and be $0$ on $[n+1, +infty[$. The formula is



$$f_n(x) = begin{cases}
1 & text{if } x le n\
1 - (x-n) & text{if } n < x le n+1 \
0 & text{if } n+1 < x\
end{cases}$$



The sequence $(f_n)_{n ge 0}$ is increasing, and converges point wise $f_n to 1$, but not uniformly (what is $|f_{n}-1|_{infty}$ ?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh right ! Thanks for spotting my mistake !
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:10
















3












$begingroup$

You may take $f_n$ to be $1$ on $[0,n]$, decrease linearly from $1$ to $0$ on $[n,n-1]$, and be $0$ on $[n+1, +infty[$. The formula is



$$f_n(x) = begin{cases}
1 & text{if } x le n\
1 - (x-n) & text{if } n < x le n+1 \
0 & text{if } n+1 < x\
end{cases}$$



The sequence $(f_n)_{n ge 0}$ is increasing, and converges point wise $f_n to 1$, but not uniformly (what is $|f_{n}-1|_{infty}$ ?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh right ! Thanks for spotting my mistake !
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:10














3












3








3





$begingroup$

You may take $f_n$ to be $1$ on $[0,n]$, decrease linearly from $1$ to $0$ on $[n,n-1]$, and be $0$ on $[n+1, +infty[$. The formula is



$$f_n(x) = begin{cases}
1 & text{if } x le n\
1 - (x-n) & text{if } n < x le n+1 \
0 & text{if } n+1 < x\
end{cases}$$



The sequence $(f_n)_{n ge 0}$ is increasing, and converges point wise $f_n to 1$, but not uniformly (what is $|f_{n}-1|_{infty}$ ?).






share|cite|improve this answer











$endgroup$



You may take $f_n$ to be $1$ on $[0,n]$, decrease linearly from $1$ to $0$ on $[n,n-1]$, and be $0$ on $[n+1, +infty[$. The formula is



$$f_n(x) = begin{cases}
1 & text{if } x le n\
1 - (x-n) & text{if } n < x le n+1 \
0 & text{if } n+1 < x\
end{cases}$$



The sequence $(f_n)_{n ge 0}$ is increasing, and converges point wise $f_n to 1$, but not uniformly (what is $|f_{n}-1|_{infty}$ ?).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 12:19

























answered Nov 30 '18 at 12:02









Joel CohenJoel Cohen

7,32412137




7,32412137












  • $begingroup$
    Oh right ! Thanks for spotting my mistake !
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:10


















  • $begingroup$
    Oh right ! Thanks for spotting my mistake !
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:10
















$begingroup$
Oh right ! Thanks for spotting my mistake !
$endgroup$
– Joel Cohen
Nov 30 '18 at 12:10




$begingroup$
Oh right ! Thanks for spotting my mistake !
$endgroup$
– Joel Cohen
Nov 30 '18 at 12:10











1












$begingroup$

Let $$f_n(x)=begin{cases}cos {pi xover 2n}&,quad 0le xle n\0& ,quad x>nend{cases}$$and you're done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:19


















1












$begingroup$

Let $$f_n(x)=begin{cases}cos {pi xover 2n}&,quad 0le xle n\0& ,quad x>nend{cases}$$and you're done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:19
















1












1








1





$begingroup$

Let $$f_n(x)=begin{cases}cos {pi xover 2n}&,quad 0le xle n\0& ,quad x>nend{cases}$$and you're done.






share|cite|improve this answer











$endgroup$



Let $$f_n(x)=begin{cases}cos {pi xover 2n}&,quad 0le xle n\0& ,quad x>nend{cases}$$and you're done.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 11:47

























answered Nov 30 '18 at 11:40









Mostafa AyazMostafa Ayaz

15.6k3939




15.6k3939












  • $begingroup$
    Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:19




















  • $begingroup$
    Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
    $endgroup$
    – Joel Cohen
    Nov 30 '18 at 12:19


















$begingroup$
Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
$endgroup$
– Joel Cohen
Nov 30 '18 at 12:19






$begingroup$
Nice idea ! I think it would also work by replacing $x mapsto cos(pi x /2)$ with any continuous function $f : [0,1] to [0,1]$ that decreases from $1$ do $0$ (you may for example take $1-frac{x}{n}$ on $[0,n]$)
$endgroup$
– Joel Cohen
Nov 30 '18 at 12:19




















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