Show that $f_n rightarrow 0$ in $C([0, 1], mathbb{R})$












0












$begingroup$


I was given the following problem and was wondering if I was on the right track.



Let
$f_n(x) = frac{1}{n} frac{nx}{1 + nx}, : 0 le x le 1$



Show that $f_n rightarrow 0$ in $C([0, 1], mathbb{R})$.



I have this theorem that I figured I could use:



$f_k rightarrow f$ uniformly on A $iff$ $f_k rightarrow f$ in $C_b$.



In this case, $C_b$ is the collection of all continuous functions on $[0,1]$. So if I can prove the function is uniformly continuous, this would prove that $f_n rightarrow 0$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?



Thanks






real-analysis convergence continuity uniform-convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
    $endgroup$
    – Ian
    Dec 7 '18 at 1:07
















0












$begingroup$


I was given the following problem and was wondering if I was on the right track.



Let
$f_n(x) = frac{1}{n} frac{nx}{1 + nx}, : 0 le x le 1$



Show that $f_n rightarrow 0$ in $C([0, 1], mathbb{R})$.



I have this theorem that I figured I could use:



$f_k rightarrow f$ uniformly on A $iff$ $f_k rightarrow f$ in $C_b$.



In this case, $C_b$ is the collection of all continuous functions on $[0,1]$. So if I can prove the function is uniformly continuous, this would prove that $f_n rightarrow 0$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?



Thanks






real-analysis convergence continuity uniform-convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
    $endgroup$
    – Ian
    Dec 7 '18 at 1:07














0












0








0





$begingroup$


I was given the following problem and was wondering if I was on the right track.



Let
$f_n(x) = frac{1}{n} frac{nx}{1 + nx}, : 0 le x le 1$



Show that $f_n rightarrow 0$ in $C([0, 1], mathbb{R})$.



I have this theorem that I figured I could use:



$f_k rightarrow f$ uniformly on A $iff$ $f_k rightarrow f$ in $C_b$.



In this case, $C_b$ is the collection of all continuous functions on $[0,1]$. So if I can prove the function is uniformly continuous, this would prove that $f_n rightarrow 0$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?



Thanks






real-analysis convergence continuity uniform-convergence







share|cite|improve this question









$endgroup$




I was given the following problem and was wondering if I was on the right track.



Let
$f_n(x) = frac{1}{n} frac{nx}{1 + nx}, : 0 le x le 1$



Show that $f_n rightarrow 0$ in $C([0, 1], mathbb{R})$.



I have this theorem that I figured I could use:



$f_k rightarrow f$ uniformly on A $iff$ $f_k rightarrow f$ in $C_b$.



In this case, $C_b$ is the collection of all continuous functions on $[0,1]$. So if I can prove the function is uniformly continuous, this would prove that $f_n rightarrow 0$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?



Thanks






real-analysis convergence continuity uniform-convergence




real-analysis convergence continuity uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 1:02









user591271user591271

876




876












  • $begingroup$
    The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
    $endgroup$
    – Ian
    Dec 7 '18 at 1:07


















  • $begingroup$
    The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
    $endgroup$
    – Ian
    Dec 7 '18 at 1:07
















$begingroup$
The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
$endgroup$
– Ian
Dec 7 '18 at 1:07




$begingroup$
The M test is useless here because it's not a series. In any case, you should be able to simply compute $| f_n |_infty$...
$endgroup$
– Ian
Dec 7 '18 at 1:07










2 Answers
2






active

oldest

votes


















3












$begingroup$

note that for $ x in [0,1] $ we have $|f_n(x)| =|frac{1}{n}frac{x}{1+nx}| le frac{1}{n}$, so $||f_n||_infty le frac{1}{n}$ which implies $f_n to 0 $ in $C_b[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice job. It's exactly what I would put. Well done.
    $endgroup$
    – ncmathsadist
    Dec 7 '18 at 1:27










  • $begingroup$
    So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
    $endgroup$
    – user591271
    Dec 7 '18 at 1:52










  • $begingroup$
    Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
    $endgroup$
    – Maksim
    Dec 7 '18 at 2:28



















1












$begingroup$

You'll notice that you can cancel to get $f_n(x) = frac{x}{1+nx}$. For $x =0$, clearly $f_n(x)= 0$. Otherwise, $nx$ gets arbitrarily large. This will imply that $f_n(x) to 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
    $endgroup$
    – Ian
    Dec 7 '18 at 14:38













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

note that for $ x in [0,1] $ we have $|f_n(x)| =|frac{1}{n}frac{x}{1+nx}| le frac{1}{n}$, so $||f_n||_infty le frac{1}{n}$ which implies $f_n to 0 $ in $C_b[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice job. It's exactly what I would put. Well done.
    $endgroup$
    – ncmathsadist
    Dec 7 '18 at 1:27










  • $begingroup$
    So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
    $endgroup$
    – user591271
    Dec 7 '18 at 1:52










  • $begingroup$
    Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
    $endgroup$
    – Maksim
    Dec 7 '18 at 2:28
















3












$begingroup$

note that for $ x in [0,1] $ we have $|f_n(x)| =|frac{1}{n}frac{x}{1+nx}| le frac{1}{n}$, so $||f_n||_infty le frac{1}{n}$ which implies $f_n to 0 $ in $C_b[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice job. It's exactly what I would put. Well done.
    $endgroup$
    – ncmathsadist
    Dec 7 '18 at 1:27










  • $begingroup$
    So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
    $endgroup$
    – user591271
    Dec 7 '18 at 1:52










  • $begingroup$
    Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
    $endgroup$
    – Maksim
    Dec 7 '18 at 2:28














3












3








3





$begingroup$

note that for $ x in [0,1] $ we have $|f_n(x)| =|frac{1}{n}frac{x}{1+nx}| le frac{1}{n}$, so $||f_n||_infty le frac{1}{n}$ which implies $f_n to 0 $ in $C_b[0,1]$






share|cite|improve this answer









$endgroup$



note that for $ x in [0,1] $ we have $|f_n(x)| =|frac{1}{n}frac{x}{1+nx}| le frac{1}{n}$, so $||f_n||_infty le frac{1}{n}$ which implies $f_n to 0 $ in $C_b[0,1]$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 1:23









MaksimMaksim

2215




2215












  • $begingroup$
    Nice job. It's exactly what I would put. Well done.
    $endgroup$
    – ncmathsadist
    Dec 7 '18 at 1:27










  • $begingroup$
    So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
    $endgroup$
    – user591271
    Dec 7 '18 at 1:52










  • $begingroup$
    Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
    $endgroup$
    – Maksim
    Dec 7 '18 at 2:28


















  • $begingroup$
    Nice job. It's exactly what I would put. Well done.
    $endgroup$
    – ncmathsadist
    Dec 7 '18 at 1:27










  • $begingroup$
    So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
    $endgroup$
    – user591271
    Dec 7 '18 at 1:52










  • $begingroup$
    Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
    $endgroup$
    – Maksim
    Dec 7 '18 at 2:28
















$begingroup$
Nice job. It's exactly what I would put. Well done.
$endgroup$
– ncmathsadist
Dec 7 '18 at 1:27




$begingroup$
Nice job. It's exactly what I would put. Well done.
$endgroup$
– ncmathsadist
Dec 7 '18 at 1:27












$begingroup$
So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
$endgroup$
– user591271
Dec 7 '18 at 1:52




$begingroup$
So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence?
$endgroup$
– user591271
Dec 7 '18 at 1:52












$begingroup$
Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
$endgroup$
– Maksim
Dec 7 '18 at 2:28




$begingroup$
Yes. However the inequality shown above, stating that $|f(x)|le frac{1}{n} $ holds for all $x in [0,1]$ directly renders uniform convergence as well.
$endgroup$
– Maksim
Dec 7 '18 at 2:28











1












$begingroup$

You'll notice that you can cancel to get $f_n(x) = frac{x}{1+nx}$. For $x =0$, clearly $f_n(x)= 0$. Otherwise, $nx$ gets arbitrarily large. This will imply that $f_n(x) to 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
    $endgroup$
    – Ian
    Dec 7 '18 at 14:38


















1












$begingroup$

You'll notice that you can cancel to get $f_n(x) = frac{x}{1+nx}$. For $x =0$, clearly $f_n(x)= 0$. Otherwise, $nx$ gets arbitrarily large. This will imply that $f_n(x) to 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
    $endgroup$
    – Ian
    Dec 7 '18 at 14:38
















1












1








1





$begingroup$

You'll notice that you can cancel to get $f_n(x) = frac{x}{1+nx}$. For $x =0$, clearly $f_n(x)= 0$. Otherwise, $nx$ gets arbitrarily large. This will imply that $f_n(x) to 0$.






share|cite|improve this answer









$endgroup$



You'll notice that you can cancel to get $f_n(x) = frac{x}{1+nx}$. For $x =0$, clearly $f_n(x)= 0$. Otherwise, $nx$ gets arbitrarily large. This will imply that $f_n(x) to 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 1:16









msmmsm

1,243515




1,243515












  • $begingroup$
    I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
    $endgroup$
    – Ian
    Dec 7 '18 at 14:38




















  • $begingroup$
    I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
    $endgroup$
    – Ian
    Dec 7 '18 at 14:38


















$begingroup$
I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
$endgroup$
– Ian
Dec 7 '18 at 14:38






$begingroup$
I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=frac{y}{1+y}$ is a bounded function on $[0,infty)$, and then the $1/n$ outside creates the decay.
$endgroup$
– Ian
Dec 7 '18 at 14:38




















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