Deciding convergence of a sequence of functions
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I am given the sequence of functions $f_n(x)=x^n - x^{2n}$ on $[0,1]$. I must define a function $f(x)$ as a pointwise limit function on the indicated interval. If it is uniform, I must then find a sequence of real numbers $B_n rightarrow 0$ such that $|f_n(x) - f(x)| leq B_n$ for all $x$ in the interval. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$.
My attempted solution: I deduced that $f(x) = 0$ for $x in [0,1]$. I looked to see if such a $B_n$ existed, so I took the derivative of $f_n(x)$ and found the maximum to occur at $x=(frac{1}{2})^{frac{1}{n}}$. Using this as $B_n$ we get that $B_n rightarrow 1$ as $n rightarrow infty$. Therefore, $f_n(x)$ should not uniformly converge (I am not sure if this is a correct conclusion).
Assuming my previous conclusion is correct, I am having a hard time choosing an $epsilon$ where I can show $|f_n(x_n) - f(x_n)| = epsilon$ to show it is not uniformly convergent. Algebra is and has always been my weakest skill, so I am not sure if I am missing a clever choice, or if I did something wrong in my problem. Any help is appreciated, please let me know if I need to elaborate more on something. Thanks.
real-analysis convergence uniform-convergence sequence-of-function
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
asked Dec 2 at 20:11
hkj447
153
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hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
I am given the sequence of functions $f_n(x)=x^n - x^{2n}$ on $[0,1]$. I must define a function $f(x)$ as a pointwise limit function on the indicated interval. If it is uniform, I must then find a sequence of real numbers $B_n rightarrow 0$ such that $|f_n(x) - f(x)| leq B_n$ for all $x$ in the interval. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$.
My attempted solution: I deduced that $f(x) = 0$ for $x in [0,1]$. I looked to see if such a $B_n$ existed, so I took the derivative of $f_n(x)$ and found the maximum to occur at $x=(frac{1}{2})^{frac{1}{n}}$. Using this as $B_n$ we get that $B_n rightarrow 1$ as $n rightarrow infty$. Therefore, $f_n(x)$ should not uniformly converge (I am not sure if this is a correct conclusion).
Assuming my previous conclusion is correct, I am having a hard time choosing an $epsilon$ where I can show $|f_n(x_n) - f(x_n)| = epsilon$ to show it is not uniformly convergent. Algebra is and has always been my weakest skill, so I am not sure if I am missing a clever choice, or if I did something wrong in my problem. Any help is appreciated, please let me know if I need to elaborate more on something. Thanks.
real-analysis convergence uniform-convergence sequence-of-function
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
asked Dec 2 at 20:11
hkj447
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the sequence of functions $f_n(x)=x^n - x^{2n}$ on $[0,1]$. I must define a function $f(x)$ as a pointwise limit function on the indicated interval. If it is uniform, I must then find a sequence of real numbers $B_n rightarrow 0$ such that $|f_n(x) - f(x)| leq B_n$ for all $x$ in the interval. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$.
My attempted solution: I deduced that $f(x) = 0$ for $x in [0,1]$. I looked to see if such a $B_n$ existed, so I took the derivative of $f_n(x)$ and found the maximum to occur at $x=(frac{1}{2})^{frac{1}{n}}$. Using this as $B_n$ we get that $B_n rightarrow 1$ as $n rightarrow infty$. Therefore, $f_n(x)$ should not uniformly converge (I am not sure if this is a correct conclusion).
Assuming my previous conclusion is correct, I am having a hard time choosing an $epsilon$ where I can show $|f_n(x_n) - f(x_n)| = epsilon$ to show it is not uniformly convergent. Algebra is and has always been my weakest skill, so I am not sure if I am missing a clever choice, or if I did something wrong in my problem. Any help is appreciated, please let me know if I need to elaborate more on something. Thanks.
real-analysis convergence uniform-convergence sequence-of-function
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
asked Dec 2 at 20:11
hkj447
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am given the sequence of functions $f_n(x)=x^n - x^{2n}$ on $[0,1]$. I must define a function $f(x)$ as a pointwise limit function on the indicated interval. If it is uniform, I must then find a sequence of real numbers $B_n rightarrow 0$ such that $|f_n(x) - f(x)| leq B_n$ for all $x$ in the interval. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$.
My attempted solution: I deduced that $f(x) = 0$ for $x in [0,1]$. I looked to see if such a $B_n$ existed, so I took the derivative of $f_n(x)$ and found the maximum to occur at $x=(frac{1}{2})^{frac{1}{n}}$. Using this as $B_n$ we get that $B_n rightarrow 1$ as $n rightarrow infty$. Therefore, $f_n(x)$ should not uniformly converge (I am not sure if this is a correct conclusion).
Assuming my previous conclusion is correct, I am having a hard time choosing an $epsilon$ where I can show $|f_n(x_n) - f(x_n)| = epsilon$ to show it is not uniformly convergent. Algebra is and has always been my weakest skill, so I am not sure if I am missing a clever choice, or if I did something wrong in my problem. Any help is appreciated, please let me know if I need to elaborate more on something. Thanks.
real-analysis convergence uniform-convergence sequence-of-function
real-analysis convergence uniform-convergence sequence-of-function
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
asked Dec 2 at 20:11
hkj447
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
asked Dec 2 at 20:11
hkj447
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
edited Dec 2 at 20:28
José Carlos Santos
144k20114214
144k20114214
asked Dec 2 at 20:11
hkj447
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 20:11
hkj447
153
asked Dec 2 at 20:11
hkj447
153
153
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27
add a comment |
Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27
Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It's more simple than that. Since$$f_nleft(sqrt[n]{frac12}right)=frac14,$$just use $varepsilon=frac14$ to prove that the convergence is not uniform.
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's more simple than that. Since$$f_nleft(sqrt[n]{frac12}right)=frac14,$$just use $varepsilon=frac14$ to prove that the convergence is not uniform.
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
add a comment |
up vote
2
down vote
accepted
It's more simple than that. Since$$f_nleft(sqrt[n]{frac12}right)=frac14,$$just use $varepsilon=frac14$ to prove that the convergence is not uniform.
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's more simple than that. Since$$f_nleft(sqrt[n]{frac12}right)=frac14,$$just use $varepsilon=frac14$ to prove that the convergence is not uniform.
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
It's more simple than that. Since$$f_nleft(sqrt[n]{frac12}right)=frac14,$$just use $varepsilon=frac14$ to prove that the convergence is not uniform.
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
answered Dec 2 at 20:18
José Carlos Santos
144k20114214
144k20114214
add a comment |
add a comment |
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
hkj447 is a new contributor. Be nice, and check out our Code of Conduct.
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Take $epsilon=1/2$.
– hamam_Abdallah
Dec 2 at 20:17
@hamam_Abdallah That will not work.
– José Carlos Santos
Dec 2 at 20:27