Does this sequence of function converges uniformly?
$begingroup$
I was practicing for my final and stumbled on this problem.
Let $f_n$:$[0,1]$ $mapsto$ $R$ such that
$f(x)_n$ =
begin{cases}
nx, & 0leq text{ $x$ } lt1/n\
1, & 1/nleqtext{ $x$ }leq 1
end{cases}
I could find its piecewise limit, which is
$f(x)$ =
begin{cases}
0, & text{ $x$ =0} \
1, & text{ $otherwise$ }
end{cases}
I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$
then I realized there were more than one interval of both $f_n$ and $f$.
So I used $x$ to indicate the supremum of each interval
for $x=0$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $0$
for $1/nleqtext{ $x$ }leq 1$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $1$
Now I am stuck here
for $0lt text{ $x$ } lt1/n$
What should i do with this interval?
And does $f(x)_n$ converges uniformly?
real-analysis uniform-convergence
$endgroup$
add a comment |
$begingroup$
I was practicing for my final and stumbled on this problem.
Let $f_n$:$[0,1]$ $mapsto$ $R$ such that
$f(x)_n$ =
begin{cases}
nx, & 0leq text{ $x$ } lt1/n\
1, & 1/nleqtext{ $x$ }leq 1
end{cases}
I could find its piecewise limit, which is
$f(x)$ =
begin{cases}
0, & text{ $x$ =0} \
1, & text{ $otherwise$ }
end{cases}
I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$
then I realized there were more than one interval of both $f_n$ and $f$.
So I used $x$ to indicate the supremum of each interval
for $x=0$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $0$
for $1/nleqtext{ $x$ }leq 1$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $1$
Now I am stuck here
for $0lt text{ $x$ } lt1/n$
What should i do with this interval?
And does $f(x)_n$ converges uniformly?
real-analysis uniform-convergence
$endgroup$
add a comment |
$begingroup$
I was practicing for my final and stumbled on this problem.
Let $f_n$:$[0,1]$ $mapsto$ $R$ such that
$f(x)_n$ =
begin{cases}
nx, & 0leq text{ $x$ } lt1/n\
1, & 1/nleqtext{ $x$ }leq 1
end{cases}
I could find its piecewise limit, which is
$f(x)$ =
begin{cases}
0, & text{ $x$ =0} \
1, & text{ $otherwise$ }
end{cases}
I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$
then I realized there were more than one interval of both $f_n$ and $f$.
So I used $x$ to indicate the supremum of each interval
for $x=0$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $0$
for $1/nleqtext{ $x$ }leq 1$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $1$
Now I am stuck here
for $0lt text{ $x$ } lt1/n$
What should i do with this interval?
And does $f(x)_n$ converges uniformly?
real-analysis uniform-convergence
$endgroup$
I was practicing for my final and stumbled on this problem.
Let $f_n$:$[0,1]$ $mapsto$ $R$ such that
$f(x)_n$ =
begin{cases}
nx, & 0leq text{ $x$ } lt1/n\
1, & 1/nleqtext{ $x$ }leq 1
end{cases}
I could find its piecewise limit, which is
$f(x)$ =
begin{cases}
0, & text{ $x$ =0} \
1, & text{ $otherwise$ }
end{cases}
I was thinking about using the $sup$$|f(x)_n-f(x)|<epsilon$
then I realized there were more than one interval of both $f_n$ and $f$.
So I used $x$ to indicate the supremum of each interval
for $x=0$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $0$
for $1/nleqtext{ $x$ }leq 1$
Clearly $sup$$|f(x)_n-f(x)|=0<epsilon$
Since both function yield $1$
Now I am stuck here
for $0lt text{ $x$ } lt1/n$
What should i do with this interval?
And does $f(x)_n$ converges uniformly?
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Dec 9 '18 at 17:21
AtchanaphongAtchanaphong
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.
$endgroup$
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
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One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
add a comment |
$begingroup$
The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.
$endgroup$
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.
$endgroup$
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
add a comment |
$begingroup$
If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.
$endgroup$
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
add a comment |
$begingroup$
If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.
$endgroup$
If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.
answered Dec 9 '18 at 17:27
GenericMathematicianGenericMathematician
863
863
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
add a comment |
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
Oh! I totally forgot about that! Thank you, sir!
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:31
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
@Atchanaphong You're welcome!
$endgroup$
– GenericMathematician
Dec 9 '18 at 17:33
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
One more question, if you would answer. On that subinterval $(0,ϵ/n^2)$ I find that its supremum of difference would be $1$ since I can choose $x$ to be as close to $0$ as possible, no matter how large $n$ is, so $nx$ goes to $0$. Do I understand that correctly?
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:43
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
$begingroup$
@Atchanaphong Yup!
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:11
add a comment |
$begingroup$
The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.
$endgroup$
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
add a comment |
$begingroup$
The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.
$endgroup$
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
add a comment |
$begingroup$
The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.
$endgroup$
The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.
answered Dec 9 '18 at 17:24
José Carlos SantosJosé Carlos Santos
162k22128232
162k22128232
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
add a comment |
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
$begingroup$
Thank you. I totally forgot about this since I was focusing on the supremum.
$endgroup$
– Atchanaphong
Dec 9 '18 at 17:33
add a comment |
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