Mixing
x/n uniformly convergent?
$begingroup$
Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?
With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
analysis uniform-convergence
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
$endgroup$
add a comment |
$begingroup$
Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?
With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
analysis uniform-convergence
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
$endgroup$
$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50
add a comment |
$begingroup$
Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?
With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
analysis uniform-convergence
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
$endgroup$
Let $f_n(x) : mathbb{R} rightarrow mathbb{R}$ be defined as $dfrac{x}{n}$. Is $f_n$ uniformly convergent?
With "$(a_n)$ is limited and $(b_n)$ converges to 0 $Rightarrow (a_nb_n)$ converges also to 0" i would guess that $xcdot dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
analysis uniform-convergence
analysis uniform-convergence
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
edited Jan 30 '17 at 13:58
Arjihad
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
asked Jan 30 '17 at 13:43
ArjihadArjihad
390112
390112
$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50
add a comment |
$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50
$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50
$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
$endgroup$
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
add a comment |
$begingroup$
If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
$endgroup$
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
$begingroup$
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
$endgroup$
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
add a comment |
$begingroup$
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
$endgroup$
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
add a comment |
$begingroup$
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
$endgroup$
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
answered Jan 30 '17 at 13:50
b00n heTb00n heT
10.4k12235
10.4k12235
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
add a comment |
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
Is $mathbb{R}$ limited?
$endgroup$
– Arjihad
Jan 30 '17 at 13:59
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
$begingroup$
of course not (if by limited you mean bounded)
$endgroup$
– b00n heT
Jan 30 '17 at 15:01
add a comment |
$begingroup$
If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
$endgroup$
add a comment |
$begingroup$
If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
$endgroup$
add a comment |
$begingroup$
If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
$endgroup$
If you are taking $x in [a, b]$ a fixed, closed sub-interval of $mathbb{R}$, then $|frac{x}{n}| leq |frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x in mathbb{R}$, then given any $epsilon > 0$, there exists $x in mathbb{R}$ such that $|frac{x}{n}| geq epsilon$ (it is enough to choose $x geq n epsilon$). So you cannot have $|frac{x}{n}| leq epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $mathbb{R}$.
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
answered Jan 30 '17 at 13:57
Nicolas PelletierNicolas Pelletier
18116
18116
add a comment |
add a comment |
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$begingroup$
You need to specify an interval for the values of $x$.
$endgroup$
– GaC
Jan 30 '17 at 13:50