Determine whether the convergence is uniform or almost uniform
Let $f_n(x) = frac{nx}{n^2x^2 + 1}$.
For each domain X, given below, how do you determine whether this sequence converges pointwise.
Then if it does is it possible find the limit function and determine whether the convergence is uniform or almost uniform.
X = [0,∞)
X = [1,∞)
convergence uniform-convergence pointwise-convergence
edited Nov 18 at 14:26
user328442
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
add a comment |
Let $f_n(x) = frac{nx}{n^2x^2 + 1}$.
For each domain X, given below, how do you determine whether this sequence converges pointwise.
Then if it does is it possible find the limit function and determine whether the convergence is uniform or almost uniform.
X = [0,∞)
X = [1,∞)
convergence uniform-convergence pointwise-convergence
edited Nov 18 at 14:26
user328442
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37
add a comment |
Let $f_n(x) = frac{nx}{n^2x^2 + 1}$.
For each domain X, given below, how do you determine whether this sequence converges pointwise.
Then if it does is it possible find the limit function and determine whether the convergence is uniform or almost uniform.
X = [0,∞)
X = [1,∞)
convergence uniform-convergence pointwise-convergence
edited Nov 18 at 14:26
user328442
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
Let $f_n(x) = frac{nx}{n^2x^2 + 1}$.
For each domain X, given below, how do you determine whether this sequence converges pointwise.
Then if it does is it possible find the limit function and determine whether the convergence is uniform or almost uniform.
X = [0,∞)
X = [1,∞)
convergence uniform-convergence pointwise-convergence
convergence uniform-convergence pointwise-convergence
edited Nov 18 at 14:26
user328442
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
edited Nov 18 at 14:26
user328442
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
edited Nov 18 at 14:26
user328442
1,7981516
edited Nov 18 at 14:26
user328442
1,7981516
edited Nov 18 at 14:26
user328442
1,7981516
1,7981516
asked Nov 18 at 14:21
Johnmallu
266
asked Nov 18 at 14:21
Johnmallu
266
asked Nov 18 at 14:21
Johnmallu
266
266
it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37
add a comment |
it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37
it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37
add a comment |
1 Answer
1
active
oldest
votes
For $xin[1,infty)$ we have
$$|f_n(x)|le frac1n$$
and the convergence is uniform on $[1,infty)$.
For $xin[0,1]$, there exists an $epsilon>0$ ($epsilon=1/2$) and a number $xin[0,1]$ ( $x=1/n$)
$$|f_n(1/n)|=frac12=epsilon$$
and the convergence fails to be uniform.
But on any compact subset of $(0,1]$ the convergence is uniform. In particular, for all $delta>0$, the convergence is uniform on $[delta,1]$. So, $f_n(x)$ is almost uniformly convergent on $[0,1]$.
answered Nov 18 at 14:52
Mark Viola
130k1274170
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
For $xin[1,infty)$ we have
$$|f_n(x)|le frac1n$$
and the convergence is uniform on $[1,infty)$.
For $xin[0,1]$, there exists an $epsilon>0$ ($epsilon=1/2$) and a number $xin[0,1]$ ( $x=1/n$)
$$|f_n(1/n)|=frac12=epsilon$$
and the convergence fails to be uniform.
But on any compact subset of $(0,1]$ the convergence is uniform. In particular, for all $delta>0$, the convergence is uniform on $[delta,1]$. So, $f_n(x)$ is almost uniformly convergent on $[0,1]$.
answered Nov 18 at 14:52
Mark Viola
130k1274170
add a comment |
For $xin[1,infty)$ we have
$$|f_n(x)|le frac1n$$
and the convergence is uniform on $[1,infty)$.
For $xin[0,1]$, there exists an $epsilon>0$ ($epsilon=1/2$) and a number $xin[0,1]$ ( $x=1/n$)
$$|f_n(1/n)|=frac12=epsilon$$
and the convergence fails to be uniform.
But on any compact subset of $(0,1]$ the convergence is uniform. In particular, for all $delta>0$, the convergence is uniform on $[delta,1]$. So, $f_n(x)$ is almost uniformly convergent on $[0,1]$.
answered Nov 18 at 14:52
Mark Viola
130k1274170
add a comment |
For $xin[1,infty)$ we have
$$|f_n(x)|le frac1n$$
and the convergence is uniform on $[1,infty)$.
For $xin[0,1]$, there exists an $epsilon>0$ ($epsilon=1/2$) and a number $xin[0,1]$ ( $x=1/n$)
$$|f_n(1/n)|=frac12=epsilon$$
and the convergence fails to be uniform.
But on any compact subset of $(0,1]$ the convergence is uniform. In particular, for all $delta>0$, the convergence is uniform on $[delta,1]$. So, $f_n(x)$ is almost uniformly convergent on $[0,1]$.
answered Nov 18 at 14:52
Mark Viola
130k1274170
For $xin[1,infty)$ we have
$$|f_n(x)|le frac1n$$
and the convergence is uniform on $[1,infty)$.
For $xin[0,1]$, there exists an $epsilon>0$ ($epsilon=1/2$) and a number $xin[0,1]$ ( $x=1/n$)
$$|f_n(1/n)|=frac12=epsilon$$
and the convergence fails to be uniform.
But on any compact subset of $(0,1]$ the convergence is uniform. In particular, for all $delta>0$, the convergence is uniform on $[delta,1]$. So, $f_n(x)$ is almost uniformly convergent on $[0,1]$.
answered Nov 18 at 14:52
Mark Viola
130k1274170
edited Nov 18 at 15:03
answered Nov 18 at 14:52
Mark Viola
130k1274170
answered Nov 18 at 14:52
Mark Viola
130k1274170
answered Nov 18 at 14:52
Mark Viola
130k1274170
130k1274170
add a comment |
add a comment |
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it converges pointwise to 0. You just fix x and look at the limit as n goes to infinity.
– mathnoob
Nov 18 at 14:30
For each domain it converges point wise to 0? And I can understand the limit suggestion but after this how do you interpret if it's uniform convergence or almost uniform? @mathnoob
– Johnmallu
Nov 18 at 14:37