Mixing
Can we deduce that $f(A)⊂A$
$begingroup$
Let $(f_{n})_{n≥1}$ be a sequence of continuous functions verifying $$f_{n}(A)⊆A$$ for all $n≥1$. Here $A$ is a compact set. Assuming that this sequence converges uniformly to a continuous function $f$
My question is: Can we deduce that $$f(A)⊆A$$
If no, Is there some conditions on the functions $(f_{n})_{n≥1}$ in which we can get the inclusion for the limiting function $f$.
real-analysis uniform-convergence
asked Feb 3 at 10:30
ChinaChina
1,4191029
$endgroup$
add a comment |
$begingroup$
Let $(f_{n})_{n≥1}$ be a sequence of continuous functions verifying $$f_{n}(A)⊆A$$ for all $n≥1$. Here $A$ is a compact set. Assuming that this sequence converges uniformly to a continuous function $f$
My question is: Can we deduce that $$f(A)⊆A$$
If no, Is there some conditions on the functions $(f_{n})_{n≥1}$ in which we can get the inclusion for the limiting function $f$.
real-analysis uniform-convergence
asked Feb 3 at 10:30
ChinaChina
1,4191029
$endgroup$
$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
1
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35
add a comment |
$begingroup$
Let $(f_{n})_{n≥1}$ be a sequence of continuous functions verifying $$f_{n}(A)⊆A$$ for all $n≥1$. Here $A$ is a compact set. Assuming that this sequence converges uniformly to a continuous function $f$
My question is: Can we deduce that $$f(A)⊆A$$
If no, Is there some conditions on the functions $(f_{n})_{n≥1}$ in which we can get the inclusion for the limiting function $f$.
real-analysis uniform-convergence
asked Feb 3 at 10:30
ChinaChina
1,4191029
$endgroup$
Let $(f_{n})_{n≥1}$ be a sequence of continuous functions verifying $$f_{n}(A)⊆A$$ for all $n≥1$. Here $A$ is a compact set. Assuming that this sequence converges uniformly to a continuous function $f$
My question is: Can we deduce that $$f(A)⊆A$$
If no, Is there some conditions on the functions $(f_{n})_{n≥1}$ in which we can get the inclusion for the limiting function $f$.
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Feb 3 at 10:30
ChinaChina
1,4191029
asked Feb 3 at 10:30
ChinaChina
1,4191029
edited Feb 3 at 10:42
China
asked Feb 3 at 10:30
ChinaChina
1,4191029
asked Feb 3 at 10:30
ChinaChina
1,4191029
asked Feb 3 at 10:30
ChinaChina
1,4191029
1,4191029
$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
1
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35
add a comment |
$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
1
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35
$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
1
1
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $A$ is compact then $A$ is closed. Thus ${f_n(x)}$ is a sequence in $A$ that converges to $f(x)$. Since $A$ is closed, $f(x) in A$ for all $x in A$. Thus, $f(A) subseteq A$.
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
$endgroup$
add a comment |
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$begingroup$
If $A$ is compact then $A$ is closed. Thus ${f_n(x)}$ is a sequence in $A$ that converges to $f(x)$. Since $A$ is closed, $f(x) in A$ for all $x in A$. Thus, $f(A) subseteq A$.
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
$endgroup$
add a comment |
$begingroup$
If $A$ is compact then $A$ is closed. Thus ${f_n(x)}$ is a sequence in $A$ that converges to $f(x)$. Since $A$ is closed, $f(x) in A$ for all $x in A$. Thus, $f(A) subseteq A$.
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
$endgroup$
add a comment |
$begingroup$
If $A$ is compact then $A$ is closed. Thus ${f_n(x)}$ is a sequence in $A$ that converges to $f(x)$. Since $A$ is closed, $f(x) in A$ for all $x in A$. Thus, $f(A) subseteq A$.
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
$endgroup$
If $A$ is compact then $A$ is closed. Thus ${f_n(x)}$ is a sequence in $A$ that converges to $f(x)$. Since $A$ is closed, $f(x) in A$ for all $x in A$. Thus, $f(A) subseteq A$.
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
answered Feb 3 at 10:55
Anthony TerAnthony Ter
39616
39616
add a comment |
add a comment |
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$begingroup$
Does $subset$ mean $subsetneq $ or $subseteq$?
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:33
$begingroup$
@HagenvonEitzen: It is a strict inclusion without equality.
$endgroup$
– China
Feb 3 at 10:34
1
$begingroup$
Then consider $A=[-1,1]$ and $f_n(x)=(1-frac1n)x$
$endgroup$
– Hagen von Eitzen
Feb 3 at 10:35