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Prove $A_m$ is closed in $C[0,1]$
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Let $minmathbbN$ arbitrary and $A_m subset C[0,1]$, the subset of all function $f$ verify: Exists $x_0in[0,1-frac1m]$ such that: $fracf(x_0+h-f(x_0)hleq m forall hin(0,frac1m$
Prove $A_m$ is closed in $C[0,1]$
My idea is:
Prove $A_m$ is bounded uniformly and equicontinuous then by Ascoli theorem $A_m$ be compact.
This implies $A_m$ is closed set.
But i don't know how to prove this, can someone help me?
general-topology
asked Jul 15 at 20:51
Bvss12
1,609516
add a comment |Â
up vote
0
down vote
favorite
Let $minmathbbN$ arbitrary and $A_m subset C[0,1]$, the subset of all function $f$ verify: Exists $x_0in[0,1-frac1m]$ such that: $fracf(x_0+h-f(x_0)hleq m forall hin(0,frac1m$
Prove $A_m$ is closed in $C[0,1]$
My idea is:
Prove $A_m$ is bounded uniformly and equicontinuous then by Ascoli theorem $A_m$ be compact.
This implies $A_m$ is closed set.
But i don't know how to prove this, can someone help me?
general-topology
asked Jul 15 at 20:51
Bvss12
1,609516
Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $minmathbbN$ arbitrary and $A_m subset C[0,1]$, the subset of all function $f$ verify: Exists $x_0in[0,1-frac1m]$ such that: $fracf(x_0+h-f(x_0)hleq m forall hin(0,frac1m$
Prove $A_m$ is closed in $C[0,1]$
My idea is:
Prove $A_m$ is bounded uniformly and equicontinuous then by Ascoli theorem $A_m$ be compact.
This implies $A_m$ is closed set.
But i don't know how to prove this, can someone help me?
general-topology
asked Jul 15 at 20:51
Bvss12
1,609516
Let $minmathbbN$ arbitrary and $A_m subset C[0,1]$, the subset of all function $f$ verify: Exists $x_0in[0,1-frac1m]$ such that: $fracf(x_0+h-f(x_0)hleq m forall hin(0,frac1m$
Prove $A_m$ is closed in $C[0,1]$
My idea is:
Prove $A_m$ is bounded uniformly and equicontinuous then by Ascoli theorem $A_m$ be compact.
This implies $A_m$ is closed set.
But i don't know how to prove this, can someone help me?
general-topology
asked Jul 15 at 20:51
Bvss12
1,609516
asked Jul 15 at 20:51
Bvss12
1,609516
asked Jul 15 at 20:51
Bvss12
1,609516
asked Jul 15 at 20:51
Bvss12
1,609516
1,609516
Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08
add a comment |Â
Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08
Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08
add a comment |Â
1 Answer
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Fix $h$. $fto inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence ${fin C[0,1]:inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Fix $h$. $fto inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence ${fin C[0,1]:inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
Fix $h$. $fto inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence ${fin C[0,1]:inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Fix $h$. $fto inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence ${fin C[0,1]:inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
Fix $h$. $fto inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence ${fin C[0,1]:inf_xin [0,1-frac 1 m] frac f(x+h)-f(x) h leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
answered Jul 15 at 23:40
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
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Do you mean that $A_m$ is the set of all functions in $mathcalC[0, 1]$ which satisfy the condition "exists $x_0 in [0, 1-frac1m]$ such that..." ?
– Adayah
Jul 15 at 22:21
Yes @Adayah is that
– Bvss12
Jul 15 at 22:24
Given the attempt you used on this question, one remark I think relevant is: the set $A_m$ is not uniformly bounded. To see this, for each $fin A_m$, and each $cinmathbbR$, $f+cin A_m$. Thus, $A_m$ is not uniformly bounded (nor is it compact, either). Moreover, I'd believe its rather extraordinary to find a family of functions in $C[0,1]$ that is equicontinuous and uniformly bounded. So, more often than not, I'd asume this is not the case.
– Nate River
Jul 16 at 13:08