Mixing
Compact subset of space of Continuous functions
$begingroup$
Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?
I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.
real-analysis general-topology functional-analysis
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
$endgroup$
add a comment |
$begingroup$
Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?
I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.
real-analysis general-topology functional-analysis
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
$endgroup$
$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
1
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
2
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30
add a comment |
$begingroup$
Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?
I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.
real-analysis general-topology functional-analysis
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
$endgroup$
Let $mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K={finmathscr{C}[0,1]|int_0^1f(t)dt=1}$. Then is $K$ compact in the space $mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?
I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.
real-analysis general-topology functional-analysis
real-analysis general-topology functional-analysis
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
asked Jan 8 at 11:33
vidyarthividyarthi
2,9421832
2,9421832
$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
1
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
2
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30
add a comment |
$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
1
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
2
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30
$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
1
1
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
2
2
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$
Hence $K$ isn't bounded so it cannot be compact.
An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.
If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
$endgroup$
add a comment |
$begingroup$
Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?
answered Jan 8 at 11:42
FredFred
45.3k1847
$endgroup$
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$
Hence $K$ isn't bounded so it cannot be compact.
An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.
If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
$endgroup$
add a comment |
$begingroup$
Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$
Hence $K$ isn't bounded so it cannot be compact.
An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.
If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
$endgroup$
add a comment |
$begingroup$
Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$
Hence $K$ isn't bounded so it cannot be compact.
An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.
If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
$endgroup$
Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+nsin(2pi x)$.
We have $(f_n)_n subseteq K$ but $$|f_n|_infty ge f_nleft(frac14right) = 1+nsinleft(fracpi2right) = 1+n $$
Hence $K$ isn't bounded so it cannot be compact.
An alternative argument: define a linear functional $phi : C[0,1] to mathbb{R}$ as $phi(f) = int_0^{1/2}f(t),dt$. We have that $phi$ is bounded and hence continuous with respect to the supremum norm.
If $K$ were compact, $phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have
$$phi(f_n) = int_0^{1/2}f_n(t),dt = frac12 + frac{n}pi$$
which is a contradiction.
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
answered Jan 8 at 14:02
mechanodroidmechanodroid
27.4k62447
27.4k62447
add a comment |
add a comment |
$begingroup$
Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?
answered Jan 8 at 11:42
FredFred
45.3k1847
$endgroup$
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
add a comment |
$begingroup$
Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?
answered Jan 8 at 11:42
FredFred
45.3k1847
$endgroup$
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
add a comment |
$begingroup$
Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?
answered Jan 8 at 11:42
FredFred
45.3k1847
$endgroup$
Let $f_n(t)=(n+1)t^n$. Then $f_n in K$ for all $n$. Can you proceed ?
answered Jan 8 at 11:42
FredFred
45.3k1847
answered Jan 8 at 11:42
FredFred
45.3k1847
answered Jan 8 at 11:42
FredFred
45.3k1847
answered Jan 8 at 11:42
FredFred
45.3k1847
45.3k1847
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
add a comment |
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
$begingroup$
The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $lim_{ntoinfty}int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way?
$endgroup$
– vidyarthi
Jan 8 at 11:47
2
2
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
$begingroup$
@vidyarthi No, just note that $|f_n|_{text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact.
$endgroup$
– Henno Brandsma
Jan 8 at 13:01
add a comment |
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$begingroup$
Have you heard of Arzela-Ascoli theorem?
$endgroup$
– Thomas Shelby
Jan 8 at 11:34
$begingroup$
Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work.
$endgroup$
– Giuseppe Negro
Jan 8 at 11:39
$begingroup$
@ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right
$endgroup$
– vidyarthi
Jan 8 at 11:39
1
$begingroup$
@ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer.
$endgroup$
– daw
Jan 8 at 12:56
2
$begingroup$
@ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded).
$endgroup$
– MaoWao
Jan 8 at 13:30