Mixing
Is $i: (C^1, ||·||_{W^{1,2}}) → (C^0, ||·||_∞)$ a linear, continuous, compact map?
$begingroup$
Consider the map
$$i: (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$
which maps every function to itself, and with Sobolev norm defined as
$$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$
Is $i$ linear, continuous, compact?
Linearity
Consider $u,vin C^1[0,1]$ and $a,bin mathbb{R}$:
$i(au+bv)=au+bv=ai(u)+bi(v)$
Continuity
By fundamental theorem of calculus: $u(x)=u(0)+int_0^x u'(t)dt$. Then:
$begin{align*}
|u(x)| &le |u(0)|+int_0^x |u'(t)|dt \
&le |u(0)|+int_0^1 |u'(t)|dt \
&le C|u(0)|^2+Cint_0^1 |u'(t)|^2dt \
&le Cint_0^1 |u(t)|^2dt+Cint_0^1 |u'(t)|^2dt \
&le C(||u||_{L^2}+||u'||_{L^2}) \
&=C||u||_{W^{1,2}}
end{align*}$
for $C$ large enough and by mean value theorem.
Since $i$ is linear and bounded, it is also continuous.
Compactness
$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.
Are the computations for linearity and continuity correct?
How to check the compactness?
functional-analysis operator-theory
asked Jan 21 at 14:52
sound wavesound wave
28819
$endgroup$
|
show 4 more comments
$begingroup$
Consider the map
$$i: (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$
which maps every function to itself, and with Sobolev norm defined as
$$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$
Is $i$ linear, continuous, compact?
Linearity
Consider $u,vin C^1[0,1]$ and $a,bin mathbb{R}$:
$i(au+bv)=au+bv=ai(u)+bi(v)$
Continuity
By fundamental theorem of calculus: $u(x)=u(0)+int_0^x u'(t)dt$. Then:
$begin{align*}
|u(x)| &le |u(0)|+int_0^x |u'(t)|dt \
&le |u(0)|+int_0^1 |u'(t)|dt \
&le C|u(0)|^2+Cint_0^1 |u'(t)|^2dt \
&le Cint_0^1 |u(t)|^2dt+Cint_0^1 |u'(t)|^2dt \
&le C(||u||_{L^2}+||u'||_{L^2}) \
&=C||u||_{W^{1,2}}
end{align*}$
for $C$ large enough and by mean value theorem.
Since $i$ is linear and bounded, it is also continuous.
Compactness
$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.
Are the computations for linearity and continuity correct?
How to check the compactness?
functional-analysis operator-theory
asked Jan 21 at 14:52
sound wavesound wave
28819
$endgroup$
2
$begingroup$
Do you know the Ascoli theorem?
$endgroup$
– Mindlack
Jan 21 at 14:59
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
$endgroup$
– sound wave
Jan 21 at 15:11
2
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
2
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
2
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59
|
show 4 more comments
$begingroup$
Consider the map
$$i: (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$
which maps every function to itself, and with Sobolev norm defined as
$$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$
Is $i$ linear, continuous, compact?
Linearity
Consider $u,vin C^1[0,1]$ and $a,bin mathbb{R}$:
$i(au+bv)=au+bv=ai(u)+bi(v)$
Continuity
By fundamental theorem of calculus: $u(x)=u(0)+int_0^x u'(t)dt$. Then:
$begin{align*}
|u(x)| &le |u(0)|+int_0^x |u'(t)|dt \
&le |u(0)|+int_0^1 |u'(t)|dt \
&le C|u(0)|^2+Cint_0^1 |u'(t)|^2dt \
&le Cint_0^1 |u(t)|^2dt+Cint_0^1 |u'(t)|^2dt \
&le C(||u||_{L^2}+||u'||_{L^2}) \
&=C||u||_{W^{1,2}}
end{align*}$
for $C$ large enough and by mean value theorem.
Since $i$ is linear and bounded, it is also continuous.
Compactness
$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.
Are the computations for linearity and continuity correct?
How to check the compactness?
functional-analysis operator-theory
asked Jan 21 at 14:52
sound wavesound wave
28819
$endgroup$
Consider the map
$$i: (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$
which maps every function to itself, and with Sobolev norm defined as
$$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$
Is $i$ linear, continuous, compact?
Linearity
Consider $u,vin C^1[0,1]$ and $a,bin mathbb{R}$:
$i(au+bv)=au+bv=ai(u)+bi(v)$
Continuity
By fundamental theorem of calculus: $u(x)=u(0)+int_0^x u'(t)dt$. Then:
$begin{align*}
|u(x)| &le |u(0)|+int_0^x |u'(t)|dt \
&le |u(0)|+int_0^1 |u'(t)|dt \
&le C|u(0)|^2+Cint_0^1 |u'(t)|^2dt \
&le Cint_0^1 |u(t)|^2dt+Cint_0^1 |u'(t)|^2dt \
&le C(||u||_{L^2}+||u'||_{L^2}) \
&=C||u||_{W^{1,2}}
end{align*}$
for $C$ large enough and by mean value theorem.
Since $i$ is linear and bounded, it is also continuous.
Compactness
$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.
Are the computations for linearity and continuity correct?
How to check the compactness?
functional-analysis operator-theory
functional-analysis operator-theory
asked Jan 21 at 14:52
sound wavesound wave
28819
asked Jan 21 at 14:52
sound wavesound wave
28819
asked Jan 21 at 14:52
sound wavesound wave
28819
asked Jan 21 at 14:52
sound wavesound wave
28819
asked Jan 21 at 14:52
sound wavesound wave
28819
28819
2
$begingroup$
Do you know the Ascoli theorem?
$endgroup$
– Mindlack
Jan 21 at 14:59
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
$endgroup$
– sound wave
Jan 21 at 15:11
2
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
2
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
2
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59
|
show 4 more comments
2
$begingroup$
Do you know the Ascoli theorem?
$endgroup$
– Mindlack
Jan 21 at 14:59
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
$endgroup$
– sound wave
Jan 21 at 15:11
2
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
2
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
2
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59
2
2
$begingroup$
Do you know the Ascoli theorem?
$endgroup$
– Mindlack
Jan 21 at 14:59
$begingroup$
Do you know the Ascoli theorem?
$endgroup$
– Mindlack
Jan 21 at 14:59
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
$endgroup$
– sound wave
Jan 21 at 15:11
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
$endgroup$
– sound wave
Jan 21 at 15:11
2
2
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
2
2
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
2
2
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b in [0,1]$, and $f in G $. Then we have $ |f(a)-f(b)| = |int_a^b f' | = |int_a^b (f' times 1) | le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} le ||f'||_{L^2}|a-b|^{1/2} le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_infty le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $rin [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $xin [0,1]$ we have $|u(x)| = |int^x_r u'(t)dt +u(r)| le |int^x_r u'(t)dt| +|u(r)| = |int^x_r u'(t) times 1 dt| +|int_0^1 |u(r)| times 1 dt| le ||u'||_{L^2} + |int_0^1 |u(t)| times 1 dt|le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
answered Jan 21 at 16:39
MaksimMaksim
60718
$endgroup$
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
1
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
1
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b in [0,1]$, and $f in G $. Then we have $ |f(a)-f(b)| = |int_a^b f' | = |int_a^b (f' times 1) | le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} le ||f'||_{L^2}|a-b|^{1/2} le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_infty le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $rin [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $xin [0,1]$ we have $|u(x)| = |int^x_r u'(t)dt +u(r)| le |int^x_r u'(t)dt| +|u(r)| = |int^x_r u'(t) times 1 dt| +|int_0^1 |u(r)| times 1 dt| le ||u'||_{L^2} + |int_0^1 |u(t)| times 1 dt|le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
answered Jan 21 at 16:39
MaksimMaksim
60718
$endgroup$
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
1
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
1
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
|
show 3 more comments
$begingroup$
For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b in [0,1]$, and $f in G $. Then we have $ |f(a)-f(b)| = |int_a^b f' | = |int_a^b (f' times 1) | le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} le ||f'||_{L^2}|a-b|^{1/2} le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_infty le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $rin [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $xin [0,1]$ we have $|u(x)| = |int^x_r u'(t)dt +u(r)| le |int^x_r u'(t)dt| +|u(r)| = |int^x_r u'(t) times 1 dt| +|int_0^1 |u(r)| times 1 dt| le ||u'||_{L^2} + |int_0^1 |u(t)| times 1 dt|le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
answered Jan 21 at 16:39
MaksimMaksim
60718
$endgroup$
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
1
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
1
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
|
show 3 more comments
$begingroup$
For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b in [0,1]$, and $f in G $. Then we have $ |f(a)-f(b)| = |int_a^b f' | = |int_a^b (f' times 1) | le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} le ||f'||_{L^2}|a-b|^{1/2} le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_infty le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $rin [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $xin [0,1]$ we have $|u(x)| = |int^x_r u'(t)dt +u(r)| le |int^x_r u'(t)dt| +|u(r)| = |int^x_r u'(t) times 1 dt| +|int_0^1 |u(r)| times 1 dt| le ||u'||_{L^2} + |int_0^1 |u(t)| times 1 dt|le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
answered Jan 21 at 16:39
MaksimMaksim
60718
$endgroup$
For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b in [0,1]$, and $f in G $. Then we have $ |f(a)-f(b)| = |int_a^b f' | = |int_a^b (f' times 1) | le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} le ||f'||_{L^2}|a-b|^{1/2} le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_infty le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $rin [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $xin [0,1]$ we have $|u(x)| = |int^x_r u'(t)dt +u(r)| le |int^x_r u'(t)dt| +|u(r)| = |int^x_r u'(t) times 1 dt| +|int_0^1 |u(r)| times 1 dt| le ||u'||_{L^2} + |int_0^1 |u(t)| times 1 dt|le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
answered Jan 21 at 16:39
MaksimMaksim
60718
edited Jan 21 at 17:58
answered Jan 21 at 16:39
MaksimMaksim
60718
answered Jan 21 at 16:39
MaksimMaksim
60718
answered Jan 21 at 16:39
MaksimMaksim
60718
60718
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Thank you very much! Could you also say if the computation about continuity is correct?
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– sound wave
Jan 21 at 17:13
1
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Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
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– Maksim
Jan 21 at 17:17
1
$begingroup$
See edit above.
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– Maksim
Jan 21 at 17:53
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
|
show 3 more comments
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
1
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
1
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
$begingroup$
Thank you very much! Could you also say if the computation about continuity is correct?
$endgroup$
– sound wave
Jan 21 at 17:13
1
1
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
$begingroup$
Linearity is Ok, but the way you show the continuity seems not ok. E.g. the constant $C$ is dependant on the choice of $u$.
$endgroup$
– Maksim
Jan 21 at 17:17
1
1
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
$begingroup$
See edit above.
$endgroup$
– Maksim
Jan 21 at 17:53
1
1
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
$begingroup$
That is needed for the inequality $|int_0^1 |u(r)| times 1 dt| le |int_0^1 |u(t)| times 1 dt|$. Note that the first integrand is constant but the second is not.
$endgroup$
– Maksim
Jan 21 at 18:05
1
1
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
$begingroup$
$|u(r)| =|u(r)| times (1-0) = |u(r)|int_0^1 dt= int_0^1 |u(r)| dt =|int_0^1 |u(r)| dt|$
$endgroup$
– Maksim
Jan 28 at 10:49
|
show 3 more comments
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2
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Do you know the Ascoli theorem?
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– Mindlack
Jan 21 at 14:59
$begingroup$
It says that if a sequence $u_n$ of continuous functions is equicontinuous, then $u_n$ has a subsequence which converge uniformly. Right?
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– sound wave
Jan 21 at 15:11
2
$begingroup$
The sequence must also be bounded, but right. So prove the unit ball in your origin space is bounded in $C^0$ and equicontinuous.
$endgroup$
– Mindlack
Jan 21 at 15:18
2
$begingroup$
The unit ball of $C^0$ for the $C^0$ norm is not. However, a smaller set (such as the unit ball of $C^1$ for the $H^1$ norm) could be compact for the $C^0$ norm.
$endgroup$
– Mindlack
Jan 21 at 15:40
2
$begingroup$
Could, as in : not forbidden by Riesz theorem.
$endgroup$
– Mindlack
Jan 21 at 15:59