Mixing
a question about Differential quotient operator in sobolev spaces
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Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
We use $|cdot|$ to denote the norm in $L^2(R^n)$.
We have the following lemma:
Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.
Then we will have the following two statements:
- From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.
My question is:
Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.
I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
$$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
But I can't figure out how to prove this.
functional-analysis pde sobolev-spaces
asked 7 hours ago
chloe hj
414
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Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
We use $|cdot|$ to denote the norm in $L^2(R^n)$.
We have the following lemma:
Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.
Then we will have the following two statements:
- From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.
My question is:
Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.
I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
$$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
But I can't figure out how to prove this.
functional-analysis pde sobolev-spaces
asked 7 hours ago
chloe hj
414
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
We use $|cdot|$ to denote the norm in $L^2(R^n)$.
We have the following lemma:
Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.
Then we will have the following two statements:
- From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.
My question is:
Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.
I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
$$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
But I can't figure out how to prove this.
functional-analysis pde sobolev-spaces
asked 7 hours ago
chloe hj
414
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
We use $|cdot|$ to denote the norm in $L^2(R^n)$.
We have the following lemma:
Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.
Then we will have the following two statements:
- From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.
My question is:
Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.
I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
$$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
But I can't figure out how to prove this.
functional-analysis pde sobolev-spaces
functional-analysis pde sobolev-spaces
asked 7 hours ago
chloe hj
414
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
chloe hj
414
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 6 hours ago
asked 7 hours ago
chloe hj
414
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 7 hours ago
chloe hj
414
asked 7 hours ago
chloe hj
414
414
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1 Answer
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Regarding the first question:
$$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $hto0$.
answered 6 hours ago
maxmilgram
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Regarding the first question:
$$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $hto0$.
answered 6 hours ago
maxmilgram
3927
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maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
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Regarding the first question:
$$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $hto0$.
answered 6 hours ago
maxmilgram
3927
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
up vote
0
down vote
Regarding the first question:
$$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $hto0$.
answered 6 hours ago
maxmilgram
3927
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Regarding the first question:
$$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $hto0$.
answered 6 hours ago
maxmilgram
3927
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
maxmilgram
3927
New contributor
maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 6 hours ago
maxmilgram
3927
answered 6 hours ago
maxmilgram
3927
3927
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New contributor
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chloe hj is a new contributor. Be nice, and check out our Code of Conduct.
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