Mixing
Continuous map between $L^p$ spaces
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The following theorem appears in Appendix B of Rabinowitz' book Minimax Methods in Critical Point Theory:
Let $Omega subset Bbb{R}^n$ be bounded and $gin C(overline{Omega}times Bbb {R},Bbb {R})$ such that there exist constants $r,sge 1$ and $a_1,a_2ge 0$ such that for all $x in overline{Omega}, yin Bbb{R}$ $$|g(x,y)|le a_1 + a_2|y|^{r/s}$$
Then the map $varphi(x)mapsto g(x,varphi(x))$ belongs to $C(L^r(Omega),L^s(Omega))$.
In the proof, he says "To prove the continuity of this map, observe that it is continuous at $varphi$ if and only if $f(x,z(x)) = g(x,z(x)+varphi(x))-g(x,varphi(x))$ is continuous at $z=0$. Therefore we can assume $varphi = 0$ and $g(x,0)=0$."
I don't understand how this assumpion can be made without loss of generality, and was unable to finish the proof without it. Any help would be appreciated.
Edit: I have found a partial answer in a different thread:
Continuity proof of a function between $L^p$ spaces
In the post it says:
Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x))|leq A_1+A_2 |phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
This would solve my problem, since the former two constants don't depend on $z$ and can be thrown together, leaving the case that was already proven. However, I couldn't derive this estimate.
edit2: I was wrong in assuming this solves the problem since, as pointed out by the users supinf and Peter Melech, the constant may not depend on x.
real-analysis functional-analysis analysis lp-spaces
asked yesterday
J. Snow
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J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
6
down vote
favorite
The following theorem appears in Appendix B of Rabinowitz' book Minimax Methods in Critical Point Theory:
Let $Omega subset Bbb{R}^n$ be bounded and $gin C(overline{Omega}times Bbb {R},Bbb {R})$ such that there exist constants $r,sge 1$ and $a_1,a_2ge 0$ such that for all $x in overline{Omega}, yin Bbb{R}$ $$|g(x,y)|le a_1 + a_2|y|^{r/s}$$
Then the map $varphi(x)mapsto g(x,varphi(x))$ belongs to $C(L^r(Omega),L^s(Omega))$.
In the proof, he says "To prove the continuity of this map, observe that it is continuous at $varphi$ if and only if $f(x,z(x)) = g(x,z(x)+varphi(x))-g(x,varphi(x))$ is continuous at $z=0$. Therefore we can assume $varphi = 0$ and $g(x,0)=0$."
I don't understand how this assumpion can be made without loss of generality, and was unable to finish the proof without it. Any help would be appreciated.
Edit: I have found a partial answer in a different thread:
Continuity proof of a function between $L^p$ spaces
In the post it says:
Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x))|leq A_1+A_2 |phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
This would solve my problem, since the former two constants don't depend on $z$ and can be thrown together, leaving the case that was already proven. However, I couldn't derive this estimate.
edit2: I was wrong in assuming this solves the problem since, as pointed out by the users supinf and Peter Melech, the constant may not depend on x.
real-analysis functional-analysis analysis lp-spaces
asked yesterday
J. Snow
314
New contributor
J. Snow 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
6
down vote
favorite
up vote
6
down vote
favorite
The following theorem appears in Appendix B of Rabinowitz' book Minimax Methods in Critical Point Theory:
Let $Omega subset Bbb{R}^n$ be bounded and $gin C(overline{Omega}times Bbb {R},Bbb {R})$ such that there exist constants $r,sge 1$ and $a_1,a_2ge 0$ such that for all $x in overline{Omega}, yin Bbb{R}$ $$|g(x,y)|le a_1 + a_2|y|^{r/s}$$
Then the map $varphi(x)mapsto g(x,varphi(x))$ belongs to $C(L^r(Omega),L^s(Omega))$.
In the proof, he says "To prove the continuity of this map, observe that it is continuous at $varphi$ if and only if $f(x,z(x)) = g(x,z(x)+varphi(x))-g(x,varphi(x))$ is continuous at $z=0$. Therefore we can assume $varphi = 0$ and $g(x,0)=0$."
I don't understand how this assumpion can be made without loss of generality, and was unable to finish the proof without it. Any help would be appreciated.
Edit: I have found a partial answer in a different thread:
Continuity proof of a function between $L^p$ spaces
In the post it says:
Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x))|leq A_1+A_2 |phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
This would solve my problem, since the former two constants don't depend on $z$ and can be thrown together, leaving the case that was already proven. However, I couldn't derive this estimate.
edit2: I was wrong in assuming this solves the problem since, as pointed out by the users supinf and Peter Melech, the constant may not depend on x.
real-analysis functional-analysis analysis lp-spaces
asked yesterday
J. Snow
314
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The following theorem appears in Appendix B of Rabinowitz' book Minimax Methods in Critical Point Theory:
Let $Omega subset Bbb{R}^n$ be bounded and $gin C(overline{Omega}times Bbb {R},Bbb {R})$ such that there exist constants $r,sge 1$ and $a_1,a_2ge 0$ such that for all $x in overline{Omega}, yin Bbb{R}$ $$|g(x,y)|le a_1 + a_2|y|^{r/s}$$
Then the map $varphi(x)mapsto g(x,varphi(x))$ belongs to $C(L^r(Omega),L^s(Omega))$.
In the proof, he says "To prove the continuity of this map, observe that it is continuous at $varphi$ if and only if $f(x,z(x)) = g(x,z(x)+varphi(x))-g(x,varphi(x))$ is continuous at $z=0$. Therefore we can assume $varphi = 0$ and $g(x,0)=0$."
I don't understand how this assumpion can be made without loss of generality, and was unable to finish the proof without it. Any help would be appreciated.
Edit: I have found a partial answer in a different thread:
Continuity proof of a function between $L^p$ spaces
In the post it says:
Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x))|leq A_1+A_2 |phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
This would solve my problem, since the former two constants don't depend on $z$ and can be thrown together, leaving the case that was already proven. However, I couldn't derive this estimate.
edit2: I was wrong in assuming this solves the problem since, as pointed out by the users supinf and Peter Melech, the constant may not depend on x.
real-analysis functional-analysis analysis lp-spaces
real-analysis functional-analysis analysis lp-spaces
asked yesterday
J. Snow
314
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
J. Snow
314
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
J. Snow
314
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
J. Snow
314
asked yesterday
J. Snow
314
314
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J. Snow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J. Snow 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 |
add a comment |
1 Answer
1
active
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up vote
4
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For $xinoverline{Omega}$ and $frac{r}{s}geq 1$ You get
$$|f(x,z(x))|=|g(x,z(x)+varphi(x))-g(x,phi(x))|leq a_1+a_2|z(x)+varphi|^{frac{r}{s}}+a_1+a_2|varphi(x)|^{frac{r}{s}}leq$$
$$2a_1+a_22^{frac{r}{s}-1}(|z(x)|^{frac{r}{s}}+|varphi(x)|^{frac{r}{s}})+a_2|varphi(x)|^{frac{r}{s}}leq $$
$$underbrace{2a_1}_{=A_1}+underbrace{(a_22^{frac{r}{s}-1}+a_2)}_{=A_2}|varphi(x)|^{frac{r}{s}}+underbrace{a_22^{frac{r}{s}-1}}_{=A_3}|z(x)|^frac{r}{s}$$
where You used that $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ and Jensen's inequality. In the case $frac{r}{s}<1$ You can proceed analogously using $(a+b)^{frac{r}{s}}leq a^{frac{r}{s}}+b^{frac{r}{s}}$ valid in this range.
answered yesterday
Peter Melech
2,464813
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
For $xinoverline{Omega}$ and $frac{r}{s}geq 1$ You get
$$|f(x,z(x))|=|g(x,z(x)+varphi(x))-g(x,phi(x))|leq a_1+a_2|z(x)+varphi|^{frac{r}{s}}+a_1+a_2|varphi(x)|^{frac{r}{s}}leq$$
$$2a_1+a_22^{frac{r}{s}-1}(|z(x)|^{frac{r}{s}}+|varphi(x)|^{frac{r}{s}})+a_2|varphi(x)|^{frac{r}{s}}leq $$
$$underbrace{2a_1}_{=A_1}+underbrace{(a_22^{frac{r}{s}-1}+a_2)}_{=A_2}|varphi(x)|^{frac{r}{s}}+underbrace{a_22^{frac{r}{s}-1}}_{=A_3}|z(x)|^frac{r}{s}$$
where You used that $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ and Jensen's inequality. In the case $frac{r}{s}<1$ You can proceed analogously using $(a+b)^{frac{r}{s}}leq a^{frac{r}{s}}+b^{frac{r}{s}}$ valid in this range.
answered yesterday
Peter Melech
2,464813
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
add a comment |
up vote
4
down vote
For $xinoverline{Omega}$ and $frac{r}{s}geq 1$ You get
$$|f(x,z(x))|=|g(x,z(x)+varphi(x))-g(x,phi(x))|leq a_1+a_2|z(x)+varphi|^{frac{r}{s}}+a_1+a_2|varphi(x)|^{frac{r}{s}}leq$$
$$2a_1+a_22^{frac{r}{s}-1}(|z(x)|^{frac{r}{s}}+|varphi(x)|^{frac{r}{s}})+a_2|varphi(x)|^{frac{r}{s}}leq $$
$$underbrace{2a_1}_{=A_1}+underbrace{(a_22^{frac{r}{s}-1}+a_2)}_{=A_2}|varphi(x)|^{frac{r}{s}}+underbrace{a_22^{frac{r}{s}-1}}_{=A_3}|z(x)|^frac{r}{s}$$
where You used that $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ and Jensen's inequality. In the case $frac{r}{s}<1$ You can proceed analogously using $(a+b)^{frac{r}{s}}leq a^{frac{r}{s}}+b^{frac{r}{s}}$ valid in this range.
answered yesterday
Peter Melech
2,464813
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
add a comment |
up vote
4
down vote
up vote
4
down vote
For $xinoverline{Omega}$ and $frac{r}{s}geq 1$ You get
$$|f(x,z(x))|=|g(x,z(x)+varphi(x))-g(x,phi(x))|leq a_1+a_2|z(x)+varphi|^{frac{r}{s}}+a_1+a_2|varphi(x)|^{frac{r}{s}}leq$$
$$2a_1+a_22^{frac{r}{s}-1}(|z(x)|^{frac{r}{s}}+|varphi(x)|^{frac{r}{s}})+a_2|varphi(x)|^{frac{r}{s}}leq $$
$$underbrace{2a_1}_{=A_1}+underbrace{(a_22^{frac{r}{s}-1}+a_2)}_{=A_2}|varphi(x)|^{frac{r}{s}}+underbrace{a_22^{frac{r}{s}-1}}_{=A_3}|z(x)|^frac{r}{s}$$
where You used that $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ and Jensen's inequality. In the case $frac{r}{s}<1$ You can proceed analogously using $(a+b)^{frac{r}{s}}leq a^{frac{r}{s}}+b^{frac{r}{s}}$ valid in this range.
answered yesterday
Peter Melech
2,464813
For $xinoverline{Omega}$ and $frac{r}{s}geq 1$ You get
$$|f(x,z(x))|=|g(x,z(x)+varphi(x))-g(x,phi(x))|leq a_1+a_2|z(x)+varphi|^{frac{r}{s}}+a_1+a_2|varphi(x)|^{frac{r}{s}}leq$$
$$2a_1+a_22^{frac{r}{s}-1}(|z(x)|^{frac{r}{s}}+|varphi(x)|^{frac{r}{s}})+a_2|varphi(x)|^{frac{r}{s}}leq $$
$$underbrace{2a_1}_{=A_1}+underbrace{(a_22^{frac{r}{s}-1}+a_2)}_{=A_2}|varphi(x)|^{frac{r}{s}}+underbrace{a_22^{frac{r}{s}-1}}_{=A_3}|z(x)|^frac{r}{s}$$
where You used that $(frac{a+b}{2})^{frac{r}{s}}leqfrac{a^frac{r}{s}+b^frac{r}{s}}{2}$ which holds by the convexity of $xmapsto x^{frac{r}{s}}$ and Jensen's inequality. In the case $frac{r}{s}<1$ You can proceed analogously using $(a+b)^{frac{r}{s}}leq a^{frac{r}{s}}+b^{frac{r}{s}}$ valid in this range.
answered yesterday
Peter Melech
2,464813
answered yesterday
Peter Melech
2,464813
answered yesterday
Peter Melech
2,464813
answered yesterday
Peter Melech
2,464813
2,464813
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
add a comment |
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
2
2
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
How would this help? You still have dependence on $x$ in the inequality for $f$, whereas you dont have dependence on $x$ in the assumption for $g$.
– supinf
yesterday
1
1
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
Yes, I think so too and this is just an answer to the OP's question concerning the derivation of this estimate. To proceed further one has to argue via dominated convergence as in the question quoted by the OP and " Therefor we can assume $varphi=0$... in Rabinowitz's book seems to be not correct or at least quite sloppy
– Peter Melech
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
You are correct, this alone doesn't solve the problem.
– J. Snow
yesterday
add a comment |
J. Snow is a new contributor. Be nice, and check out our Code of Conduct.
J. Snow is a new contributor. Be nice, and check out our Code of Conduct.
J. Snow is a new contributor. Be nice, and check out our Code of Conduct.
J. Snow is a new contributor. Be nice, and check out our Code of Conduct.
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