Show $X = {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$
$begingroup$
Given $(X,mathcal{A}, mu)$ a measurable space. Let $ p in ]1,infty[$, and $(f_{n})_{n}subseteq L^{p}(X,mu)$ so that $sup_{n}||f_{n}||_{p}<infty$ and $f_{n}to 0$, $n to infty$ $mu-$a.e.
i) Prove $ forall epsilon > 0$ and $n in mathbb N$, $X = {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$
ii) Show $lim_{nto infty}int_{X}|f_{n}|^{p-1}|g|dmu = 0 = lim_{n to infty}int_{X}|g|^{p-1}|f_{n}|dmu$
My ideas:
i) The inclusion ${f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon} subseteq X$ is trivial. I honestly do not know where to begin on $X subseteq {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$.
ii) I believe that i) is then used to split up $g$, and then hopefully get
$lim_{epsilon to 0}lim_{ntoinfty}(int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|leqepsilon}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|geqepsilon^2}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|g|leqepsilon}}dmu)=0$
Any guidance is greatly appreciated.
real-analysis integration measure-theory
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
$endgroup$
add a comment |
$begingroup$
Given $(X,mathcal{A}, mu)$ a measurable space. Let $ p in ]1,infty[$, and $(f_{n})_{n}subseteq L^{p}(X,mu)$ so that $sup_{n}||f_{n}||_{p}<infty$ and $f_{n}to 0$, $n to infty$ $mu-$a.e.
i) Prove $ forall epsilon > 0$ and $n in mathbb N$, $X = {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$
ii) Show $lim_{nto infty}int_{X}|f_{n}|^{p-1}|g|dmu = 0 = lim_{n to infty}int_{X}|g|^{p-1}|f_{n}|dmu$
My ideas:
i) The inclusion ${f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon} subseteq X$ is trivial. I honestly do not know where to begin on $X subseteq {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$.
ii) I believe that i) is then used to split up $g$, and then hopefully get
$lim_{epsilon to 0}lim_{ntoinfty}(int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|leqepsilon}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|geqepsilon^2}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|g|leqepsilon}}dmu)=0$
Any guidance is greatly appreciated.
real-analysis integration measure-theory
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
$endgroup$
add a comment |
$begingroup$
Given $(X,mathcal{A}, mu)$ a measurable space. Let $ p in ]1,infty[$, and $(f_{n})_{n}subseteq L^{p}(X,mu)$ so that $sup_{n}||f_{n}||_{p}<infty$ and $f_{n}to 0$, $n to infty$ $mu-$a.e.
i) Prove $ forall epsilon > 0$ and $n in mathbb N$, $X = {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$
ii) Show $lim_{nto infty}int_{X}|f_{n}|^{p-1}|g|dmu = 0 = lim_{n to infty}int_{X}|g|^{p-1}|f_{n}|dmu$
My ideas:
i) The inclusion ${f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon} subseteq X$ is trivial. I honestly do not know where to begin on $X subseteq {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$.
ii) I believe that i) is then used to split up $g$, and then hopefully get
$lim_{epsilon to 0}lim_{ntoinfty}(int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|leqepsilon}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|geqepsilon^2}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|g|leqepsilon}}dmu)=0$
Any guidance is greatly appreciated.
real-analysis integration measure-theory
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
$endgroup$
Given $(X,mathcal{A}, mu)$ a measurable space. Let $ p in ]1,infty[$, and $(f_{n})_{n}subseteq L^{p}(X,mu)$ so that $sup_{n}||f_{n}||_{p}<infty$ and $f_{n}to 0$, $n to infty$ $mu-$a.e.
i) Prove $ forall epsilon > 0$ and $n in mathbb N$, $X = {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$
ii) Show $lim_{nto infty}int_{X}|f_{n}|^{p-1}|g|dmu = 0 = lim_{n to infty}int_{X}|g|^{p-1}|f_{n}|dmu$
My ideas:
i) The inclusion ${f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon} subseteq X$ is trivial. I honestly do not know where to begin on $X subseteq {f_{n} leq epsilon |g|} cup {|f_{n}|geq epsilon^{2}}cup{|g|leq epsilon}$.
ii) I believe that i) is then used to split up $g$, and then hopefully get
$lim_{epsilon to 0}lim_{ntoinfty}(int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|leqepsilon}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|f_{n}|geqepsilon^2}}dmu+int_{X}|f_{n}|^{p-1}|g|chi_{{|g|leqepsilon}}dmu)=0$
Any guidance is greatly appreciated.
real-analysis integration measure-theory
real-analysis integration measure-theory
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
asked Dec 7 '18 at 22:08
SABOYSABOY
684311
684311
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hypothesis on $g$ is missing. The result is true if $g in L^{p}$. For i) just observe that if $|f_n| <epsilon ^{2}$ and $|g|>epsilon$ then $|f_n| <epsilon ^{2} <epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar.
So let us show that $int_A |f_n|^{p-1} |g|dmu $ can be made small (by suitable choice of $epsilon$) if $A$ is one of the sets $A_1={|f_n| leq epsilon |g|}, A_2={|f_n| geq epsilon ^{2}}$ and $A_3={|g|leq epsilon}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $frac p {p-1}$ and $p$) to see that the integral is less than or equal to $|f_n|_p(int_{A_2} |g|^{p}dmu)$. Note that $|f_n|$ is bounded and $mu(A_2) to 0$ as $n to infty$.
Finally for $A=A_3$ use a similar argument and note that $int |g|^{p} I_{{|g| leq epsilon}} to 0$ as $epsilon to 0$.
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
$endgroup$
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
add a comment |
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1 Answer
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$begingroup$
Hypothesis on $g$ is missing. The result is true if $g in L^{p}$. For i) just observe that if $|f_n| <epsilon ^{2}$ and $|g|>epsilon$ then $|f_n| <epsilon ^{2} <epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar.
So let us show that $int_A |f_n|^{p-1} |g|dmu $ can be made small (by suitable choice of $epsilon$) if $A$ is one of the sets $A_1={|f_n| leq epsilon |g|}, A_2={|f_n| geq epsilon ^{2}}$ and $A_3={|g|leq epsilon}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $frac p {p-1}$ and $p$) to see that the integral is less than or equal to $|f_n|_p(int_{A_2} |g|^{p}dmu)$. Note that $|f_n|$ is bounded and $mu(A_2) to 0$ as $n to infty$.
Finally for $A=A_3$ use a similar argument and note that $int |g|^{p} I_{{|g| leq epsilon}} to 0$ as $epsilon to 0$.
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
$endgroup$
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
add a comment |
$begingroup$
Hypothesis on $g$ is missing. The result is true if $g in L^{p}$. For i) just observe that if $|f_n| <epsilon ^{2}$ and $|g|>epsilon$ then $|f_n| <epsilon ^{2} <epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar.
So let us show that $int_A |f_n|^{p-1} |g|dmu $ can be made small (by suitable choice of $epsilon$) if $A$ is one of the sets $A_1={|f_n| leq epsilon |g|}, A_2={|f_n| geq epsilon ^{2}}$ and $A_3={|g|leq epsilon}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $frac p {p-1}$ and $p$) to see that the integral is less than or equal to $|f_n|_p(int_{A_2} |g|^{p}dmu)$. Note that $|f_n|$ is bounded and $mu(A_2) to 0$ as $n to infty$.
Finally for $A=A_3$ use a similar argument and note that $int |g|^{p} I_{{|g| leq epsilon}} to 0$ as $epsilon to 0$.
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
$endgroup$
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
add a comment |
$begingroup$
Hypothesis on $g$ is missing. The result is true if $g in L^{p}$. For i) just observe that if $|f_n| <epsilon ^{2}$ and $|g|>epsilon$ then $|f_n| <epsilon ^{2} <epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar.
So let us show that $int_A |f_n|^{p-1} |g|dmu $ can be made small (by suitable choice of $epsilon$) if $A$ is one of the sets $A_1={|f_n| leq epsilon |g|}, A_2={|f_n| geq epsilon ^{2}}$ and $A_3={|g|leq epsilon}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $frac p {p-1}$ and $p$) to see that the integral is less than or equal to $|f_n|_p(int_{A_2} |g|^{p}dmu)$. Note that $|f_n|$ is bounded and $mu(A_2) to 0$ as $n to infty$.
Finally for $A=A_3$ use a similar argument and note that $int |g|^{p} I_{{|g| leq epsilon}} to 0$ as $epsilon to 0$.
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
$endgroup$
Hypothesis on $g$ is missing. The result is true if $g in L^{p}$. For i) just observe that if $|f_n| <epsilon ^{2}$ and $|g|>epsilon$ then $|f_n| <epsilon ^{2} <epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar.
So let us show that $int_A |f_n|^{p-1} |g|dmu $ can be made small (by suitable choice of $epsilon$) if $A$ is one of the sets $A_1={|f_n| leq epsilon |g|}, A_2={|f_n| geq epsilon ^{2}}$ and $A_3={|g|leq epsilon}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $frac p {p-1}$ and $p$) to see that the integral is less than or equal to $|f_n|_p(int_{A_2} |g|^{p}dmu)$. Note that $|f_n|$ is bounded and $mu(A_2) to 0$ as $n to infty$.
Finally for $A=A_3$ use a similar argument and note that $int |g|^{p} I_{{|g| leq epsilon}} to 0$ as $epsilon to 0$.
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
answered Dec 7 '18 at 23:56
Kavi Rama MurthyKavi Rama Murthy
58.7k42161
58.7k42161
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
add a comment |
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
$begingroup$
I do not understand what you are actually proving in i)
$endgroup$
– SABOY
Dec 9 '18 at 19:07
1
1
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
@SABOY You know that RHS is contained in LHS. To prove the reverse inclusion I am trying to prove that $(RHS)^{c}$ (complement of RHS) is contained in $(LHS)^{c}$ (which is empty, of course). I hope this clears up things a bit.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:17
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
$begingroup$
Very smart. Thank you!
$endgroup$
– SABOY
Dec 10 '18 at 8:10
add a comment |
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