Mixing
Does this conclude that $ int_E f_jg_j dx to int_E fgdx$?
$begingroup$
Suppose $f_j to f$ strongly in $L^2(E)$ and $g_j to g $ weakly in $L^2(E)$, where $E subset mathbb{R}^d$ is measurable. Show that $ int_Ef_jg_jdx to int_E fgdx$.
Answer:
I am quoting the following theorem:
$text{Product of weak-strong converging sequence}:$
Let $1 < p < infty$, $ u_n: Omega to mathbb{R}$ be a sequence in $L^p(Omega)$, and $u in L^p(Omega)$. Let $v_n: Omega to mathbb{R}$ be a sequence in $L^{p'}(Omega)$, $ frac{1}{p}+frac{1}{p'}=1$. Suppose,
$u_n$ converges to $u$ weakly in $L^p(Omega)$,
$v_n$ converges to $v$ strongly in $L^{p'}(Omega)$,
Then $u_n v_n to uv$ weakly in $L^{1}(Omega)$
This is the theorem.
By applying the above theorem, we get
$f_jg_j$ converges to $fg$ weakly in $L^{1}(Omega)$.
Does this conclude that $ int_E f_jg_j dx to int_E fgdx$?
Help me
measure-theory weak-convergence strong-convergence
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
$endgroup$
add a comment |
$begingroup$
Suppose $f_j to f$ strongly in $L^2(E)$ and $g_j to g $ weakly in $L^2(E)$, where $E subset mathbb{R}^d$ is measurable. Show that $ int_Ef_jg_jdx to int_E fgdx$.
Answer:
I am quoting the following theorem:
$text{Product of weak-strong converging sequence}:$
Let $1 < p < infty$, $ u_n: Omega to mathbb{R}$ be a sequence in $L^p(Omega)$, and $u in L^p(Omega)$. Let $v_n: Omega to mathbb{R}$ be a sequence in $L^{p'}(Omega)$, $ frac{1}{p}+frac{1}{p'}=1$. Suppose,
$u_n$ converges to $u$ weakly in $L^p(Omega)$,
$v_n$ converges to $v$ strongly in $L^{p'}(Omega)$,
Then $u_n v_n to uv$ weakly in $L^{1}(Omega)$
This is the theorem.
By applying the above theorem, we get
$f_jg_j$ converges to $fg$ weakly in $L^{1}(Omega)$.
Does this conclude that $ int_E f_jg_j dx to int_E fgdx$?
Help me
measure-theory weak-convergence strong-convergence
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
$endgroup$
1
$begingroup$
Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31
add a comment |
$begingroup$
Suppose $f_j to f$ strongly in $L^2(E)$ and $g_j to g $ weakly in $L^2(E)$, where $E subset mathbb{R}^d$ is measurable. Show that $ int_Ef_jg_jdx to int_E fgdx$.
Answer:
I am quoting the following theorem:
$text{Product of weak-strong converging sequence}:$
Let $1 < p < infty$, $ u_n: Omega to mathbb{R}$ be a sequence in $L^p(Omega)$, and $u in L^p(Omega)$. Let $v_n: Omega to mathbb{R}$ be a sequence in $L^{p'}(Omega)$, $ frac{1}{p}+frac{1}{p'}=1$. Suppose,
$u_n$ converges to $u$ weakly in $L^p(Omega)$,
$v_n$ converges to $v$ strongly in $L^{p'}(Omega)$,
Then $u_n v_n to uv$ weakly in $L^{1}(Omega)$
This is the theorem.
By applying the above theorem, we get
$f_jg_j$ converges to $fg$ weakly in $L^{1}(Omega)$.
Does this conclude that $ int_E f_jg_j dx to int_E fgdx$?
Help me
measure-theory weak-convergence strong-convergence
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
$endgroup$
Suppose $f_j to f$ strongly in $L^2(E)$ and $g_j to g $ weakly in $L^2(E)$, where $E subset mathbb{R}^d$ is measurable. Show that $ int_Ef_jg_jdx to int_E fgdx$.
Answer:
I am quoting the following theorem:
$text{Product of weak-strong converging sequence}:$
Let $1 < p < infty$, $ u_n: Omega to mathbb{R}$ be a sequence in $L^p(Omega)$, and $u in L^p(Omega)$. Let $v_n: Omega to mathbb{R}$ be a sequence in $L^{p'}(Omega)$, $ frac{1}{p}+frac{1}{p'}=1$. Suppose,
$u_n$ converges to $u$ weakly in $L^p(Omega)$,
$v_n$ converges to $v$ strongly in $L^{p'}(Omega)$,
Then $u_n v_n to uv$ weakly in $L^{1}(Omega)$
This is the theorem.
By applying the above theorem, we get
$f_jg_j$ converges to $fg$ weakly in $L^{1}(Omega)$.
Does this conclude that $ int_E f_jg_j dx to int_E fgdx$?
Help me
measure-theory weak-convergence strong-convergence
measure-theory weak-convergence strong-convergence
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
asked Jan 21 at 19:15
M. A. SARKARM. A. SARKAR
2,3221719
2,3221719
1
$begingroup$
Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31
add a comment |
1
$begingroup$
Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31
1
1
$begingroup$
Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31
$begingroup$
Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes: weak convergence of a sequence $ left(h_nright)_{ngeqslant 1}$ in $mathbb L^1$ to a function $h$ means that for each linear continuous functional $Phicolonmathbb L^1tomathbb R$, the sequence $left(Phileft(h_nright)right)_{ngeqslant 1}$ converges to $Phileft(hright)$. This definition is applied to $Phicolon hmapsto int_E h(x)mathrm dx$, which is linear and continuous.
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
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$begingroup$
Yes: weak convergence of a sequence $ left(h_nright)_{ngeqslant 1}$ in $mathbb L^1$ to a function $h$ means that for each linear continuous functional $Phicolonmathbb L^1tomathbb R$, the sequence $left(Phileft(h_nright)right)_{ngeqslant 1}$ converges to $Phileft(hright)$. This definition is applied to $Phicolon hmapsto int_E h(x)mathrm dx$, which is linear and continuous.
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Yes: weak convergence of a sequence $ left(h_nright)_{ngeqslant 1}$ in $mathbb L^1$ to a function $h$ means that for each linear continuous functional $Phicolonmathbb L^1tomathbb R$, the sequence $left(Phileft(h_nright)right)_{ngeqslant 1}$ converges to $Phileft(hright)$. This definition is applied to $Phicolon hmapsto int_E h(x)mathrm dx$, which is linear and continuous.
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Yes: weak convergence of a sequence $ left(h_nright)_{ngeqslant 1}$ in $mathbb L^1$ to a function $h$ means that for each linear continuous functional $Phicolonmathbb L^1tomathbb R$, the sequence $left(Phileft(h_nright)right)_{ngeqslant 1}$ converges to $Phileft(hright)$. This definition is applied to $Phicolon hmapsto int_E h(x)mathrm dx$, which is linear and continuous.
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
Yes: weak convergence of a sequence $ left(h_nright)_{ngeqslant 1}$ in $mathbb L^1$ to a function $h$ means that for each linear continuous functional $Phicolonmathbb L^1tomathbb R$, the sequence $left(Phileft(h_nright)right)_{ngeqslant 1}$ converges to $Phileft(hright)$. This definition is applied to $Phicolon hmapsto int_E h(x)mathrm dx$, which is linear and continuous.
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
answered Jan 21 at 22:20
Davide GiraudoDavide Giraudo
127k17154268
127k17154268
add a comment |
add a comment |
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Yes, since $1_E$ is in $(L^1(Omega))^*=L^infty(Omega)$.
$endgroup$
– Song
Jan 21 at 19:31