Mixing
Excercise 4.3 (3) in Brezis. Convergence in $L_p$
$begingroup$
Let $(f_n)$ in $L_p(Omega)$ $1leq p< infty$ and $(g_n)$ bounded in $L_{infty}(Omega)$ assume that $f_n rightarrow f$ in $L_p(Omega)$ and $g_n rightarrow g$ a.e. Prove that $$f_ng_nrightarrow fg$$ in $L_p(Omega)$.
I have done the following:
Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_pleq||f_n-f||_p||g_n||_{infty}+||f(g_n-g)||_p$$
I know the first term is less than $epsilon$ (because $f_n$ converges in $L_p(Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)rightarrow 0$ in $L_p(Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.
functional-analysis convergence lp-spaces
asked Jan 9 at 19:36
AlfdavAlfdav
787
$endgroup$
add a comment |
$begingroup$
Let $(f_n)$ in $L_p(Omega)$ $1leq p< infty$ and $(g_n)$ bounded in $L_{infty}(Omega)$ assume that $f_n rightarrow f$ in $L_p(Omega)$ and $g_n rightarrow g$ a.e. Prove that $$f_ng_nrightarrow fg$$ in $L_p(Omega)$.
I have done the following:
Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_pleq||f_n-f||_p||g_n||_{infty}+||f(g_n-g)||_p$$
I know the first term is less than $epsilon$ (because $f_n$ converges in $L_p(Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)rightarrow 0$ in $L_p(Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.
functional-analysis convergence lp-spaces
asked Jan 9 at 19:36
AlfdavAlfdav
787
$endgroup$
1
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43
add a comment |
$begingroup$
Let $(f_n)$ in $L_p(Omega)$ $1leq p< infty$ and $(g_n)$ bounded in $L_{infty}(Omega)$ assume that $f_n rightarrow f$ in $L_p(Omega)$ and $g_n rightarrow g$ a.e. Prove that $$f_ng_nrightarrow fg$$ in $L_p(Omega)$.
I have done the following:
Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_pleq||f_n-f||_p||g_n||_{infty}+||f(g_n-g)||_p$$
I know the first term is less than $epsilon$ (because $f_n$ converges in $L_p(Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)rightarrow 0$ in $L_p(Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.
functional-analysis convergence lp-spaces
asked Jan 9 at 19:36
AlfdavAlfdav
787
$endgroup$
Let $(f_n)$ in $L_p(Omega)$ $1leq p< infty$ and $(g_n)$ bounded in $L_{infty}(Omega)$ assume that $f_n rightarrow f$ in $L_p(Omega)$ and $g_n rightarrow g$ a.e. Prove that $$f_ng_nrightarrow fg$$ in $L_p(Omega)$.
I have done the following:
Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_pleq||f_n-f||_p||g_n||_{infty}+||f(g_n-g)||_p$$
I know the first term is less than $epsilon$ (because $f_n$ converges in $L_p(Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)rightarrow 0$ in $L_p(Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.
functional-analysis convergence lp-spaces
functional-analysis convergence lp-spaces
asked Jan 9 at 19:36
AlfdavAlfdav
787
asked Jan 9 at 19:36
AlfdavAlfdav
787
asked Jan 9 at 19:36
AlfdavAlfdav
787
asked Jan 9 at 19:36
AlfdavAlfdav
787
asked Jan 9 at 19:36
AlfdavAlfdav
787
787
1
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43
add a comment |
1
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43
1
1
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43
add a comment |
1 Answer
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$begingroup$
The only remaining thing to is show that $int_Omega h_nto 0$, where $h_n(x)=leftlvert f(x)left(g_n(x)-g(x)right)rightrvert^p$.
Since $f$ belongs to $mathbb L^p$, the quantity $leftlvert f(x) rightrvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)to 0$ follows from the implication (if $cgeqslant 0$ and $a_nto 0$) then $ccdot a_nto 0$.
For the domination condition, let $M:=sup_{ngeqslant 1}leftlVert g_nrightrVert_infty$. Then $leftlvert g(x)rightrvertleqslant M$ for almost every $x$ hence
$$leftlvert h_n(x) rightrvertleqslant left(2Mright)^pleftlvert f(x) rightrvert^p mbox{ a.e.}$$
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The only remaining thing to is show that $int_Omega h_nto 0$, where $h_n(x)=leftlvert f(x)left(g_n(x)-g(x)right)rightrvert^p$.
Since $f$ belongs to $mathbb L^p$, the quantity $leftlvert f(x) rightrvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)to 0$ follows from the implication (if $cgeqslant 0$ and $a_nto 0$) then $ccdot a_nto 0$.
For the domination condition, let $M:=sup_{ngeqslant 1}leftlVert g_nrightrVert_infty$. Then $leftlvert g(x)rightrvertleqslant M$ for almost every $x$ hence
$$leftlvert h_n(x) rightrvertleqslant left(2Mright)^pleftlvert f(x) rightrvert^p mbox{ a.e.}$$
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
The only remaining thing to is show that $int_Omega h_nto 0$, where $h_n(x)=leftlvert f(x)left(g_n(x)-g(x)right)rightrvert^p$.
Since $f$ belongs to $mathbb L^p$, the quantity $leftlvert f(x) rightrvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)to 0$ follows from the implication (if $cgeqslant 0$ and $a_nto 0$) then $ccdot a_nto 0$.
For the domination condition, let $M:=sup_{ngeqslant 1}leftlVert g_nrightrVert_infty$. Then $leftlvert g(x)rightrvertleqslant M$ for almost every $x$ hence
$$leftlvert h_n(x) rightrvertleqslant left(2Mright)^pleftlvert f(x) rightrvert^p mbox{ a.e.}$$
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
The only remaining thing to is show that $int_Omega h_nto 0$, where $h_n(x)=leftlvert f(x)left(g_n(x)-g(x)right)rightrvert^p$.
Since $f$ belongs to $mathbb L^p$, the quantity $leftlvert f(x) rightrvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)to 0$ follows from the implication (if $cgeqslant 0$ and $a_nto 0$) then $ccdot a_nto 0$.
For the domination condition, let $M:=sup_{ngeqslant 1}leftlVert g_nrightrVert_infty$. Then $leftlvert g(x)rightrvertleqslant M$ for almost every $x$ hence
$$leftlvert h_n(x) rightrvertleqslant left(2Mright)^pleftlvert f(x) rightrvert^p mbox{ a.e.}$$
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
The only remaining thing to is show that $int_Omega h_nto 0$, where $h_n(x)=leftlvert f(x)left(g_n(x)-g(x)right)rightrvert^p$.
Since $f$ belongs to $mathbb L^p$, the quantity $leftlvert f(x) rightrvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)to 0$ follows from the implication (if $cgeqslant 0$ and $a_nto 0$) then $ccdot a_nto 0$.
For the domination condition, let $M:=sup_{ngeqslant 1}leftlVert g_nrightrVert_infty$. Then $leftlvert g(x)rightrvertleqslant M$ for almost every $x$ hence
$$leftlvert h_n(x) rightrvertleqslant left(2Mright)^pleftlvert f(x) rightrvert^p mbox{ a.e.}$$
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 10 at 20:04
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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1
$begingroup$
Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $epsilon$ for $n$ large.
$endgroup$
– Keen-ameteur
Jan 9 at 20:02
$begingroup$
yes but i made another advance, since $g_nrightarrow g$ a.e. can i say that $fg_nrightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^pleq M^p|f|^p$ then I can use dominated convergence.
$endgroup$
– Alfdav
Jan 9 at 20:16
$begingroup$
The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $vert f(g-g_n) vert^p $ instead of $vert fg_nvert^p$.
$endgroup$
– Keen-ameteur
Jan 9 at 20:24
$begingroup$
$f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^pleq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<epsilon$
$endgroup$
– Alfdav
Jan 9 at 20:41
$begingroup$
I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it
$endgroup$
– Alfdav
Jan 9 at 20:43