Mixing
Does convergence in $L^2$ imply convergence of composition?
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Let ${f_n}$ be a sequence in $L^2(mathbb{R})$ such that $f_n to 0$ in $L^2(mathbb{R})$ as $n to infty$. Let $u in L^2(mathbb{R})$ such that $f_n(u) in L^2(mathbb{R})$. Can we conclude $f_n(u) to 0$ in $ L^2(mathbb{R})$ as $n to infty$.
I'm guessing this result is not true, but every sequence I take seems to verify the claim.
functional-analysis analysis
asked Jan 16 at 9:29
Goal123Goal123
502212
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add a comment |
$begingroup$
Let ${f_n}$ be a sequence in $L^2(mathbb{R})$ such that $f_n to 0$ in $L^2(mathbb{R})$ as $n to infty$. Let $u in L^2(mathbb{R})$ such that $f_n(u) in L^2(mathbb{R})$. Can we conclude $f_n(u) to 0$ in $ L^2(mathbb{R})$ as $n to infty$.
I'm guessing this result is not true, but every sequence I take seems to verify the claim.
functional-analysis analysis
asked Jan 16 at 9:29
Goal123Goal123
502212
$endgroup$
add a comment |
$begingroup$
Let ${f_n}$ be a sequence in $L^2(mathbb{R})$ such that $f_n to 0$ in $L^2(mathbb{R})$ as $n to infty$. Let $u in L^2(mathbb{R})$ such that $f_n(u) in L^2(mathbb{R})$. Can we conclude $f_n(u) to 0$ in $ L^2(mathbb{R})$ as $n to infty$.
I'm guessing this result is not true, but every sequence I take seems to verify the claim.
functional-analysis analysis
asked Jan 16 at 9:29
Goal123Goal123
502212
$endgroup$
Let ${f_n}$ be a sequence in $L^2(mathbb{R})$ such that $f_n to 0$ in $L^2(mathbb{R})$ as $n to infty$. Let $u in L^2(mathbb{R})$ such that $f_n(u) in L^2(mathbb{R})$. Can we conclude $f_n(u) to 0$ in $ L^2(mathbb{R})$ as $n to infty$.
I'm guessing this result is not true, but every sequence I take seems to verify the claim.
functional-analysis analysis
functional-analysis analysis
asked Jan 16 at 9:29
Goal123Goal123
502212
asked Jan 16 at 9:29
Goal123Goal123
502212
asked Jan 16 at 9:29
Goal123Goal123
502212
asked Jan 16 at 9:29
Goal123Goal123
502212
asked Jan 16 at 9:29
Goal123Goal123
502212
502212
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Let $f_n(x)=nI_{(0,frac 1 {n^{3}})}$ and $u(x)=x^{2}I_{(0,1)}$. Then the conditions are all satisfied but $|f_ncirc u|_2^{2} to infty$,
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
Unfortunately, $fin L^2(Bbb R)$ is not an actual function, but an equivalence class of functions. As a result, the composition map $fcirc u$ is not well-defined in $L^2(Bbb R)$ in general. (This is so even if we disregard measurability issues.) This makes the question as non-sense. For example, $1_{Bbb Q}$ is just $ 0$ in $L^2(Bbb R)$ but $$(1_Bbb Qcirc 1_Bbb Q)(x) =1$$ while
$$
(0circ 0)(x) = 0.
$$
answered Jan 16 at 9:41
SongSong
14.2k1633
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Let $f_n(x)=nI_{(0,frac 1 {n^{3}})}$ and $u(x)=x^{2}I_{(0,1)}$. Then the conditions are all satisfied but $|f_ncirc u|_2^{2} to infty$,
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
Let $f_n(x)=nI_{(0,frac 1 {n^{3}})}$ and $u(x)=x^{2}I_{(0,1)}$. Then the conditions are all satisfied but $|f_ncirc u|_2^{2} to infty$,
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
Let $f_n(x)=nI_{(0,frac 1 {n^{3}})}$ and $u(x)=x^{2}I_{(0,1)}$. Then the conditions are all satisfied but $|f_ncirc u|_2^{2} to infty$,
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
Let $f_n(x)=nI_{(0,frac 1 {n^{3}})}$ and $u(x)=x^{2}I_{(0,1)}$. Then the conditions are all satisfied but $|f_ncirc u|_2^{2} to infty$,
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
answered Jan 16 at 9:35
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
61.7k42262
add a comment |
add a comment |
$begingroup$
Unfortunately, $fin L^2(Bbb R)$ is not an actual function, but an equivalence class of functions. As a result, the composition map $fcirc u$ is not well-defined in $L^2(Bbb R)$ in general. (This is so even if we disregard measurability issues.) This makes the question as non-sense. For example, $1_{Bbb Q}$ is just $ 0$ in $L^2(Bbb R)$ but $$(1_Bbb Qcirc 1_Bbb Q)(x) =1$$ while
$$
(0circ 0)(x) = 0.
$$
answered Jan 16 at 9:41
SongSong
14.2k1633
$endgroup$
add a comment |
$begingroup$
Unfortunately, $fin L^2(Bbb R)$ is not an actual function, but an equivalence class of functions. As a result, the composition map $fcirc u$ is not well-defined in $L^2(Bbb R)$ in general. (This is so even if we disregard measurability issues.) This makes the question as non-sense. For example, $1_{Bbb Q}$ is just $ 0$ in $L^2(Bbb R)$ but $$(1_Bbb Qcirc 1_Bbb Q)(x) =1$$ while
$$
(0circ 0)(x) = 0.
$$
answered Jan 16 at 9:41
SongSong
14.2k1633
$endgroup$
add a comment |
$begingroup$
Unfortunately, $fin L^2(Bbb R)$ is not an actual function, but an equivalence class of functions. As a result, the composition map $fcirc u$ is not well-defined in $L^2(Bbb R)$ in general. (This is so even if we disregard measurability issues.) This makes the question as non-sense. For example, $1_{Bbb Q}$ is just $ 0$ in $L^2(Bbb R)$ but $$(1_Bbb Qcirc 1_Bbb Q)(x) =1$$ while
$$
(0circ 0)(x) = 0.
$$
answered Jan 16 at 9:41
SongSong
14.2k1633
$endgroup$
Unfortunately, $fin L^2(Bbb R)$ is not an actual function, but an equivalence class of functions. As a result, the composition map $fcirc u$ is not well-defined in $L^2(Bbb R)$ in general. (This is so even if we disregard measurability issues.) This makes the question as non-sense. For example, $1_{Bbb Q}$ is just $ 0$ in $L^2(Bbb R)$ but $$(1_Bbb Qcirc 1_Bbb Q)(x) =1$$ while
$$
(0circ 0)(x) = 0.
$$
answered Jan 16 at 9:41
SongSong
14.2k1633
edited Jan 16 at 9:47
answered Jan 16 at 9:41
SongSong
14.2k1633
answered Jan 16 at 9:41
SongSong
14.2k1633
answered Jan 16 at 9:41
SongSong
14.2k1633
14.2k1633
add a comment |
add a comment |
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