$F(L_2([0,1],mu))$ is dense in $L_2([0,1],mu)$, where $F$ is the multiplication operator
$begingroup$
Consider $A=L_2([0,1],mu)$ where $mu$ is the Lebesgue measure. Let $F:Arightarrow A$ be linear given by $Ff(x)=xf(x)$ for $fin A$ and $xin[0,1]$.
I want to show that $F(A)$ (the range of $F$) is a proper dense subspace of $A$. I've managed to show that $F(A)$ is a proper subspace of $A$, but I'm having trouble showing it's dense.
real-analysis operator-theory hilbert-spaces
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
$endgroup$
add a comment |
$begingroup$
Consider $A=L_2([0,1],mu)$ where $mu$ is the Lebesgue measure. Let $F:Arightarrow A$ be linear given by $Ff(x)=xf(x)$ for $fin A$ and $xin[0,1]$.
I want to show that $F(A)$ (the range of $F$) is a proper dense subspace of $A$. I've managed to show that $F(A)$ is a proper subspace of $A$, but I'm having trouble showing it's dense.
real-analysis operator-theory hilbert-spaces
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
$endgroup$
add a comment |
$begingroup$
Consider $A=L_2([0,1],mu)$ where $mu$ is the Lebesgue measure. Let $F:Arightarrow A$ be linear given by $Ff(x)=xf(x)$ for $fin A$ and $xin[0,1]$.
I want to show that $F(A)$ (the range of $F$) is a proper dense subspace of $A$. I've managed to show that $F(A)$ is a proper subspace of $A$, but I'm having trouble showing it's dense.
real-analysis operator-theory hilbert-spaces
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
$endgroup$
Consider $A=L_2([0,1],mu)$ where $mu$ is the Lebesgue measure. Let $F:Arightarrow A$ be linear given by $Ff(x)=xf(x)$ for $fin A$ and $xin[0,1]$.
I want to show that $F(A)$ (the range of $F$) is a proper dense subspace of $A$. I've managed to show that $F(A)$ is a proper subspace of $A$, but I'm having trouble showing it's dense.
real-analysis operator-theory hilbert-spaces
real-analysis operator-theory hilbert-spaces
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
edited Jan 8 at 14:19
Davide Giraudo
125k16150261
125k16150261
asked Jan 8 at 13:59
Kane JKane J
715
asked Jan 8 at 13:59
Kane JKane J
715
asked Jan 8 at 13:59
Kane JKane J
715
715
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A space $S$ of a Hilbert space $mathcal H$ is dense if its orthogonal is reduced to ${0}$ (because $left(S^perpright)^perp$ is the closure of $S$).
Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $fin A$,
$$int_{[0,1]}xf(x)g(x)mathrm dlambda(x)=0;$$
in particular, choosing $fcolon xmapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
add a comment |
$begingroup$
If you are lazy (just like me), then you can use the identity $ker F=(operatorname{ran}F^ast)^perp$. Clearly $F=F^ast$ and $ker F=0$, so $(mathrm{ran}F)^perp={0}$.
If you are slightly less lazy, note that $fin C_c((0,1))$ implies $(xmapsto f(x)/x)in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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$begingroup$
A space $S$ of a Hilbert space $mathcal H$ is dense if its orthogonal is reduced to ${0}$ (because $left(S^perpright)^perp$ is the closure of $S$).
Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $fin A$,
$$int_{[0,1]}xf(x)g(x)mathrm dlambda(x)=0;$$
in particular, choosing $fcolon xmapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
add a comment |
$begingroup$
A space $S$ of a Hilbert space $mathcal H$ is dense if its orthogonal is reduced to ${0}$ (because $left(S^perpright)^perp$ is the closure of $S$).
Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $fin A$,
$$int_{[0,1]}xf(x)g(x)mathrm dlambda(x)=0;$$
in particular, choosing $fcolon xmapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
add a comment |
$begingroup$
A space $S$ of a Hilbert space $mathcal H$ is dense if its orthogonal is reduced to ${0}$ (because $left(S^perpright)^perp$ is the closure of $S$).
Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $fin A$,
$$int_{[0,1]}xf(x)g(x)mathrm dlambda(x)=0;$$
in particular, choosing $fcolon xmapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
A space $S$ of a Hilbert space $mathcal H$ is dense if its orthogonal is reduced to ${0}$ (because $left(S^perpright)^perp$ is the closure of $S$).
Let $g$ be a function in the orthogonal of $F(A)$. This means that for all $fin A$,
$$int_{[0,1]}xf(x)g(x)mathrm dlambda(x)=0;$$
in particular, choosing $fcolon xmapsto xg(x)$ gives that $x^2g(x)^2=0$ almost everywhere hence $g=0$ almost everywhere.
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 8 at 14:14
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
add a comment |
add a comment |
$begingroup$
If you are lazy (just like me), then you can use the identity $ker F=(operatorname{ran}F^ast)^perp$. Clearly $F=F^ast$ and $ker F=0$, so $(mathrm{ran}F)^perp={0}$.
If you are slightly less lazy, note that $fin C_c((0,1))$ implies $(xmapsto f(x)/x)in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
$endgroup$
add a comment |
$begingroup$
If you are lazy (just like me), then you can use the identity $ker F=(operatorname{ran}F^ast)^perp$. Clearly $F=F^ast$ and $ker F=0$, so $(mathrm{ran}F)^perp={0}$.
If you are slightly less lazy, note that $fin C_c((0,1))$ implies $(xmapsto f(x)/x)in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
$endgroup$
add a comment |
$begingroup$
If you are lazy (just like me), then you can use the identity $ker F=(operatorname{ran}F^ast)^perp$. Clearly $F=F^ast$ and $ker F=0$, so $(mathrm{ran}F)^perp={0}$.
If you are slightly less lazy, note that $fin C_c((0,1))$ implies $(xmapsto f(x)/x)in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
$endgroup$
If you are lazy (just like me), then you can use the identity $ker F=(operatorname{ran}F^ast)^perp$. Clearly $F=F^ast$ and $ker F=0$, so $(mathrm{ran}F)^perp={0}$.
If you are slightly less lazy, note that $fin C_c((0,1))$ implies $(xmapsto f(x)/x)in C_c((0,1))$. Since $C_c((0,1))$ is dense in $A$, you are done.
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
answered Jan 8 at 14:15
MaoWaoMaoWao
2,543616
2,543616
add a comment |
add a comment |
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