Mixing
Multiplication operator by $x$ on $L^2(0,1)$ is isomorphic to its square
$begingroup$
The following is Exercise II.8.7 from J. B. Conway's A Course in Functional Analysis.
Let $A:L^2(0,1)to L^2(0,1)$ be defined by $(Af)(x)=xf(x)$ for $f$ in $L^2(0,1)$ and $x$ in $(0,1)$. Show that $Acong A^2$.
Here $Acong A^2$ means there is an unitary isomorphism $U:L^2(0,1)to L^2(0,1)$ such that $UAU^{-1}=A^2$. I guess I should construct an explicit isomorphism, because $A$ is not compact, and the author has not proved any structure theorem for non-compact operators yet. However, I don't know how I can find such an isomorphism. I understand that $L^2(0,1)$ has Hilbert basis $e^{2pi inx}$, and I computed the matrix of $A$ with respect to this basis, but I can't see how $Acong A^2$.
Any hints will be appreciated!
functional-analysis operator-theory hilbert-spaces
asked Jan 31 at 2:09
ColescuColescu
3,23211136
$endgroup$
add a comment |
$begingroup$
The following is Exercise II.8.7 from J. B. Conway's A Course in Functional Analysis.
Let $A:L^2(0,1)to L^2(0,1)$ be defined by $(Af)(x)=xf(x)$ for $f$ in $L^2(0,1)$ and $x$ in $(0,1)$. Show that $Acong A^2$.
Here $Acong A^2$ means there is an unitary isomorphism $U:L^2(0,1)to L^2(0,1)$ such that $UAU^{-1}=A^2$. I guess I should construct an explicit isomorphism, because $A$ is not compact, and the author has not proved any structure theorem for non-compact operators yet. However, I don't know how I can find such an isomorphism. I understand that $L^2(0,1)$ has Hilbert basis $e^{2pi inx}$, and I computed the matrix of $A$ with respect to this basis, but I can't see how $Acong A^2$.
Any hints will be appreciated!
functional-analysis operator-theory hilbert-spaces
asked Jan 31 at 2:09
ColescuColescu
3,23211136
$endgroup$
add a comment |
$begingroup$
The following is Exercise II.8.7 from J. B. Conway's A Course in Functional Analysis.
Let $A:L^2(0,1)to L^2(0,1)$ be defined by $(Af)(x)=xf(x)$ for $f$ in $L^2(0,1)$ and $x$ in $(0,1)$. Show that $Acong A^2$.
Here $Acong A^2$ means there is an unitary isomorphism $U:L^2(0,1)to L^2(0,1)$ such that $UAU^{-1}=A^2$. I guess I should construct an explicit isomorphism, because $A$ is not compact, and the author has not proved any structure theorem for non-compact operators yet. However, I don't know how I can find such an isomorphism. I understand that $L^2(0,1)$ has Hilbert basis $e^{2pi inx}$, and I computed the matrix of $A$ with respect to this basis, but I can't see how $Acong A^2$.
Any hints will be appreciated!
functional-analysis operator-theory hilbert-spaces
asked Jan 31 at 2:09
ColescuColescu
3,23211136
$endgroup$
The following is Exercise II.8.7 from J. B. Conway's A Course in Functional Analysis.
Let $A:L^2(0,1)to L^2(0,1)$ be defined by $(Af)(x)=xf(x)$ for $f$ in $L^2(0,1)$ and $x$ in $(0,1)$. Show that $Acong A^2$.
Here $Acong A^2$ means there is an unitary isomorphism $U:L^2(0,1)to L^2(0,1)$ such that $UAU^{-1}=A^2$. I guess I should construct an explicit isomorphism, because $A$ is not compact, and the author has not proved any structure theorem for non-compact operators yet. However, I don't know how I can find such an isomorphism. I understand that $L^2(0,1)$ has Hilbert basis $e^{2pi inx}$, and I computed the matrix of $A$ with respect to this basis, but I can't see how $Acong A^2$.
Any hints will be appreciated!
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Jan 31 at 2:09
ColescuColescu
3,23211136
asked Jan 31 at 2:09
ColescuColescu
3,23211136
asked Jan 31 at 2:09
ColescuColescu
3,23211136
asked Jan 31 at 2:09
ColescuColescu
3,23211136
asked Jan 31 at 2:09
ColescuColescu
3,23211136
3,23211136
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let $Vin B(L^2(0,1))$ be given by
$$
Vf,(x)=frac{f(sqrt x)}{sqrt2,x^{1/4}}.
$$
This map is invertible: if
$$
Wg,(x)=sqrt{2x},g(x^2).
$$
Then $VW=WV=I$, so $V$ is invertible. Also,
$$tag1
|Vf|^2=int_0^1frac{|f(sqrt x)|^2}{2,x^{1/2}},dx=int_0^1|f(u)|^2,du=|f|^2.
$$
So $V$ is a unitary. Also,
$$
VA^2f,(x)=V(x^2f),(x)=frac{x,f(sqrt x)}{sqrt2,x^{1/4}}=AVf,(x).
$$
So $VA^2=AV$, and then $VA^2V^*=A$.
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $Vin B(L^2(0,1))$ be given by
$$
Vf,(x)=frac{f(sqrt x)}{sqrt2,x^{1/4}}.
$$
This map is invertible: if
$$
Wg,(x)=sqrt{2x},g(x^2).
$$
Then $VW=WV=I$, so $V$ is invertible. Also,
$$tag1
|Vf|^2=int_0^1frac{|f(sqrt x)|^2}{2,x^{1/2}},dx=int_0^1|f(u)|^2,du=|f|^2.
$$
So $V$ is a unitary. Also,
$$
VA^2f,(x)=V(x^2f),(x)=frac{x,f(sqrt x)}{sqrt2,x^{1/4}}=AVf,(x).
$$
So $VA^2=AV$, and then $VA^2V^*=A$.
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
add a comment |
$begingroup$
Let $Vin B(L^2(0,1))$ be given by
$$
Vf,(x)=frac{f(sqrt x)}{sqrt2,x^{1/4}}.
$$
This map is invertible: if
$$
Wg,(x)=sqrt{2x},g(x^2).
$$
Then $VW=WV=I$, so $V$ is invertible. Also,
$$tag1
|Vf|^2=int_0^1frac{|f(sqrt x)|^2}{2,x^{1/2}},dx=int_0^1|f(u)|^2,du=|f|^2.
$$
So $V$ is a unitary. Also,
$$
VA^2f,(x)=V(x^2f),(x)=frac{x,f(sqrt x)}{sqrt2,x^{1/4}}=AVf,(x).
$$
So $VA^2=AV$, and then $VA^2V^*=A$.
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
add a comment |
$begingroup$
Let $Vin B(L^2(0,1))$ be given by
$$
Vf,(x)=frac{f(sqrt x)}{sqrt2,x^{1/4}}.
$$
This map is invertible: if
$$
Wg,(x)=sqrt{2x},g(x^2).
$$
Then $VW=WV=I$, so $V$ is invertible. Also,
$$tag1
|Vf|^2=int_0^1frac{|f(sqrt x)|^2}{2,x^{1/2}},dx=int_0^1|f(u)|^2,du=|f|^2.
$$
So $V$ is a unitary. Also,
$$
VA^2f,(x)=V(x^2f),(x)=frac{x,f(sqrt x)}{sqrt2,x^{1/4}}=AVf,(x).
$$
So $VA^2=AV$, and then $VA^2V^*=A$.
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
Let $Vin B(L^2(0,1))$ be given by
$$
Vf,(x)=frac{f(sqrt x)}{sqrt2,x^{1/4}}.
$$
This map is invertible: if
$$
Wg,(x)=sqrt{2x},g(x^2).
$$
Then $VW=WV=I$, so $V$ is invertible. Also,
$$tag1
|Vf|^2=int_0^1frac{|f(sqrt x)|^2}{2,x^{1/2}},dx=int_0^1|f(u)|^2,du=|f|^2.
$$
So $V$ is a unitary. Also,
$$
VA^2f,(x)=V(x^2f),(x)=frac{x,f(sqrt x)}{sqrt2,x^{1/4}}=AVf,(x).
$$
So $VA^2=AV$, and then $VA^2V^*=A$.
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
edited Jan 31 at 6:21
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 4:09
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
add a comment |
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
1
1
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
$begingroup$
Thanks! May I ask how did you find such a map?
$endgroup$
– Colescu
Jan 31 at 5:02
2
2
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
$begingroup$
The unitary couldn't be multiplication, because it would commute. Next thing to try was a change of variable, since unitaries are isometric. Natural change of variables were $xlongmapsto x^2$ and $xlongmapsto sqrt x $, but they need to preserve the equality $(1) $, so I included the (square root of the) change of variable.
$endgroup$
– Martin Argerami
Jan 31 at 6:21
add a comment |
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