Find the orthogonal complement of a subset of $L^2$
I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.
By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$
Any hint or else would be really appreciated
functional-analysis hilbert-spaces lp-spaces
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
add a comment |
I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.
By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$
Any hint or else would be really appreciated
functional-analysis hilbert-spaces lp-spaces
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50
add a comment |
I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.
By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$
Any hint or else would be really appreciated
functional-analysis hilbert-spaces lp-spaces
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.
By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$
Any hint or else would be really appreciated
functional-analysis hilbert-spaces lp-spaces
functional-analysis hilbert-spaces lp-spaces
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
edited Dec 28 '18 at 14:34
Davide Giraudo
125k16150260
125k16150260
asked Dec 27 '18 at 18:16
VoB
673213
asked Dec 27 '18 at 18:16
VoB
673213
asked Dec 27 '18 at 18:16
VoB
673213
673213
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50
add a comment |
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50
add a comment |
1 Answer
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The orthogonal is trivial.
Indeed, assume $v in C^perp$.
If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.
If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
add a comment |
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1 Answer
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1 Answer
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The orthogonal is trivial.
Indeed, assume $v in C^perp$.
If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.
If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
add a comment |
The orthogonal is trivial.
Indeed, assume $v in C^perp$.
If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.
If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
add a comment |
The orthogonal is trivial.
Indeed, assume $v in C^perp$.
If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.
If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
The orthogonal is trivial.
Indeed, assume $v in C^perp$.
If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.
If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
answered Dec 27 '18 at 18:44
mechanodroid
26.8k62347
26.8k62347
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
add a comment |
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
– VoB
Dec 27 '18 at 19:11
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
– VoB
Dec 27 '18 at 19:12
add a comment |
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I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
– Dog_69
Dec 27 '18 at 18:38
Sorry but I don't know anything about Legendre Polynomials yet
– VoB
Dec 27 '18 at 18:50