Find the orthogonal complement of a subset of $L^2$












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 14:34









Davide Giraudo

128k17156268




128k17156268










asked Dec 27 '18 at 18:16









VoBVoB

801513




801513












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50


















  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50
















$begingroup$
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
$endgroup$
– Dog_69
Dec 27 '18 at 18:38






$begingroup$
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
$endgroup$
– Dog_69
Dec 27 '18 at 18:38














$begingroup$
Sorry but I don't know anything about Legendre Polynomials yet
$endgroup$
– VoB
Dec 27 '18 at 18:50




$begingroup$
Sorry but I don't know anything about Legendre Polynomials yet
$endgroup$
– VoB
Dec 27 '18 at 18:50










1 Answer
1






active

oldest

votes


















5












$begingroup$

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 $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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 $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12
















5












$begingroup$

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 $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12














5












5








5





$begingroup$

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 $v = 0$.






share|cite|improve this answer









$endgroup$



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 $v = 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 18:44









mechanodroidmechanodroid

28.9k62648




28.9k62648












  • $begingroup$
    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 $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12


















  • $begingroup$
    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 $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12
















$begingroup$
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 $v in C^{perp}$.
$endgroup$
– VoB
Dec 27 '18 at 19:11






$begingroup$
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 $v in C^{perp}$.
$endgroup$
– VoB
Dec 27 '18 at 19:11














$begingroup$
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
$endgroup$
– VoB
Dec 27 '18 at 19:12




$begingroup$
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
$endgroup$
– VoB
Dec 27 '18 at 19:12


















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