$L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$












0














Let $L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$. For every $f in L^2[-a,a]$, I want to find its projection and its distance from $L_d$.



Now, I think the fact that $f(x) = frac{f(x) + f(-x)}{2} + frac{f(x) - f(-x)}{2}$ might be relevant to the problem.



Of course, If f is indeed an odd function, then its distance from $L_d$ is actually 0 and it doesn't have a projection. But otherwise I do not know how to proceed.



I guess the projection on $L_d$ is the "odd part" of the function, but I am not sure how to prove that formally.



EDIT: My solution is the following:



Projection of f on $L_d$ = -f(-x) because I just put f(x) = -f(-x) in the expression of f that I outlined before. Is that right?






functional-analysis







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    0














    Let $L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$. For every $f in L^2[-a,a]$, I want to find its projection and its distance from $L_d$.



    Now, I think the fact that $f(x) = frac{f(x) + f(-x)}{2} + frac{f(x) - f(-x)}{2}$ might be relevant to the problem.



    Of course, If f is indeed an odd function, then its distance from $L_d$ is actually 0 and it doesn't have a projection. But otherwise I do not know how to proceed.



    I guess the projection on $L_d$ is the "odd part" of the function, but I am not sure how to prove that formally.



    EDIT: My solution is the following:



    Projection of f on $L_d$ = -f(-x) because I just put f(x) = -f(-x) in the expression of f that I outlined before. Is that right?






    functional-analysis







    share|cite|improve this question



























      0












      0








      0







      Let $L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$. For every $f in L^2[-a,a]$, I want to find its projection and its distance from $L_d$.



      Now, I think the fact that $f(x) = frac{f(x) + f(-x)}{2} + frac{f(x) - f(-x)}{2}$ might be relevant to the problem.



      Of course, If f is indeed an odd function, then its distance from $L_d$ is actually 0 and it doesn't have a projection. But otherwise I do not know how to proceed.



      I guess the projection on $L_d$ is the "odd part" of the function, but I am not sure how to prove that formally.



      EDIT: My solution is the following:



      Projection of f on $L_d$ = -f(-x) because I just put f(x) = -f(-x) in the expression of f that I outlined before. Is that right?






      functional-analysis







      share|cite|improve this question















      Let $L_d = { phi in L_{[-a,a]}^2$ $vert$ $phi(t) = -phi(-t)$ $forall t in [-a,a] }$. For every $f in L^2[-a,a]$, I want to find its projection and its distance from $L_d$.



      Now, I think the fact that $f(x) = frac{f(x) + f(-x)}{2} + frac{f(x) - f(-x)}{2}$ might be relevant to the problem.



      Of course, If f is indeed an odd function, then its distance from $L_d$ is actually 0 and it doesn't have a projection. But otherwise I do not know how to proceed.



      I guess the projection on $L_d$ is the "odd part" of the function, but I am not sure how to prove that formally.



      EDIT: My solution is the following:



      Projection of f on $L_d$ = -f(-x) because I just put f(x) = -f(-x) in the expression of f that I outlined before. Is that right?






      functional-analysis




      functional-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 20:36

























      asked Nov 27 '18 at 20:01









      qcc101

      474113




      474113






















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          The orthogonal projection $Pphi$ onto $L_d$ is the unique $psi in L_d$ such that $(phi -psi)perp L_d$. So,
          $$
          Pphi = frac{phi(t)-phi(-t)}{2},
          $$



          because $(phi - Pphi)$ is even, which makes it orthogonal to $L_d$. So your guess is correct: The projection into the odd functions is the odd part.






          share|cite|improve this answer





















          • Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
            – qcc101
            Nov 28 '18 at 7:05










          • The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
            – DisintegratingByParts
            Nov 28 '18 at 12:36











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          The orthogonal projection $Pphi$ onto $L_d$ is the unique $psi in L_d$ such that $(phi -psi)perp L_d$. So,
          $$
          Pphi = frac{phi(t)-phi(-t)}{2},
          $$



          because $(phi - Pphi)$ is even, which makes it orthogonal to $L_d$. So your guess is correct: The projection into the odd functions is the odd part.






          share|cite|improve this answer





















          • Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
            – qcc101
            Nov 28 '18 at 7:05










          • The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
            – DisintegratingByParts
            Nov 28 '18 at 12:36
















          1














          The orthogonal projection $Pphi$ onto $L_d$ is the unique $psi in L_d$ such that $(phi -psi)perp L_d$. So,
          $$
          Pphi = frac{phi(t)-phi(-t)}{2},
          $$



          because $(phi - Pphi)$ is even, which makes it orthogonal to $L_d$. So your guess is correct: The projection into the odd functions is the odd part.






          share|cite|improve this answer





















          • Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
            – qcc101
            Nov 28 '18 at 7:05










          • The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
            – DisintegratingByParts
            Nov 28 '18 at 12:36














          1












          1








          1






          The orthogonal projection $Pphi$ onto $L_d$ is the unique $psi in L_d$ such that $(phi -psi)perp L_d$. So,
          $$
          Pphi = frac{phi(t)-phi(-t)}{2},
          $$



          because $(phi - Pphi)$ is even, which makes it orthogonal to $L_d$. So your guess is correct: The projection into the odd functions is the odd part.






          share|cite|improve this answer












          The orthogonal projection $Pphi$ onto $L_d$ is the unique $psi in L_d$ such that $(phi -psi)perp L_d$. So,
          $$
          Pphi = frac{phi(t)-phi(-t)}{2},
          $$



          because $(phi - Pphi)$ is even, which makes it orthogonal to $L_d$. So your guess is correct: The projection into the odd functions is the odd part.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 22:09









          DisintegratingByParts

          58.7k42579




          58.7k42579












          • Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
            – qcc101
            Nov 28 '18 at 7:05










          • The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
            – DisintegratingByParts
            Nov 28 '18 at 12:36


















          • Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
            – qcc101
            Nov 28 '18 at 7:05










          • The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
            – DisintegratingByParts
            Nov 28 '18 at 12:36
















          Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
          – qcc101
          Nov 28 '18 at 7:05




          Ok, so is the projection of $f(x)$ just $-f(-x)$ or is it $frac{f(x) - f(-x)}{2}$?
          – qcc101
          Nov 28 '18 at 7:05












          The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
          – DisintegratingByParts
          Nov 28 '18 at 12:36




          The projection of $f$ onto $L_d$ must be an odd function because $L_d$ consists of odd functions. The projection of $f$ onto $L_d$ is $frac{f(x)-f(-x)}{2}$. And, indeed, $f-frac{f(x)-f(-x)}{2}=frac{f(x)+f(-x)}{2}$ is an even function, which is orthogonal to all the odd functions; so that checks out.
          – DisintegratingByParts
          Nov 28 '18 at 12:36


















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