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Orthornormal basis in L^([a,b])

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data
    [itex](e_n)[/itex] is orthonormal basis for [itex]L^2([0,1])[/itex].

    Want to show that [itex](f_n)[/itex] is basis for [itex]L^2([a,b])[/itex] when [itex]f_n(u) = (b-a)^{-1/2}e_n(\frac{u-a}{b-a})[/itex]


    2. Relevant equations
    [itex]f_n(u) = (b-a)^{-1/2}e_n(\frac{u-a}{b-a})[/itex]


    3. The attempt at a solution
    I did show that [itex] (f_n) [/itex] is an orthonormal sequence. But how can I show that it is also a basis...
     
  2. jcsd
  3. Nov 3, 2008 #2

    HallsofIvy

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    Since they are orthonormal they are necessarily independent so you only need to show that they span the space. Given any f in L2([a, b]) you need to show that f is equal to [itex]\sum a_n f_n[/itex]. Use your "relevant equation" to rewrite that in terms of en and use the fact that {en} spans L2([0, 1]).
     
  4. Nov 3, 2008 #3
    but then i get
    [itex] \sum a_n f_n(u) = (b-a)^{-1/2}\sum a_n e_n( (u-a)/(b-a) ) [/itex]

    Can I then just say that since [itex]e_n[/itex] spans [itex]L^2([0,1])[/itex] then [itex] e_n( (u-a)/(b-a)) [/itex] spans [itex] L^2([a,b]) [/itex] so the above equals [itex]f[/itex] for the right choice of [itex]a_n[/itex].

    Or am I missing something

    Thanks
     
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