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Injective operator in L^2(0,1)

  1. Jan 6, 2010 #1
    Hi!

    Define an integral operator K: L^2 (0,1) -> L^2(0,1) by:

    Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from t=0 to t=1].

    Why is "obvious" that K is a one-to-one operator?

    I know K is one to one if Kx(t) = 0 implies x(t) = 0 but I don't see why this is true. Can you please explain why?
     
  2. jcsd
  3. Jan 6, 2010 #2

    Landau

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    Please use Latex. Do you mean the following? (you wrote 'from t=0 to t=1' while s is your integration variable)

    [tex]Kx(t)=\int_0^1 (1+ts)\exp(ts)x(s)\mbox{d}s[/tex]
     
  4. Jan 6, 2010 #3
    Yes, sorry about that. It is from s=0 to s=1.
     
  5. Jan 12, 2010 #4
    [tex]L^2(0,1)[/tex] is a Hilbert separable space with inner product
    [tex] <u,v> = \int_0^1 uv dx [/tex]​
    and the set of Legendre polynomials is a Hilbert base of [tex]L^2(0,1)[/tex]
    [tex]P_n (x) = \frac{\sqrt{2n+1}}{ 2^{n+1/2} n!} \frac{d^n}{dx^n} \left[ (x^2 - 1)^n \right] \; \; , \; n = 1,2,3,... [/tex]​
    It mean
    [tex]\forall v \in L^2(0,1) , v = \sum_{n=1}^{\infty} v_n P_n [/tex]​
    where [tex] v_n = <v, P_n> [/tex].

    Now we assume [tex] x \in L^2 (0,1) \, , \, Kx = 0 [/tex] and try to prove that [tex]x = 0[/tex].

    First put [tex] k(t,s) =(1+ts)\exp(ts) [/tex], by applying the dominate convergence theorem, we have
    [tex] \frac{d}{dt} \int _0^1 k(s,t) x(s) ds = \int _0^1 \frac{\partial}{\partial t}k(s,t) x(s) ds [/tex]​
    and further we have
    [tex] \frac{d^n Kx}{dt^n} (0) = \left. \left( \frac{d^n}{dt^n} \int _0^1 k(s,t) x(s) ds \right) \right|_{t= 0}= \int _0^1 \left. \left( \frac{\partial^n}{\partial t^n}k(s,t) \right) \right|_{t=0} x(s) ds \;\;\; (1) [/tex]​
    by applying the induction principle we get
    [tex] \frac{\partial^n}{\partial t^n}k(s,t) = [(n+1)s^n + ts^{n+1} ] \exp(ts) \; ,\; \forall n = 1,2,3,... \;\;\; (2)[/tex]​
    Combining (1)(2) and assumption Kx = 0, we get then
    [tex] \int_{0}^{1} s^n x(s) = 0 \; , \; \forall n = 1,2,3, ... \;\;\; \;\;\; (*)[/tex]​
    From [tex](*) [/tex] we have
    [tex] x_n = <x, P_n> = 0 \; ,\; \forall n = 1,2,3,... [/tex]​
    finally
    [tex] x = \sum_{n=1}^{\infty} x_n P_n= 0 [/tex]​
     
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