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Thread: Adjoint of T

  1. #1
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    Adjoint of T

    If a bounded operator T:L^2([0,1],\mathbb{C})\rightarrow L^2([0,1],\mathbb{C}) is defined by

    Tf(x)=\int^{1}_{0}(x-y)f(y)dy

    What is its adjoint?

    I'm assuming the usual inner product on L^2

    \langle f,g\rangle=\int^{1}_{0}f(x)\overline{g(x)}dx

    Can anyone give me some help finding this? I'm not too good at manipulating integrals. (Not asking for the definition of adjoint)
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    If a bounded operator T:L^2([0,1],\mathbb{C})\rightarrow L^2([0,1],\mathbb{C}) is defined by

    Tf(x)=\int^{1}_{0}(x-y)f(y)dy

    What is its adjoint?

    I'm assuming the usual inner product on L^2

    \langle f,g\rangle=\int^{1}_{0}f(x)\overline{g(x)}dx

    Can anyone give me some help finding this? I'm not too good at manipulating integrals. (Not asking for the definition of adjoint)
    Start with the adjoint equation: \langle T^*f,g\rangle = \langle f,Tg\rangle.

    The left side is \langle T^*f,g\rangle = \int_0^1 (T^*f)(y)\overline{g(y)}\,dy.\qquad(1)

    The right side is

    \begin{aligned}\langle f,Tg\rangle = \int_0^1f(x)\overline{(Tg)(x)}\,dx &= \int_0^1f(x)\int_0^1(x-y)\overline{g(y)}\,dydx \\ &= \int_0^1\!\!\!\int_0^1(x-y)f(x)\overline{g(y)}\,dxdy \\ &= \int_0^1\biggl(\int_0^1(x-y)f(x)\,dx\biggr)\overline{g(y)}\,dy\qquad(2)\end{  aligned}

    (changing the order of integration). Compare (1) and (2) to see that (T^*f)(y) = \int_0^1(x-y)f(x)\,dx. Notice that this is equal to -(Tf)(y), so that T^* = -T, in other words T is skew-adjoint.
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  3. #3
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    Thanks for that, if it is skew self adjoint is it still normal?
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  4. #4
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    Quote Originally Posted by ejgmath View Post
    Thanks for that, if it is skew self adjoint is it still normal?
    Yes, obviously, because T^*T = TT^* = -T^2.
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