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Thread: The adjoint of unbounded operator

  1. #1
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    The adjoint of unbounded operator

    Hello. I'm trying to find the adjoint of A, where A is unbounded operator in L^2[0,1], D(A)=C[0,1], (Af)t=f(0).
    I think, A^* is operator A^*g=g. Is this correct? And what is D(A^*)?
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  2. #2
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    Quote Originally Posted by karkusha View Post
    Hello. I'm trying to find the adjoint of A, where A is unbounded operator in L^2[0,1], D(A)=C[0,1], (Af)t=f(0).
    I think, A^* is operator A^*g=g. Is this correct? And what is D(A^*)?
    If g\in D(A^*), with A^*(g) = h, then \langle Af,g\rangle = \langle f,h\rangle for all f in C[0,1]. Writing the inner products as integrals, \displaystyle\int_0^1\!\!\!f(0)\overline{g(t)}\,dt = \int_0^1\!\!\! f(t)\overline{h(t)}\,dt. In particular, if f(0) = 0 then \displaystyle\int_0^1\!\!\! f(t)\overline{h(t)}\,dt = 0. But continuous functions vanishing at 0 are dense in L^2[0,1], so if h is orthogonal to all such functions then h = 0.

    So I reckon that A^* is the zero operator. Next, what is the domain of A^*? Well, if g\in D(A^*) and A^*(g) = 0 then \displaystyle\int_0^1\!\!\!f(0)\overline{g(t)}\,dt = 0 for all f in C[0,1]. That condition is equivalent to \displaystyle\int_0^1\!\!\!g(t)\,dt = 0. So it looks as though the domain of A^* is \{g\in L^2[0,1] : \langle g,1\rangle = 0\} (where 1 denotes the constant function 1 on [0,1]).
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  3. #3
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    Yes, you are right! Thank you very much!
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