Finding the adjoint

**anegligibleperson** · August 14th 2012, 11:55 AM

Given the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined by
$(Tf)(x) = \int^x_0 f(t) \ dt$
How do I find adjoint of $T$ ? I'm totally lost as to how to go about doing this. Thanks in advance!

**HallsofIvy** · August 14th 2012, 01:43 PM

In an inner product space, the "adjoint" of a linear transformation, A, is the linear transformation A* such that <Au, v>= <u, A*v>. Here the inner product is $<u(x), v(x)>= \int_0^1 u(x)v(x)dx$ . So you are looking for A* such that $<Au, v>= \int_0^1 \left(\int_0^x u(t)dt\right)v(x)dx= \int_0^1 u(x)A*(v(x))dx$ . Do the integrals on the left and find the function A* that makes the right side equal to that. I would suggest using "integration by parts".

**anegligibleperson** · August 14th 2012, 02:12 PM

@HallsofIvy, could you be more explicit? I get $\int_0^xu(t)dtV(x)|_0^1 - \int_0^1u(x)V(x)dx$ then I don't know how to proceed.

**Vlasev** · August 15th 2012, 04:19 AM

You actually obtain

$\int_0^1 U(x)v(x)dx = U(x)V(x)|_0^1-\int_0^1u(x)V(x)dx$

That is

$\langle u, T^*v\rangle = \langle Tu,v\rangle = U(x)V(x)|_0^1 - \langle u,Tv\rangle$

This suggest that you need to write the product $U(x)V(x)$ in a form with $u$ in it. That is, you need to make it look more like $\langle u,Tv\rangle$ and somehow combine the two. I hope this helps.

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