Given the operator $\displaystyle T: L^2([0,1]) \rightarrow L^2([0,1]) $ defined by

$\displaystyle (Tf)(x) = \int^x_0 f(t) \ dt$

How do I find adjoint of $\displaystyle T$? I'm totally lost as to how to go about doing this. Thanks in advance!

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- Aug 14th 2012, 11:55 AManegligiblepersonFinding the adjoint
Given the operator $\displaystyle T: L^2([0,1]) \rightarrow L^2([0,1]) $ defined by

$\displaystyle (Tf)(x) = \int^x_0 f(t) \ dt$

How do I find adjoint of $\displaystyle T$? I'm totally lost as to how to go about doing this. Thanks in advance! - Aug 14th 2012, 01:43 PMHallsofIvyRe: Finding the adjoint
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 $\displaystyle <u(x), v(x)>= \int_0^1 u(x)v(x)dx$. So you are looking for A* such that $\displaystyle <Au, v>= \int_0^1 \left(\int_0^x u(t)dt\right)v(x)dx= \int_0^1 u(x)A*(v(x))dx$. 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".

- Aug 14th 2012, 02:12 PManegligiblepersonRe: Finding the adjoint
@HallsofIvy, could you be more explicit? I get $\displaystyle \int_0^xu(t)dtV(x)|_0^1 - \int_0^1u(x)V(x)dx$ then I don't know how to proceed.

- Aug 15th 2012, 04:19 AMVlasevRe: Finding the adjoint
You actually obtain

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

That is

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

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