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

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

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)

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

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

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.

3. Thanks for that, if it is skew self adjoint is it still normal?

4. Originally Posted by ejgmath
Thanks for that, if it is skew self adjoint is it still normal?
Yes, obviously, because $T^*T = TT^* = -T^2$.