# Showing an operator is self adjont

I am trying to show that the operator:

$$Tf(s)=5s^2\int_0^1t^2f(t)dt+2\int_0^1f(t)dt$$ is self adjoint where $H=L(0,1)$ with real scalars and $t\in \mathcal{L}(H)$.

So I can re-write this operator as:

$$Tf(s)=\int_0^1 (5s^2t^2+2)(f(t))dt$$ so this is in the form of an integral operator.

To show that this is self adjoint i just need to show that:

$\int_0^1 K(s,t)f(t)dt=\int_0^1\overline{K(s,t)}f(t)dt$

But then as $\overline{(5s^2t^2+2)}=5s^2t^2+2$ we are done?

Is the above correct, it seems to simple?

Thanks for any help

Edit:

I am using the following:

Integral operators are of the form:

$Tf(s)=\int_0^1 K(s,t)f(t)dt$

So in order to calculate the adjoint we have that:

$\langle Tf,g\rangle=\int_0^1 Tf(s)\overline{g(s)}ds=\int_0^1\int_0^1 K(s,t)f(t)\overline{g(s)}dtds=\int_0^1f(t)\overline{\int_0^1\overline{K(s,t)}g(s)dsdt}=\langle f,T^*g\rangle$

This gives that the adjoint is of the form:

$\int_0^1 \overline{K(s,t)}g(t)ds$

So once I have written my operator in the form of an integral operator to show it is self adjoint I only have to show what I have above (I think).

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I am a noob in this area, but aint you trying to show <Tf, g> = <f, Tg> where <,> is the $L^2$ inner product. I don't see any $g$ floating around and I don't know what your $K$ is. – Lost1 Apr 14 '13 at 10:49
His $K$ is clearly $K(s,t)=5s^2t^2+2$. But @Lost1 is correct. You need to multiply by $g(s)$ and integrate over $s$, too. Still, that is not much work; you are almost done. – Harald Hanche-Olsen Apr 14 '13 at 11:01
@Lost I have made a few edits explaining what I am doing. – hmmmm Apr 14 '13 at 11:05
@HaraldHanche-Olsen I have made a few edits, does it now make sense? – hmmmm Apr 14 '13 at 11:06
Looks good to me. (I see a minor TeXnical error, where \bar{g(s)} should have been using \overline.) – Harald Hanche-Olsen Apr 14 '13 at 12:40