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Let $K\in L^2([a,b]\times[a,b])$, $K(s,t)=\overline K(t,s)$,

define $Tf(s)=\int_a^bK(s,t)\bar f(t)\,dt$ for $f \in L^2([a,b])$

I need to show that the eigenfunctions of $T$ are an orthonormal basis for $L_2([a,b])$

I tried to show $T$ is self-adjoint (it isn't because of the conjugate on $f$ in the intergral) and that it is normal - $TT^*=T^*T$ (also doesnt work)

is there another way to show?

Michael Hardy
  • 1
Benuci
  • 33
  • since it's self-adjoint compact its eigenfunctions form a basis of the underlying space – Gustave Jun 26 '21 at 19:49
  • thats the issue, its not self-adjoint – Benuci Jun 26 '21 at 20:02
  • What's your definition of the inner product? – Brian Borchers Jun 26 '21 at 20:38
  • If the inner product is defined by $\displaystyle\langle f,g\rangle = \int_a^b f(s)\overline{g(s)} , ds,$ then you have $$ \begin{align} & \langle Tf,g\rangle = \int_a^b (Tf)(s)\overline{g(s)}, ds = \int_a^b\left( \int_a^b K(s,t) \overline {f(t)} , dt \right) \overline {g(s)} , ds \ {} \ = {} & \int_a^b \left( \int_a^b K(s,t)\overline{g(s)} , ds \right) \overline{f(t)} , dt = \int_a^b (Tg)(t)\overline {f(t)} , dt \ {} \ = {} & \langle Tg,f\rangle = \overline{ \langle f, Tg\rangle}. \end{align} $$ – Michael Hardy Jun 26 '21 at 21:03

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