Discussion
If K(\(t\), \(s\)) is symmetric kernel of the Fredholm integral equation. Define the operator K by :
\[ K\phi = \int_{a}^{b} K(t, s)\, \phi(s)\, ds \]
and consider the following statements :
(I) K is self-adjoint operator.
(II) Inner product \(\langle K\phi, \phi \rangle\) is always real.
Then :
Answer: C
Both (I) and (II) are true.
For a real symmetric kernel, \(K(t, s) = K(s, t)\), so
\[ \begin{aligned} \langle K\phi, \psi \rangle &= \int\!\!\int K(t, s)\phi(s)\psi(t)\, ds\, dt \\ &= \langle \phi, K\psi \rangle \end{aligned} \]
which makes K self-adjoint (I). For a self-adjoint operator, \(\langle K\phi, \phi \rangle = \langle \phi, K\phi \rangle\), which is \(\overline{\langle K\phi, \phi \rangle}\), so the inner product is real (II). This is why symmetric kernels have real characteristic values and orthogonal eigenfunctions.
Since both statements hold, the other options are wrong.
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