Let $E$ be the vector space of real valued continuous functions on an interval $[a,b]$. Let $K = K(x,y)$ be a continuous function of two variables, defined on the square $a \leq x \leq b$ and $a \leq y \leq b$. An element $f$ of $E$ is said to be an eigenfunction for $K$, with respect to a real number $r$, if $$f(y) = r\int_{a}^{b} K(x,y)f(x)\,dx.$$ We take $E$ with the $L^{2}$-norm of the hermitian product given by $$\langle f,g\rangle = \int_{a}^{b} fg.$$ Prove that if $f_{1}, \dotsc, f_{n}$ are in $E$, mutually orthogonal, and of $L^{2}$-nrom equal to 1, and if they are eigenfunctions with respect to the same number $r$, then $n$ is bounded by a number depending only on $K$ and $r$. [Hint: Aply Bessel's inequality.]
I assume that I am supposed to cleverly choose an element of $E$ and then use Bessel's inequality, but I have not found the proper element. I tried setting $g=\sum f_{i}$ and showing that $g$ is also an eigenfunction for $(K,r)$. But I wasn't able to get much further. Using the orthogonality of the $f_{i}$, I did show that $\lvert g\rvert^{2} = n$, but this follows from the Pythagoras Theorem.
If you mean the eigenfunctions of an integral operator $K$ with kernel $K(x,y)$ and if $K(x,y)$ is assumed, say, to be continuous in $(x,y)$, then the operator $K$ is compact and hence any eigenvalue $\lambda$ has finite multiplicity (because $K\lambda$ is compact and at the same time is the identity operator on the eigenspace).