Let $H = L^2[0,1]$. Define an operator $K \in B(H)$ by $Kf(x) := x^2 \cdot \int_{0}^{1} y f(y) \; \text{d} y$. Show that $K$ is compact and compute its spectrum.
I already showed that $ ||K || \leq \sqrt{\frac{1}{15}}$ and as K is an bounded integraloperator with continuous kernel $k(x,y) = x^2y$ it is also compact since it can be approximated by finite rank operators. To compute the spectrum we now want to know for which $\lambda \in \mathbb{C}$ the operator $ ( \lambda Id - K ) $ is not invertible. So I tried to solve the equation $ \lambda f(x) - x^2 \int_0^1 y f(y) \; \text{d} y = g(x) $ for $f(x)$ but I didn't get anything useful on the paper and I dont know if thats the right way to do it.
How can I continue from here? Thank you very much !
As observed by Ruy in the comment, $Kf \in \mbox{Span}(x^2)$ for all $f \in L^2([0,1])$ and thus the unique nonzero eigenvalue $\lambda=\frac 1 4$.
In fact $K x^2=x^2 \int_0^1y^3dy=x^2 \frac 1 4$. This also prove that $K$ is compact.