Define the bounded linear operator $K: L^2(0,1) \rightarrow L^2(0,1)$ by $$(Kf)(x) = \int_0^1 xy(1-xy)f(y) dy.$$ Find the spectrum of $K$ and classify it.
My attempt:
Since the kernel of $K$ is $k(x,y) = xy(1-xy)$, whicch is a real-valued function, $K$ must be self adjoint. Thus $\sigma_r(K) = \emptyset$. Then, since the norm $\|k(x,y)\|_{L^2}$ is finite, $K$ is a Hilbert-Schmidt operator and thus is compact. Then, $0 \in \sigma_{pt} (K)$ and the spectrum of $K$ only contains eigenvalues.
At this point, I'm not sure how to proceed. Clearly, if $\lambda \in \sigma_{pt}(K)$, then for some $f \in L^2 (0,1)$, $$Kf = \int_0^1 xy(1-xy)f(y) dy = \lambda f(x),$$ but I'm not sure how to proceed with this calculation.
Best Answer
Hint: $Kf(x)=\lambda f(x)$ can be written as $\lambda f(x)=ax+bx^{2}$ where $a=\int_0^{1}yf(y)\, dy$ and $b =\int_0^{1}y^{2}f(y)\, dy$. Now start with the equation $ f(x)=Ax+Bx^{2}$ where $A$ and $B$ are constants. Plug this into the equation $Kf(x)=\lambda f(x)$. Can you take over from here?
You should get the equations: $\frac A 3+\frac B 4 =\lambda A$ and $-(\frac A 4+\frac B 5 )=\lambda B$ by comparing coefficients of $x$ and $x^{2}$ on the two sides.