I am a bit stuck on the following problem, and would like some help in the right direction, not a complete solution.
Let $K:[0,1] \times [0,1] \to \mathbb{R}$ be the characteristic function of $\{(x,y) ~:~ y \geq -x+1\}$. I wish to find the spectrum of $T_K:L^2([0,1]) \to L^2([0,1])$ where $$(T_Kf)(x)=\int_0^1 K(x,y) f(y) \ dy = \int_{-x+1}^1 f(y) \ dy.$$ I know that $\sigma(T_K)=\{\lambda ~:~ T_K-\lambda I \text{ is not invertible}\}$. So I need to find all $\lambda$ such that either:
- there exists $f$ such that $T_Kf=\lambda f$ (i.e., $\lambda$ is an eigenvalue of $T_K$); or
- there exists $g$ such that $g \neq T_K f-\lambda f$ for any $f$ (i.e., $T_K-\lambda I$ is not surjective).
For the first bullet point, $T_Kf=\lambda f$ implies $\int_{-x+1}^1 f(y) \ dy = \lambda f(x)$ for all $x \in [0,1]$. The LHS is continuous, so the RHS is continuous. Then in fact the LHS is differentiable, so after differentiating we obtain: $f(-x+1)=\lambda f'(x)$. Also, plugging in $x=0$ into the integral equation gives $0=\lambda f(0)$, so either $\lambda=0$ or $f(0)=0$. If $\lambda=0$ then we have $\int_{-x+1}^1f(y) \ dy = 0$ for all $x\in [0,1]$, so in fact $f=0$. Otherwise, $0=f(0)=f(-1+1)=\lambda f'(1)$, and $\lambda \neq 0$, so $f'(1)=0$. I am not sure how to proceed in solving for possible values of $\lambda$.
EDIT: Differentiating $f(-x+1)=\lambda f'(x)$ again gives $-f'(-x+1)=\lambda f''(x)$. Also, replacing $x$ with $-x+1$ in $f(-x+1)=\lambda f'(x)$ gives $f(x)=\lambda f'(-x+1)$. Equating the two expressions for $f'(-x+1)$, we get $f''(x)=-\frac{1}{\lambda^2}f(x)$, so $f(x)=A\cos\left(\frac{x}{\lambda}\right)+B \sin\left(\frac{x}{\lambda}\right)$. Then using $f(0)=0$, we know $A=0$, and using $f'(1)=0$, we find $\frac{1}{\lambda} = \frac{\pi}{2}+k\pi$, $k \in \mathbb{Z}$.
For the second bullet point, I really have no idea where to start.
Not a complete solution:
Start by proving that $T_K$ is a compact operator. You can do this by showing that $T_K$ is a limit of finite-rank operators: Let $(f_n)$ be an ortonormal basis of $L^2([0,1])$ and then define $$ T_n(f)(x) = \int_0^1\sum_{i=1}^n\sum_{j=1}^n a_{ij} f_i(x)f_j(y)f(y) \; dy $$ where $$ a_{ij}=\int_0^1\int_0^1 K(x,y)f_i(x)f_j(y) \; dxdy. $$ See section 4.5a of Functional Analysis, an introduction by Eidelman, Milman, Tsolomis, for example.
Now, as $T_K$ is compact, from the spectral theory for compact operators, we know that $\sigma(T_K)=\{0\}\cup \sigma_p(T_K)$, where $\sigma_p(T_K)$ is the point spectrum (i.e. the set of eigenvalues) of $T_K$. So the problem reduces to find the eigenvalues of $T_K$.
I'll keep trying to complete this problem!