Let Af(x) = $\int_0^1 K(x,y)f(y)dy$, $A:L_2[0,1]\rightarrow L_2[0,1].$
Where $K(x,y) = \sinh(\min(x,y)\sinh(1-\max(x,y)). $ where $\sinh(x) = \frac{e^x - e^{-x}}{2}$
Find $\sigma(A), ||A||.$
I proved that $Ker (A - \lambda E) = 0 \hspace{2mm} \forall \lambda $. So now I need to find out when $Im(A - \lambda E) = L_2[0,1]$.
Cant solve the equation $g(x) = \int_0^1 K(x,y)f(y)dy$.
Any help welcome.
You can write unravel this definition by writing \begin{align} Af(x) & = \int_{0}^{x}K(x,y)f(y)dy+\int_{x}^{1}K(x,y)f(y)dy \\ & =\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy. \end{align} This is a typical kind of Green function expression. First note that $$ (Af)(0) = 0,\;\;\; (Af)(1)=0. $$ Then, \begin{align} (Af)'(x) = &-\cosh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(1-x)\sinh(x)f(x) \\ &+\cosh(x)\int_{x}^{1}\sinh(1-y)f(y)dy -\sinh(x)\sinh(1-x)f(x) \\ =& -\cosh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\cosh(x)\int_{x}^{1}\sinh(1-y)f(y)dy \\ (Af)''(x) = & +\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy-\cosh(1-x)\sinh(x)f(x) \\ &+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy-\cosh(x)\sinh(1-x)f(x) \\ = & (Af)(x) -\{ \cosh(1-x)\sinh(x)+\cosh(x)\sinh(1-x) \}f(x) \end{align} So you end up with $(Af)''+(Af) = Cf$ where $C=-\sinh(1)$. In other words, $$ \frac{1}{\sinh(1)}\left(-\frac{d^2}{dx^2}-I\right)Af = f \\ (Af)(0) = 0,\;\;\; (Af)(1) = 0. $$ $A$ is, therefore, the inverse of a differential operator. The differential operator $-\frac{d^2}{dx^2}$ with $0$ endpoint conditions has a complete orthogonal basis of eigenfunctions $\sin(\pi x),\sin(2\pi x),\sin(3\pi x),\ldots$ with eigenvalues $\pi^2n^2$ for $n=1,2,3,\cdots$. You can get the spectrum of $A$ through spectral mapping. $A$ is selfadjoint and Fredholm. So $\mathcal{R}(A-\lambda I)=L^2[0,1]$ iff $\mathcal{N}(A-\lambda I)= \{0\}$, and the latter condition is violated iff $\lambda$ is an eigenvalue. You'll also have to look at the cluster point of the spectrum, too.