Finding a spectrum of an operator in $C[0, 1]$

Question

Finding a spectrum of an operator in $C[0, 1]$: $$ (A f)(x) = \frac{x-1}{x-2}f(x) + f(0) $$

Could you please give me a hint on how to start solving such a problem?

Answer 1

Those $\lambda$ not in $\sigma(A)$ should be those such that $A-\lambda I$ is invertible. So let us try to invert $A-\lambda I$. Given $g\in C[0,1]$, we want to find $f\in C[0,1]$ such that $(A-\lambda I)f=g$. That is, $$\tag1 g(x)=\left(\frac{x-1}{x-2}-\lambda\right)\,f(x)+f(0). $$ Taking $x=1$, we get $$g(1)=-\lambda f(1)+f(0).$$ At $x=0$, $$\tag2 g(0)=\left(\frac12-\lambda\right)f(0)+f(0)=\left(\frac32-\lambda\right)f(0). $$ The condition $\lambda=\frac32$ would force $g(0)=0$, which is not true for most $g$: that tells us that $\frac32\in\sigma(A)$. When $\lambda\ne\frac32$, from $(2)$ we obtain $$\tag3 f(0)=\frac{g(0)}{\frac32-\lambda}. $$ Going back to $(1)$, we get $$ f(x)=\frac{g(x)-f(0)}{\frac{x-1}{x-2}-\lambda}=\frac{g(x)-\frac{g(0)}{\frac32-\lambda}}{\frac{x-1}{x-2}-\lambda}. $$ This requires that $\lambda$ is not in the range of $\frac{x-1}{x-2}$, which is $[0,\tfrac12]$.

So $$\sigma(A)\subset\left[0,\tfrac12\right]\cup\left\{\tfrac32\right\}.$$ We can confirm that the spectrum is not smaller: we already said that $\tfrac32\in\sigma(A)$. For any $\lambda\in[0,\tfrac12]$, if we look at $(1)$ we will get that there exists $x_0$ with $\tfrac{x-1}{x-2}-\lambda=0$. So $g(x_0)=f(0)$. Comparing with $(2)$, we get $$\tag4 g(0)=\left(\frac32-\lambda\right)\,g(x_0). $$ There are many functions $g\in C[0,1]$ that do not satisfy $(4)$, so $A-\lambda I$ is not invertible. Thus $$ \sigma(A)=[0,\tfrac12]\cup\left\{\tfrac32\right\}. $$