Spectrum of the follolowing operator in $C[0,1]$ I want to find a spectrum (with classification of points) of the following operator
$$
Af(x) = \frac{x-1}{x-2}f(x) + f(0),
$$
where $A:C[0,1] \to C[0,1]$.
There is only one point $\lambda = \frac{1}{2}$ with which I stuck. I give here my partial solution.
Firstly by direct calculations I proved that equation
$$
Af = \lambda f
$$
can be solved if and only if
$
\lambda = \frac{3}{2}
$.
So $$
\sigma_p(A) = \frac{3}{2} 
$$
Newxt I investigate when it is possible to solve the equation
$$
(A-\lambda I)f(x) = g(x),
$$
for arbitrary functions $g(x)\in C[0,1]$. We assume that $\lambda \neq \frac{3}{2}$. Then from
$$
\left(\frac{x-1}{x-2} - \lambda\right)f(x) + f(0) = g(x)
$$
considering $x = 0$ we obtain
$$
f(0) = \frac{g(0)}{\frac{3}{2} - \lambda}
$$
So $f(0)$ is uniquely determined. If $\lambda$ belongs to the range of $\frac{x-1}{x-2}$  then from
$$
\left(\frac{x-1}{x-2} - \lambda\right)f(x) = g(x)  - f(0)
$$
we obtain that for $x_{\lambda} = \frac{1-2\lambda}{1-\lambda}$ we have
$$
0\cdot f(x_{\lambda}) = g(x_{\lambda}) - f(0)
$$
so
$$
g\left(\frac{1-2\lambda}{1-\lambda} \right) = f(0) = \frac{g(0)}{\frac{3}{2} - \lambda}
$$
So we obtain a constration on $g(x)$. But this is a non trivial constration only for $\lambda \neq \frac{1}{2}$. So we can show that
$$
\overline{Im(A - \lambda I)} \neq C[0,1]
$$
for all $\lambda \in [0, \frac{1}{2})$.
But what can we say for $\lambda = \frac{1}{2}$? I'm stuck here...
Also I want to say that this is not a duplication of the question Finding a spectrum of an operator in $C[0, 1]$
since in my question the main task is to determine the classification of spectral points (that is only one point $\lambda = \frac{1}{2}$).
 A: $A-\frac 1 2 I$ is not surjective, hence $\frac 1 2 \in \sigma (A)$. To see this take any $g \in C[0,1]$ such that $g'(0)$ does not exist. Suppose $(A-\frac 1 2 I) f=g$. This gives $\frac {x-1}{x-2}f(x)+f(0)-\frac  12 f(x)=g(x)$. This can be written as $f(x)=\frac {2(x-2)(g(x)-f(0))} x$ for $x \neq 0$. For $f$ to be continuous at $0$ it is necessary that $f(0)=g(0)$. Also, letting $x \to 0$ we see that $g'(0)$ must exist.
Let $f$ be continuously differentiable. Let $(q_n)$ be a sequence of polynomials converging uniformly to $f'$. Let $p_n(x)=\int_0^{x}q_n(t)dt+f(0)+\frac {2A} {\sqrt n} \sqrt {x+\frac 1  n}$ where $A=-\frac 1  4[f(0+4f'(0)]$. Then $p_n'(0)=q_n(0)+A \to f'(0)+A$.  Hence $4p_n'(0)+p_n(0) \to 0$. Let $c_n=4p_n'(0)+p_n(0)$ Then $c_n \to 0$ and $p_n-c_n \in M$ where $M=\{g\in C^{1}[0,1]:4g'(0)+g(0)=0\}$.  Since $p_n-c_n \to \int_0^{x} f'(t)dt +f(0)=f(x)$ uniformly  it follows that any continuously differentiable is a uniform  limit of functions from $M$. Any continuous function is a uniform limit of functions which are continuously differentiable so $M$ is dense in $C[0,1]$. But every function in $M$ is in the range of $A-\frac 1  2 I$. It follows that $A-\frac 1  2 I$ is injective with a dense range (but not surjective). Thus, $\frac1 2$ belongs to the continuous spectrum of $A$.
