Why $\operatorname{Im}(\lambda-A)\neq C_{[0,1]}$?

Question

Example. Study the spectrum of the following operator: $Ax(t)=tx(t)$

$tx(t)=\lambda x(t)\\x(t)(t-\lambda)=0\:\:\:t\in[0,1]$

$x(t)\equiv 0$ and $\lambda\notin[0,1]$

Now let's see the image.

$(t-\lambda)x(t)=y(t)\:\:\:(t-\lambda)\neq 0$

$x(t)=\frac{y(t)}{t-\lambda}\:\:\lambda\in[0,1]$

$x(t)(t-\lambda)=y(t)$, then $y(\lambda)=0$ and $y_0(t)\equiv 1$ which implies $\rho(y,y_0)>1$.

Therefore $\operatorname{Im}(\lambda-A)\neq C_{[0,1]}$ then $\lambda$ is residual.

Question:

I am having a hard time understanding this proof, more specifically why $\rho(y,y_0)>1$. What does $\rho(y,y_0)>1$ mean? Is it a metric? What was the procedure used on this example?

Thanks in advance!

Answer 1

$\rho$ is the supremum metric on $C[0,1]$. Note that if $Ax(t)=\lambda x(t)$ and $x(t)$ is not the zero function then $\lambda \in [0,1]$ necessarily. Nor $\rho (y_0,y) \geq |y_0(\lambda)-y(\lambda)| =|1-0|=1$. This proves that any function $y$ in the image has distance at least 1 from $y_0$. In particular $y_0$ itself is not in the image.