Point spectrum of weighted shift operator on function space. Consider the space $C_{0}[0,\infty)$ and the operator $(Tf)(t)=\lambda f(t+a)$ where $a,\lambda>1$. I’m trying to find the point spectrum of $T$, and by definition I need to find values of $w$ For which $$T-wI \qquad (1)$$ is not one to one. 
I know that $g(t)= e^{\beta t}$ is an eigenfunction of $T$ where $Re(\beta)<0$ (so that $g$ remains in $C_0$) with eigenvalue $\lambda e^{\beta a}$ since $Tg=\lambda e^{\beta(t+a)}$.
In sequence spaces the idea seems straight forward for finding the point spectrum of shift operators however for function spaces it I’m having difficulty applying the definition and using (1) to find such $w$.
 A: If $\ \lambda f(t+a)=w f(t)\ $ for all $\ t\ge 0\ $, and $\ f(0)\ne 0\ $, then taking $\ t\in [0,a)\ $,   we get $\ f(t+a) = \frac{w}{\lambda} f(t)\ $, and then $\ f(t+na) = 
\left(\frac{w}{\lambda}\right)^n f(t)\ $ for $\ n=2,3,\dots $, by induction.  For $\ f\ $ to belong to $\ C_0[0,\infty)\ $ we require  $\ \left\vert w\right\vert<\lambda\ $.
Conversely, if  $\ \left\vert w\right\vert<\lambda\ $, and $\ g\ $ is any continuous function on $\ [0, a]\ $ with $\ g(0)\ne0\ $, and $\ g(a)= \frac{w}{\lambda }g(0)\ $, and is extended to $\ [0,\infty)\ $ by putting $\ g(t+na)= \left(\frac{w}{\lambda}\right)^ng(t)\ $ for all $\ t\in [0,a)\ $ and $\ n=1,2,\dots\ $, then $\ g\in C_0[0,\infty)\ $, $\lambda g(t+a)=w g(t)\ $, for all $\ t\ge0\ $, so $\ g\ $ is an eigenfunction of $\ T\ $ with eigenvalue $\ w\ $, and $\ w\ $ is in the point spectrum of $\ T\ $.
On the other hand, if $\ \left\vert w\right\vert\ge \lambda\ $ and $\ \lambda f(t+a)=w f(t)\ $, with $\ f\in C_0[0,\infty)\ $, then by the same argument as above, we have $\ \left\vert f(t)\right\vert\ = \left\vert \left(\frac{\lambda}{w}\right)^nf(t+na)\right\vert\ \le \left\vert f(t+na)\right\vert\ $, for $\ t\in[0,\infty)\ $ (not just $\ t\in[0,a)\ $), and $\ n=1,2, \dots\ $.  Since $\ \lim_\limits{n\rightarrow\infty}f(t+na)=0\ $, it follows that $\ f(t)=0\ $ for all $\ t\in[0,\infty)\ $.  So $\ T-wI\ $ is one to one, and $\ w\ $ is not in the point spectrum of $\ T\ $.
