We are in $L^2 \left[a,b\right]$ space. We define our operator $A$ as follows: $$A = \frac{1}{i} \frac{d}{dx}$$ The domain of our operator $A$ is equal to $\text{dom } A = \left\{ u \in L^2 \left[a,b\right] \: : \: u(a) = 0 \right\}$
Question: What is the spectrum of our operator $A$?
My attempt
We need to find those eigenvalues ($\lambda \in \mathbb{C}$) for which $(A - \lambda I)$ is not invertible. In order to find them, we instead try to find those for which $(A - \lambda I)$ is actually invertible. So we get: $$(A - \lambda I) \cdot u(x) = v(x)$$ We set $v(x) = 1$, so we have $$(\frac{1}{i} \frac{d}{dx} - \lambda) \cdot u(x) = 1$$ This will get us: $$\frac{1}{i} \frac{du}{dx} - \lambda \cdot u = 1$$ which is $$\frac{1}{i} u'(x) - \lambda u(x) = 1$$ This gives us the solution: $$u(x) = -\frac{1}{\lambda} + C e^{i \lambda x}$$ We also know, that $u(a) = 0$. So we get: $$0= u(a) = -\frac{1}{\lambda} + C e^{i \lambda a}$$ which is true for $C = \frac{1}{\lambda e^{i \lambda a}}$.
Ultimatelly, we get: $$u(x) = -\frac{1}{\lambda} + \frac{1}{\lambda e^{i \lambda a}} e^{i \lambda x}$$ Now the question is: Is $u(x) \in L^2 \left[a,b\right]$? Or in other words: For which $\lambda$ is $u(x) \in L^2 \left[a,b\right]$?
That's with what I'm struggling with, especially because we have the imaginary unit here. We defined that a function $u(x) \in L^2 \left[a,b\right]$, if $u(x)$ is measurable and $$\int_a^b |u(x)|^2 dx < \infty$$
So how can I show (for which $\lambda$) our $u(x) \in L^2 \left[a,b\right]$?
Maybe I should elaborate why it is important to find those $\lambda$: Those lambdas for which $u(x) \notin L^2 \left[a,b\right]$ are my spectrum, because then for those lambdas, $(A - \lambda I)$ wouldn't be invertible in the domain of $A$.
I already apologize if there are some misunderstandings here, we just started the topic of Spectral Theory and our Prof gave us this as an exercise.
The resolvent equation is found by solving $$ (A-\lambda I)f=g $$ which translates to solving the following ODE for $f$: $$ \frac{1}{i}f'-\lambda f = g,\;\; f(a)=0. $$ That is, $$ f'-i\lambda f=ig \\ e^{-i\lambda x}f'(x)-i\lambda e^{-i\lambda x}f(x)=ie^{-i\lambda x}g(x) \\ (e^{-i\lambda x}f(x))'=ie^{-i\lambda x}g(x) $$ Integrating over $[a,x]$ gives $$ e^{-i\lambda x}f(x)=i\int_a^xe^{-i\lambda x'}g(x')dx' \\ f(x)=i\int_a^x e^{i\lambda(x-x')}g(x')dx' $$ So the resolvent operator $R(\lambda)$ exists for all $\lambda\in\mathbb{C}$ and $$ R(\lambda)g= i\int_a^x e^{i\lambda(x-x')}g(x')dx'. $$ To see that $R(\lambda)$ is a bounded linear operator for all $\lambda\in\mathbb{C}$, you can apply the Cauchy-Schwarz inequality: \begin{align} |(R(\lambda)g)(x)|^2 &\le \int_a^x|e^{i\lambda(x-x')}|^2dx'\int_a^x|g(x')|^2dx' \\ & \le\int_a^x e^{-2\Im\lambda(x-x')}dx'\|g\|^2 \\ \|R(\lambda)g\|^2 &\le\left(\int_a^b\int_a^x e^{-2\Im\lambda(x-x')}dx'dx\right)\|g\|^2. \end{align} (This proves that $R(\lambda)$ is a bounded operator on $L^2[a,b]$ for all $\lambda\in\mathbb{C}$.) Therefore the spectrum of $A$ is $\sigma(A)=\emptyset$.