How can the spectrum of $A = \frac{d}{dx}$ in $C(\mathbb{R})$ be equal to $\mathbb{R}$, as for every $\lambda$ there exists a resolvent $R(\lambda)g$? Let's now find the eigenvalues of $A$ (which are the spectrum):
$$A f = \lambda f$$
$$f' = \lambda f$$
$$f' - \lambda f = 0$$
$$\Rightarrow \quad f(x) = C \cdot e^{\lambda x} \quad \Rightarrow \quad \forall \ \lambda \in \mathbb{R} \quad f \in C(\mathbb{R}) \quad \Rightarrow \quad \sigma (A) = \mathbb{R}$$
Now let's find the resolvent:
$$(A -  \lambda I)f = g$$
$$f' - \lambda f = g$$
$$e^{- \lambda x}f' - \lambda e^{- \lambda x}f = e^{- \lambda x} g$$ $$(e^{-\lambda x} f)' = e^{-\lambda x}g$$ $$e^{-\lambda x}f(x) - e^{-\lambda a} f(a) = \int_a^x e^{-\lambda x}g(x) dx$$ $$R(\lambda)g(x) = f(x) = e^{\lambda x} \int_a^x e^{-\lambda x} g(x) dx + e^{\lambda (x-a)} f(a) \quad \Rightarrow \quad \forall \ \lambda \in \mathbb{R} \quad f \in C(\mathbb{R})$$
But because our spectrum is equal to $\mathbb{R}$, this should mean that for no $\lambda \in \mathbb{R}$ there exists a resolvent $R(\lambda)g(x)$ (as the spectrum is also the set for which $(A - \lambda I)$ is not invertible i.e. for which the resolvent doesn't exist)
Am I mixing up definitions here? I'm really confused
 A: Several things to unpack here.

*

*The spectrum does not consists of the eigenvalues alone. It does in finite-dimension, but in general there is more stuff. The spectrum consists of those $\lambda$ such that $T-\lambda I$ is invertible and bounded, which is why what norm you use matters.


*You have an operator $A:C^1(\mathbb R)\to C(\mathbb R)$. That is, your operator is unbounded with domain $C^1(\mathbb R)$.


*Final comment: if you have no norm nor topology, you are not doing Functional Analysis nor Operator Theory.


*If the domain is all of $C^1(\mathbb R)$, then indeed every $\lambda\in\mathbb R$ is an eigenvalue.


*If there exist functions $f$ with $(A-\lambda I)f=0$, you cannot expect it to have an inverse.


*So, what's the problem? Your formula for $R(\lambda)$ uses the value $f(a)$. But $R(\lambda)$ has to evaluate on $g$; how are you supposed to know $f(a)$?


*The answer to the above question is, you don't. In these situations one usually restricts the domain, for example with the condition $f(0)=0$. Now the equation $f'-\lambda f$ has only the solution $C=0$, so no $\lambda$ is an eigenvalue. Your $R(\lambda)$ now works, and so the spectrum of $A$ is empty.


*More importantly, your $R(\lambda)$ requires $f(a)=0$ to be linear.


*If instead you want to leave the domain as all of $C^1(\mathbb R)$, you argue correctly that $A-\lambda I$ has nontrivial kernel, and so it cannot be invertible. The spectrum is then $\mathbb R$.

More detail: You say that "$f(a)$ is just a constant". Which it is. But you cannot relate it to $g$. So let us just write $c$ instead of $f(a)$. Then
$$\tag1
R_\lambda(g)(x)=e^{\lambda x}\int_a^xe^{-\lambda t} g(t)\,dt+ce^{\lambda(x-a)}.
$$
We immediately have a problem, because $R_\lambda$ is supposed to be linear, but here $R_\lambda (0)=ce^{\lambda(x-a)}$.
Still, let us assume we didn't notice that and we keep thinking.
$R_\lambda$  is supposed to be the inverse of $A-\lambda I$, so let us check.  It is easy to see that
$$
(A-\lambda I)\circ R_\lambda (g) = g. 
$$
But
\begin{align}
R_\lambda\circ (A-\lambda I)\,g
&=e^{\lambda x}\,\int_a^x e^{-\lambda t}\big(g'(t)-\lambda g(t)\big)\,dt+ce^{\lambda(x-a)}\\[0.3cm]
&=e^{\lambda x}\,\bigg[e^{-\lambda t}g(t)\Big|_a^x+\lambda\int_a^xe^{-\lambda t}g(t)\,dt-\lambda\int_a^xe^{-\lambda t}g(t)\,dt\bigg]+ce^{\lambda(x-a)}\\[0.3cm]
&=g(x)+\big(c-g(a)\big)\,e^{\lambda(x-a)}. 
\end{align}
To get $g(x)$ as the result, we would need to have $c=g(a)$. But this is a computation of $R_\lambda\circ (A-\lambda I)$, so $R_\lambda$ does not "see" $g$, it only "sees" $(A-\lambda I)g$. So we would need to somehow deduce $g(0)$ from knowing $g'-\lambda g$. Which is not possible unless you prescribe it, since $g'-\lambda g$ admits a solution for every choice of $g(0)$.
In summary:

*

*if you set the domain of $A$ to be all of $C^1(\mathbb R)$, then $A$ has every $\lambda\in\mathbb R$ as an eigenvalue. Which means that the resolvent does not exist. The $R_\lambda$ you wrote is not an inverse for $A-\lambda I$. You can see this very clearly if you take $\lambda=1$, $g(x)=e^x$. Then $(A-I)g=0$, and $R_\lambda\circ (A-\lambda I)g=g$ if we choose $c=1$. But if for instance we take $g(x)=\sin x$, then
$$
R_\lambda\circ(A-\lambda I)g=R_\lambda (\cos t-\sin t)=e^x+\sin x\ne g.
$$
That is, the constant $c$ can be arranged to work for some functions, but it will fail for others. $R_\lambda$ does not exist as an inverse for $A-\lambda I$. So in this situation $\sigma(A)=\mathbb R$ and the resolvent does not exist for any $\lambda$.


*If we set the domain of $A$ to be $D(A)=\{f\in C^1(\mathbb R): \ f(0)=0\}$ then $f'-\lambda f$ only has the solution $0$, so no $\lambda$ is an eigenvalue. But in this situation
$$
R_\lambda g=e^{\lambda x}\int_0^xe^{-\lambda t}\,g(t)\,dt
$$
and in this situation $R_\lambda\circ(A-\lambda I)=(A-\lambda I)\circ R_\lambda=\operatorname{id}$ on the domain of $A$. So $\sigma(A)=\emptyset$.


*If we (try to) set the domain of $A$ to be $D(A)=\{f\in C^1(\mathbb R): \ f(0)=1\}$ (or any other nonzero constant) now this is not even a subpsace.


*And the most important conclusion: operator theory is analysis, not algebra. The spectrum is close to useless as a purely algebraic object. One considers operators acting on normed spaces, so there is a topology hanging around. And the concept of "invertible" depends on the topology you are using. Those things cannot be ignored. Changing the topology will likely change the spectrum.
