In a book I'm reading has the definition:
Definition 1: Let $X$ be a Banach space. If $A : D(A) \subset X \rightarrow X$ is a linear, not necessarily bounded, operator in $X$, the resolvent set $\rho(A)$ if the set $$ \rho(A) = \{\lambda \in \mathbb{C} : (\lambda I - A)^{-1} \text{ is bounded in } X \}.$$
I wonder if the above definition is equivalent to the natural definition:
Definition 2: Let $X$ be a Banach space. If $A : D(A) \subset X \rightarrow X$ is a linear, not necessarily bounded, operator in $X$, the resolvent set $\rho(A)$ if the set $$ \rho(A) = \{\lambda \in \mathbb{C} : \lambda I - A \text{ is bijective}\}.$$
I don't if Definition 2 is in fact applied to unbounded operators. If the two definitions makes sense, how to prove they are equivalent? If they are not equivalent, why is preferable to use Definition 1?
They're not equivalent without further assumptions.
First, in the first definition it should be $\operatorname{Range}(\lambda I-A) = X$ and $(\lambda I - A)^{-1}$ is bounded.
In the second definition it should be $\operatorname{Range}(\lambda I-A) = X$ and $(\lambda I - A)$ is bijective.
The first definition obviously implies the second, as $(\lambda I - A)$ has to be injective for $(\lambda I - A)^{-1}$ to even be defined.
If $A$ is closed, the second definition implies the first. $\mathcal{D}(\lambda I- A) = \mathcal{D}(A)$ and $A$ being closed implies $(\lambda I - A)$ is closed. The self-map on $X\oplus X$ exchanging coordinates is a homeomorphism in the product topology, so if the graph of $(\lambda I-A)$ is closed, so is the graph of $(\lambda I - A)^{-1}$. Moreover, $\mathcal{D}((\lambda I - A)^{-1}) = \operatorname{Range}(\lambda I - A) = X$. By the closed graph theorem, $(\lambda I - A)^{-1}$ is bounded.
Without the assumption that $A$ is closed, the definitions are not equivalent. Let $H$ be a separable infinite dimensional Hilbert space with orthonormal basis $O=\{e_n\}_{n=1}^\infty$. By Zorn's lemma, there exists a $B$, $O\subset B\subset H$ such that $B$ is a Hamel basis for $H$ (i.e. linear basis.) Let $u\in B\setminus O$. Define a linear operator $A:H \to H$ such that $Au = 2u$ and $Av = v$ for all $v\in B\setminus\{u\}$. Since $B$ is a Hamel basis, any assignment of values on $B$ extends uniquely to a linear operator on $H$. (each member of $H$ is a unique finite linear combination of vectors in $B$.) From the definition it is evident that $A$ so defined is bijective from $H$ to $H$. The inverse is $A^{-1}u = u/2$ and $A^{-1}v = v$ for all $v\in B\setminus\{u\}$. The operator $A$ (or its inverse) cannot be bounded, because it is an identity on an orthonormal basis $\{e_n\}_{n=1}^\infty$ - and any bounded operator with this property is the identity operator on $H$!