The following question is from Linear Operators edited by Nelson Dunford and Jacob T. Schwartz, Chapter VII.5, Problem 30.
Let $T$ be a bounded linear operator defined on a Banach Space $X$. Assume each element in $\sigma(T)$ has negative real part. Then we define $l_R$ to be the curve that consists of left half part of the open unit disk with radius $R$, denoted by $C_R$, and the line segment on the imaginary axis $\{\mu \in \mathbb{C}\,\vert\,\mu = i s, s \in [-R, R]\}$, denoted by $[-iR, iR]$. When $R$ is large enough, for a fixed real number $\alpha > 0$, we have $$e^{\alpha T} = \frac{1}{2\pi i}\int_{l_R}e^{\alpha \lambda}(\lambda - T)^{-1}d\lambda = \frac{1}{2\pi i}\int_{C_R}e^{\alpha \lambda}(\lambda - T)^{-1}d\lambda + \frac{1}{2\pi i}\int_{-iR}^{iR}e^{\alpha \lambda}(\lambda - T)^{-1}d\lambda$$.
As $R \rightarrow \infty$, because each element in $\sigma(T)$ has negative real part, $\frac{1}{2\pi i}\int_{C_R}e^{\alpha \lambda}(\lambda - T)^{-1}d\lambda$ will converge to $0$ in operator norm and then we have $e^{\alpha T} = \lim_{R \rightarrow \infty}\int_{-iR}^{iR} e^{\alpha \lambda}(\lambda - T)^{-1}d \lambda$. Conversely, using this formula we can show, when $R$ is large enough and $\lambda$ has positive real part, $(\lambda - T)^{-1} = \int_{0}^{\infty}e^{-\alpha \lambda} e^{\alpha T}d \alpha$ (same $T$).
Question part: For a bounded linear operator $T$ defined on a Banach Space and each element in $\sigma(T)$ has negative real part, show that $r_T = \sup_{\lambda \in \sigma(T)} Re(\lambda) = \lim_{R \rightarrow \infty}\frac{1}{R} \log\,\|e^{R T}\|$. In a Banach Space I do not always have $\sup_{\lambda \in \sigma(T)}e^{\alpha \lambda} = \|e^{\alpha T}\|$ and cannot proceed. Any hints will be appreciated.
The result you are after holds irrespective of the assumption that each element in $\sigma(T)$ has negative real part.
Theorem. Given any bounded operator $T$ on any Banach space, let $$ α:= \lim_{R \rightarrow \infty}\frac{1}{R} \log\,\|e^{R T}\|, $$ and $$ β:= \sup\{{Re(λ)}: λ∈ σ(T)\}. $$ Then $α=β$.
Proof. One has that $$ e^α = \lim_{R \rightarrow \infty}\|e^{R T}\|^{\frac{1}{R}} = \lim_{n \rightarrow \infty}\|(e^T)^n\|^{\frac{1}{n}} = r(e^T), $$ by Gelfand's Formula for the spectral radius $r(e^T)$ of $e^T$. On the other hand, by the definition of the spectral radius and by the spectral mapping theorem (according to which $σ(e^T)=e^{σ(T)}$), we have $$ r(e^T) = \sup\{|λ|: λ∈ σ(e^T)\} = \sup\{|e^λ|: λ∈ σ(T)\} = $$$$ = \sup\{e^{Re(λ)}: λ∈ σ(T)\} = e^β. $$ This proves that $e^α=e^β$, so the conclusion follows by taking the logarithm on both sides.