If $f$ is entire and $g$ is continuous does it follow that $g\circ f$ is entire?
No, unless $f$ is constant, $g$ needs to be holomorphic too, not only continuous for the composition to be holomorphic. Consider $g(z) = \overline{z}$ to see that even a very well-behaved smooth function generally doesn't preserve complex differentiability.
If $A$ is a unital complex Banach algebra are the elements in the continuous dual $A^\ast$ entire?
For $A = \mathbb{C}$, the answer is yes, otherwise it depends on what you mean by "entire". Usually, an entire function is defined as a function defined and holomorphic on all of $\mathbb{C}$ (or $\mathbb{C}^n$). Then the elements of $A^\ast$ have the wrong domain to be called entire (generally). You can, however, in this context also define an entire function to be a function that is defined and (complex) Fréchet-differentiable on all of $A$, where $f$ being Fréchet-differentiable in $a$ means that there exists a $\varphi \in A^\ast$ such that $f(a+h) - f(a) - \varphi(h) \in o(\lVert h\rVert)$. In that sense an $f\in A^\ast$ is entire, and at each point it is its own derivative, $f'(a) = h \mapsto f(h)$ for all $a\in A$.
It is proved that the resolvent function $\lambda \mapsto (\lambda-a)^{-1}$ is analytic on $\mathbb{C}$ and then it is stated that if $\varphi \in A^\ast$ then $\lambda \mapsto \varphi((\lambda - a)^{-1})$ is analytic too. I am trying to understand why this is true.
That is because a $\varphi \in A^\ast$ is $\mathbb{C}$-linear and continuous, which implies Fréchet-differentiability. Writing $R(a,\lambda) = (\lambda - a)^{-1}$, we have
$$\begin{align}
\varphi(R(a,\lambda+h)) - \varphi(R(a,\lambda)) &= \varphi(R(a,\lambda+h) - R(a,\lambda))\\
&= \varphi\left(\frac{\partial R}{\partial \lambda}(a,\lambda)\cdot h + o(\lvert h\rvert)\right)\\
&= \varphi\left(\frac{\partial R}{\partial \lambda}(a,\lambda)\right)\cdot h + \varphi(o(\lvert h\rvert).
\end{align}$$
Since $\varphi$ is a continuous linear map, $\varphi(o(\lvert h\rvert) \in o(\lvert h\rvert)$, so the complex differentiability of $\varphi \circ R(a,\,\cdot\,)$ follows.
