Show that $\sigma^{-\lambda(x) - 1}$ is continuous on $(0,1)$. Let $V$ be an open connected subset of $\mathbb{C}$ and $A(V)$ be the set of all (complex-valued) analytic functions on $V$. If $\lambda \in A(V)$ with  $\Re \lambda(x) < 0$ for all $x \in V$ , show that $E:(0,1) \to A(V)$ defined by
$$E(\sigma,\cdot) = \sigma^{-\lambda(\cdot) - 1}$$
is continuous.
Edit: After some thought, I realized that
$$E(\sigma,\cdot) = \exp {\left[ (-\lambda(\cdot) - 1) \ln \sigma \right] }.$$
So my question actually reduces to proving that the maps $\sigma \mapsto \ln \sigma$ and $\sigma \mapsto e^\sigma$ are continuous on $(0,1)$, valued in $A(V)$. (Here, note that $\ln \sigma$ and $e^\sigma$ are constant functions, which are elements of $A(V)$.) My main problem here is that I am unfamiliar with the space $A(V)$; all I know is that it is a metric space because it is a Frechet space induced by seminorms on compact subsets of $V$. 
Since $(-\lambda(\cdot) - 1)$ is a fixed element of $A(V)$, we can conclude that $E$ is continuous using the fact that the composition of continuous functions is continuous.
 A: First, let me clarify what it means for a function $f:(0,1)\to X$ to be continuous, when $X$ is a space whose topology is given by a family of seminorms. Let $\mathscr{P}$ be a family of seminorms inducing the topology of $X$. Then, we say that $\omega\subset X$ is open if for any $y\in \omega$, there is a finitely many seminorms $p_1,\ldots,p_k\in\mathscr{P}$ and $\varepsilon>0$ such that
$$
\{x\in X:p_1(x-y)<\varepsilon,\ldots,p_k(x-y)<\varepsilon\} \subset \omega.
$$
Furthermore, the function $f:(0,1)\to X$ is continuous at $0<t<1$ if for any $\varepsilon>0$ and for any finite collection $p_1,\ldots,p_k\in\mathscr{P}$, there exists $\delta>0$ such that $|s-t|<\delta$ implies
$$
p_1(f(t)-f(s))<\varepsilon,\ldots,p_k(f(t)-f(s))<\varepsilon.
$$
Let $\{K_n\}$ be a sequence of compact subsets of $V$ satisfying $V=\bigcup_nK_n$ and $K_n\subset K_{n+1}$ for all $n$. Then a convenient family of seminorms that induces the topology of $A(V)$ consists of the seminorms given by
$$
p_n(g)= \sup_{z\in K_n}|g(z)|
\qquad \textrm{for} \quad g\in A(V),
$$
with $n=1,2,\ldots$. Here, because of the nestedness of $\{K_n\}$, to prove that $f:(0,1)\to A(V)$ is continuous at $0<t<1$, it would be sufficent to show that for any $\varepsilon>0$ and for any $n$, there exists $\delta>0$ such that $|s-t|<\delta$ implies
$$
p_n(f(t)-f(s)) < \varepsilon.
$$
In our case, the function $f$ is
$$
f(t)(z) = t^{-\lambda(z)-1}. \qquad\qquad(*)
$$
Let $n$ and $0<t<1$ be given. We have
$$
p_n(f(t)-f(s)) = \sup_{z\in K_n} |t^{-\lambda(z)-1}-s^{-\lambda(z)-1}|.
$$
The difference inside the absolute value can be written as
$$
t^{-\lambda(z)-1}-s^{-\lambda(z)-1} = -(\lambda(z)+1) \int_s^t \theta^{-\lambda(z)-2} \mathrm{d}\theta,
$$
hence
$$
|t^{-\lambda(z)-1}-s^{-\lambda(z)-1}| 
\leq |\lambda(z)+1| \,\big(\frac2t\big)^{2}|t-s|,
$$
where we have assumed that $s>t/2$.
This means that
$$
\sup_{z\in K_n} |t^{-\lambda(z)-1}-s^{-\lambda(z)-1}| 
\leq \big(\frac2t\big)^{2}|t-s| \sup_{z\in K_n} |\lambda(z)+1|,
$$
and that for any given $\varepsilon>0$, if we choose $0<\delta < \frac{t}2$ so small that
$$
\delta \big(\frac2t\big)^{2} \sup_{z\in K_n} |\lambda(z)+1| < \varepsilon,
$$
then for any $s\in(t-\delta,t+\delta)$ it is guaranteed that 
$$
p_n(f(t)-f(s)) < \varepsilon.
$$
This establishes that the function $f:(0,1)\to A(V)$ given by $(*)$ is continuous.
