Function $\Gamma$

Question

It is known that function $z\mapsto \Gamma(z)$ is holomorphic on $\Bbb{C}\setminus\{-\Bbb{N}\}$.

Let $\lambda>0$. Set $F=\{(2m+1)\lambda\mid m\in\Bbb{N}\}$.

Consider the function $h: F^c=C\setminus F \to \Bbb{C}$ defined by $ \mu\mapsto \Gamma\Big( \frac{\lambda-\mu}{2\lambda}\Big)$.

Can we extend $h$ to an holomorphic funtion.

In fact I have the following :

Let $\Omega$ denote the open set $\{z\in \Bbb{C}: \Re(z)>0\}$. Let $(L_\lambda)_{\lambda\in\Omega }$ be a family of closed unbounded operators, with domain $D_{\lambda}\subset L^2(\Bbb{R}^2)$.

For $\lambda>0$ the spectrum of $L_\lambda$ is the set $\sigma_{\lambda}=\{(2m+1)\lambda, m\in \Bbb{N}\}$, so its resolvent $R_{\lambda}(\mu)=(L_{\lambda}-\mu I)^{-1}$ is defined for $\mu\not\in \sigma_{\lambda}$.

More precisely for $\lambda>0,\quad \mu\not\in \sigma_{\lambda},f\in L^2(\Bbb{R}^2) $; the resolvent $R_{\lambda}(\mu):L^2(\Bbb{R}^2)\to D_{\lambda} $ has the form $$R_{\lambda}(\mu)(f)(x,y)= \Gamma(\frac{\lambda-\mu}{2\lambda}) T_{\lambda}(\mu)(f)(x,y)$$

where $ T_{\lambda}(\mu): L^2(\Bbb{R}^2)\to D_{\lambda} $ is an bounded operator on $L^2(\Bbb{R}^2)$.

My question:

Since the function $\lambda\to \Gamma(\frac{\lambda-\mu}{2\lambda})$ is holomorphic on $\Omega$ if $\mu\not\in \{(2m+1)z: m\in\Bbb{N}, z\in \Omega\}$ and If suppose that the map $\lambda\to T_{\lambda}(\mu)(f)$ is holomorphic on $\Omega$ with $f\in L^2(\Bbb{R}^2)$.

Can we say in this case that the spectrum of $(L_\lambda)$ with $\lambda\in\Omega $ is the set $\{(2m+1)\lambda, m\in\Bbb{N}\}$ with $\lambda\in\Omega $.

Thanks in advance.

Answer 1

If by extend to a holomorphic function you mean to all of $\mathbb{C}$, then the answer is evidently not, for by reversing the transformation $z \mapsto \frac{\lambda-z}{2\lambda}$ you would then be able to extend $\Gamma$ to a holomorphic function on all of $\mathbb{C}$.

Answer 2

No.

Our only difficulty occurs when the argument supplied to $\Gamma$ is a non-positive integer. Let $m \in -\mathbb{N}$ and note that we wish to ensure that $m$ is never the argument to $\Gamma$. Since the argument to $\Gamma$ is $\frac{\lambda - \mu}{2\lambda}$, difficulty occurs when $$ m = \frac{\lambda - \mu}{2\lambda} \text{.}$$ That is, when $$ (2(-m) + 1)\lambda = \mu \text{.}$$ But this says the difficulty occurs when $\mu \in F$, but the domain of $h$ is $\mathbb{C} \smallsetminus F$, so $h$ is already holomorphic.

Any extension of $h$ extends the domain to some element of $F$, each of which is a pole of $\Gamma$, so $h$ cannot be extended to a holomorphic function.