# Analytic continuation of a periodic function on the real line

In the study of superconformal indices for certain quantum field theories, one encounters the elliptic $$\Gamma$$ function, which can be expressed as: $$\log \Gamma(z;\tau,\sigma)=\sum^{\infty}_{l=1}\frac{1}{l}\frac{x^l-(x^{-1}pq)^l}{(1-p^l)(1-q^l)}, \quad q=e^{2\pi i\tau},p=e^{2\pi i\sigma},x=e^{2\pi i z}$$ I am interested in a certain limit of this function, where the imaginary parts of $$\tau$$ and $$\sigma$$ are taken to zero. Now, I have two methods at my disposal to compute this limit, which have differing domains of validity in $$z$$ space, but agree on the overlap. I have tried to abstract the problem in a way that does not refer to the specific details mentioned above.

Suppose $$f(x+1)=f(x)$$ is real analytic for $$x\in \mathbb{R}\setminus \mathbb{Z}$$. Moreover, it is once-differentiable at $$x\in \mathbb{Z}$$.

In addition, suppose that on the interval $$(0,1)$$, we know the analytic continuation into the complex plane. Let us call this function $$g(z)$$, where $$0<\mathrm{Re}(z)<1$$. Finally, we also know that $$g(z+n)=g(z)$$ for some integer $$n>1$$. So really, we know the analytic continuation of $$f(x)$$ for $$\mathrm{Re}(z)\in (mn,mn+1)$$ for $$m\in\mathbb{Z}$$.

I want to find the analytic continuation valid for any $$\mathrm{Re}(z)\in \mathbb{R}\setminus \mathbb{Z}$$. An obvious continuation is that one takes $$g([z])$$, where the fractional part is defined as: $$\text{[z]=z+k, k\in\mathbb{Z} such that 0<\mathrm{Re}(z)+k<1}.$$ This function agrees with $$f(x)$$ on $$\mathbb{R}\setminus \mathbb{Z}$$. For the specific function at hand, it turns out that $$g(z)$$ at $$z=0,1$$ is once differentiable, even though the fractional part is not defined there. This matches with the behaviour of $$f(x)$$.

My question: is this continuation unique?

• You mean $\lfloor z \rfloor = z + k$, $k \in \mathbb Z$ such that $0 < \text{Re}(z)+k < 1$. Apr 1 '21 at 14:29
• o yes corrected, thanks.
– sam
Apr 1 '21 at 14:53

Yes, of course. Since $$g(\lfloor z \rfloor)$$ is analytic on $$k < \text{Re}(z) < k+1$$, and agrees with $$f(z)$$ for $$z \in (k,k+1)$$, it is the unique analytic continuation of $$f$$ from $$(k,k+1)$$ to the strip $$k < \text{Re}(z) < k+1$$ by the identity theorem for analytic functions.