# Showing that the difference of these 2 meromorphic functions are bounded

I have the functions $$g(z)=\frac{\pi^2}{sin^2(\pi z)}$$ and $$f(z)=\sum_{n=0}^\infty\frac{1}{(z+n)^2}+\sum_{n=1}^\infty\frac{1}{(z-n)^2}$$. I need to show that the function $$f-g$$ is bounded and holomorphic on $$\mathbb{C}$$.

I know that both $$f,g$$ are holomorphic on $$\mathbb{C}\backslash\mathbb{Z}$$ with poles at $$\mathbb{Z}$$, and that their Laurent series at $$z=0$$ both have principal part $$\frac{1}{z^2}$$. I am also given that they satisfy properties $$f(z+1)=f(z)$$, and $$f(x+iy)\rightarrow0$$ for $$|y|\rightarrow\infty$$. (Same applies for $$g$$).

If I could show that $$f-g$$ is bounded, then by Riemann extension the poles $$z\in\mathbb{Z}$$ would be removable singularities. Hence $$f-g$$ would extend to a holomorphic function on $$\mathbb{C}$$.

But I am struggling with the boundedness part. Since $$g$$ itself is not bounded and by writing out $$g-f$$ nothing seems to simplify/suggest that it is bounded.

Both are bounded when $$z$$ is not too close to an integer.

Looking around zero, which is enough, $$f(z) =1/z^2+$$ something bounded by comparison with $$\zeta(2)$$ and $$\sin^2(\pi z) = (\pi z)^2+O(z^4)$$ so

$$\dfrac{\pi^2}{\sin^2(\pi z)} =\dfrac{\pi^2}{ (\pi z)^2+O(z^4)} =\dfrac1{z^2(1+O(z^2))} =\dfrac1{z^2}(1+O(z^2)) =1/z^2+O(1)$$

so their difference is bounded.

If $$|z| < c$$ where $$c$$ is small ($$c < \frac12$$, in particular) then
$$\begin{array}\\ |f(z)-\dfrac1{z^2}| &=|\sum_{n=1}^\infty\frac{1}{(z+n)^2}+\sum_{n=1}^\infty\frac{1}{(z-n)^2}|\\ &\le 2|\sum_{n=1}^\infty\frac{1}{(n-c)^2}|\\ &= 2|\frac1{(1-c)^2}+\sum_{n=2}^\infty\frac{1}{(n-c)^2}|\\ &\le 2|\frac1{(1-c)^2}+\sum_{n=2}^\infty\frac{1}{n(n-1)}| \qquad ((n-c)^2 > n(n-1) \text{ for } c < \frac12)\ (*)\\ &\le 2|\frac1{(1-c)^2}+\sum_{n=2}^\infty(\frac{1}{n-1}-\frac1{n})|\\ &\le 2|\frac1{(1-c)^2}+1|\\ \end{array}$$
so $$f(z)-\dfrac1{z^2}$$ is bounded for small $$z$$.
$$(*)\ (n-c)^2 > n(n-1) \iff n^2-2nc+c^2 > n^2-n \iff n(1-2c)+c^2 > 0$$
To show $$\dfrac1{1+O(z^2)} =1+O(z^2)$$, $$\dfrac1{1+cz^2}-1 =\dfrac{-cz^2}{1+cz^2}$$ so $$|\dfrac1{1+cz^2}-1| =|\dfrac{-cz^2}{1+cz^2}| \lt 2c|z^2|$$ if $$|cz^2| < \frac12$$ or $$|z| < \sqrt{\frac1{2c}}$$ so $$\dfrac1{1+O(z^2)} =1+O(z^2)$$ as $$z \to 0$$.
• What do you mean by 𝜁(2)? Also, how did you get $\frac{1}{1+O(z^2)} = 1+O(z^2)$? I am guessing this is due to geometric series, right? If so I think I understand. Lastly, how does this imply that their difference is bounded - isn't $O(1)$ just a polynomial, hence not bounded? – Azamat Bagatov Apr 18 '20 at 13:18
• I added additional explanations. Also, $O(1)$ means a bounded expression. – marty cohen Apr 19 '20 at 3:55