I'm having quite a bit of difficulty in proving the following
$f(z) = \frac{\sin(z)}{\prod_{m=1}^{\infty} (1 – \frac{z^2}{m^2\pi^2})}$ is an entire bounded function without zeros and must therefore be a constant.
I get that the last result follows by Liouville's Theorem, but I'm really stuck on proving this function is an entire and bounded.
Context: The original question is: "Consider the infinite product
$P(x) = x\pi \cdot \prod_{m \geq 1} (1 – \frac{x^2}{m^2}).$ Assume that this product converges and behaves 'nicely' (with respect to differentiation etc.) we would like to prove the product formula for $\sin(x\pi)$ which says that $\sin(x\pi) = P(x)$
Firstly, question (a) was to show that $P(x) = 0 \iff \sin (x\pi) = 0$. The next question, (b), is to show $P(x)$ and $\sin(x\pi)$ have the same derivative at $x = 0$. The typeset solutions say the following:
"It is possible to complete this to a proof of the product formula by showing that $$f(z) = \frac{\sin(z)}{\prod_{m=1}^{\infty} (1 – \frac{z^2}{m^2\pi^2})}$$ is an entire bounded function without zeros and must therefore be equal to a constant. It is non-trivial to show the boundedness but one of the key steps is to rewrite the product as $$\prod_{m \geq 1} (1 – \frac{(t + \pi)^2}{m^2\pi^2}) = \prod_{m \geq 1} \frac{1}{m^2}(m^2 – (\frac{t + \pi}{\pi})^2) = \prod_{m \geq 1} \frac{1}{m^2}(m^2 – (\frac{t}{\pi} + 1)^2) = \prod \frac{1}{m^2}(m -\frac{t}{\pi} – 1)(m + \frac{t}{\pi} + 1)$$ makes it possible to show it is periodic with period $\pi$.
So I'm stuck on showing
(i) The denominator is an entire function, I have no idea where to start with this
(ii) f(z) is entire
(iii) f(z) is bounded
(iv) f(z) has no zeros.
From that point the result obviously follows by Liouville's theorem.
Best Answer
Some hints for you to flesh out:
(i) to show the product is an entire function, use the product Weierstrass test (see Exc. 19 in these notes of Tao) and then the fact that a locally uniform limit of holomorphic functions is holomorphic (Thm. 36 in here).
(ii) use the fact that an absolutely convergent infinite product (see Lemma 16 here) is $0$ if and only if one of the multiplicands is $0$ to show that the product has a simple zero at each of $2\pi k$ for $k\in \mathbb Z$ (factor the product as $(z-2\pi k)g(z)$ for $g(z)$ having no zeroes at $2\pi k$). Then use the Riemann removable singularity theorem to show that the quotient of $\sin(z)$ by the product is entire.
(iii) by the hint in the typeset solutions, the product is periodic; $ \sin(z)$ is also periodic. Thus to prove boundedness, it suffices to prove boundedness on a strip, say $\{z \in \mathbb C : 0 \leq \Re z \leq 2\pi \}$. By the Phragmen-Lindelof theorem (Thm. 14 here), it essentially suffices to prove boundedness of $|f|$ on the line $\Re z = 0$. It suffices to prove boundedness of $\ln|f(0+iy)|$ for sufficiently large $y$. Do this by using Taylor expansion of $\ln(1+x)$ (observe that $\sin(0+iy) = \frac 12(e^y+e^{-y}) = \frac 12 e^y (1+e^{-2y})$, which can be handled easily using $\ln$ and the Taylor series of $\ln(1+x)$).
Unfortunately this doesn't quite work since Phragmen-Lindelof does still require some boundedness on $x_0+iy$ for $x_0 \in (0,2\pi)$; so basically you just have to do the estimates in the previous paragraph a little more carefully to account for the extra $x_0$ stuff (and this means you can cut out Phragmen-Lindelof entirely --- it was there just for the moral support I guess).
(iv) Liouville's theorem does not need the function to be nowhere zero? But this is true because you already know $\sin(z)=0 \iff P(z)=0$. You can apply Liouville to say that $f$ is a constant $C$, and then find $C$ by evaluating $f$ at $z=0$ using the derivative information you apparently already deduced for the problem: $f(z) = \frac{\sin z}z \cdot \frac{z}{P(z)}$, so taking the limit $z \to 0$ gives $f(0) = \sin'(0) \cdot \frac{1}{P'(0)} = 1$.