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**Hadamard’s Factorization Theorem**. Let $f$ be an entire function of order $ρ=1$. Then $f(s)=P(s)exp(Q(s))$, where $P(s)$ is a canonical product with the same zeros as $f$ and $Q$ is a polynomial of degree less than or equal to $ρ=1$. In this case we have $|f(s)|≤_{ε}exp(|s|^{ρ+ε})$ for all $ε>0$.

We have

$f(s)=exp(Q(s))∏_{k=1}^{∞}(((s_{k}-s)/(s_{k})))exp( (s/(s_{k})))$

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where $s_{k}$ are the roots of $f$.

Now, consider $h(s)=f(1-2t)$ where $s=\sigma+it$ and $0<t<1$, let Ω=(0,1). The equation $f(1-2t)=0$ implies that $t∈Ω₁$ where $Ω₁$ is a discrete set since the roots are isolated. For the case of $f$ we cannot have infinitely many zeros in $(0,1)$. The main reason here is that a nonzero analytic function can have infinitely many zeros but they must not have an accumulation point in $ℂ$. Now, if there are infinitely many zeros of $f(1-2t)=0$ in $(0,1)$ they do have an accumulation point in $[0,1]$ and since $f(1-2t)$ is extendable to an analytic function to $ℂ$ containing $[0,1]$ this must not happen.

My **question** is: What is the **Hadamard’s Factorization** of $h$ taking in acaunt that it has a finite number of zeros.

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“$h(s)=f(1-2t)$.” I take it you are using the standard convention $s=\sigma+it$ with $\sigma$ and $t$ real.

“$f(1-2t)$ is extendable to an analytic function.” Since this analytic function agrees with $f(1-2s)$ on an interval, it must *be* $f(1-2s)$, right? So, $h(s)$ is an entire function defined everywhere by $h(s)=f(1-2s)$, right?

Now if $h(s)$ has finitely many zeros, it must be a polynomial times an entire function with no zeros, right? So $h(s)=P(s)\exp(Q(s))$ where $P$ is a polynomial and $Q$ is entire. And since we are talking order $1$, we must have $$h(s)=P(s)e^{a+bs}$$ for some polynomial $P$ and some constants $a$ and $b$.

EDIT: OP has made some major changes in the question. The function $h$ is now only defined on the interval $(0,1)$. Hadamard factorization is a concept in complex analysis, and does not apply to a function defined only on some interval of real numbers.

No doubt OP will change the question again, but I have had enough.

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