Consider a rectangle $D(\epsilon,T)=\{x+iy\in\mathbb{C}\mid \frac{1}{2}-\epsilon\leq x\leq \frac{1}{2}+\epsilon,\ 0\leq y\leq T\}$ where $T>1$ and $\epsilon>0$ is arbitrarily small which depends on $T$.
Question Let $f(z)$ be a non constant entire function with no zeros on the real axis and satisfied $f(z)=f(1-z)$. Prove that there exists $\epsilon_0>0$ such that $f(z)$ has no zeros on the boundary of the rectangle $D(\epsilon,T)$ whenever $0<\epsilon<\epsilon_0$. Please note that $\epsilon$ depends on $T$
My try: Let $T>1$ be not an ordinate of a zero of $f(z)$. We prove the above question by the method of contradiction. If possible let us suppose no such $\epsilon_0$ exists. Then for each $\epsilon>0$, there must exist $\epsilon'$ with $0<\epsilon'<\epsilon$ such that the boundary of the rectangle $D(\epsilon',T)$ has atleast one zero of $f$.
In particular if we take $\epsilon_n=\epsilon_0^*/2^n$ where $n\in \mathbb{N}$ and $0<\epsilon_0^*\leq 1$, then there exists $\epsilon_{n}'$ such that $0<\epsilon_n'<\epsilon_n=\epsilon_0^*/2^n$ such that the boundary of $D(\epsilon_n',T)$ has atleast one zero $z_n$ of $f$.
Without loss of generality we may assume that $\{\epsilon_n'\}_{n\geq 1}$ is decreasing and $\epsilon_n'\to 0^+$ (to see this choose $\epsilon_1'<\epsilon_1$ and then choose $\epsilon_2'<\epsilon_1'$ and $\epsilon_2'<\epsilon_0^*/2^2$. Next we choose $\epsilon_3'$ such that $\epsilon_3'<\epsilon_2'$ and $\epsilon_3'<\epsilon_0^*/2^3$
and we continue this process).
Now we consider a set $A=\{z_n\in \mathbb{C}\mid n\in \mathbb{N}\}$ where $z_n$ are those zeros of $f$ as we have chosen above. Now we can have two cases:
Case $1$: $A$ is an infinite set: Since $\epsilon_n=\epsilon_0^*/2^n$ so we have $\epsilon_0^*>\epsilon_1>\epsilon_2>\epsilon_3>…$ and hence all the zeros $z_n$ of set $A$ lie inside the rectangle $D(\epsilon_0^*,T)$ where $0<\epsilon_0^*\leq 1$ and clearly $D(\epsilon_0^*,T)$ is a bounded set. Hence we get that $A$ is also a bounded set.
So by the Bolzano–Weierstrass theorem, set $A$ has a limit point. Since $f$ is entire so by the Identity theorem since $A\subset \mathbb{C}$ is an infinite bounded subset of the zeros of $f$ and it has a limit point so $f$ must be identically zero which is a contradiction.
Case $2$: $A$ is a finite set:
In this case there are infinitely many $D(\epsilon_n',T)$ having common zeros of $f$, thereby forcing zeros to lie on the real axis (as only the real axis can be common in $D(\epsilon_n',T)$ ). This again contradicts the fact that $f$ has no zeros on the real axis.
Is the above solution correct? Do we need any modification? Please help me with this problem.
Edit We have proved provided that $T$ is not equal to an ordinate of a zero. But this restriction is now irrelevant, as may be seen by replacing $T$ by a larger value $T'$ (distinct from the ordinate of the zeros) and making $T'\to T+0$
Best Answer
Assume that $f$ does not vanish on the closed half-line $$E=\left\{z=x+iy\,:\,x={1\over 2}, y\ge 0\right\}$$ Fix $T>0$ and let $$E_T=\left \{z=x+iy\,:\,x={1\over 2},\ 0\le y\le T\right \}$$ As the segment $E_T$ is compact (bounded and closed), there exists $\varepsilon_0>0$ such that $f$ does not vanish on $$V_{\varepsilon_0}=\left \{z=x+iy\,:\, {1\over 2}-\varepsilon_0 \le x\le {1\over 2}+\varepsilon_0,\ 0\le y\le T\right \}$$ Indeed, assume by contradiction that $f$ vanishes in every $V_{1/n}$ say at $z_n.$ Then the sequence $z_n$ has an accumulation point $z_0$ in the segment $E_T.$ Thus $f\equiv 0$ which contradicts the assumptions.
Hence $f$ does not vanish on the boundary of any subrectangle of $V_{\varepsilon_0}.$
Assume now that $f$ has roots on the half-line $E.$ Let $T$ be chosen so that $f({1\over 2}+iT)\neq 0.$ The function $f$ may have finitely many roots in the segment $E_T,$ as the roots of a nonzero entire function cannot accumulate. Let $z_1,z_2,\ldots, z_n$ denote these roots (if there are any) and $m_1,m_2,\ldots, m_n$ denote their multiplicities. The function $$g(z)=(z-z_1)^{-m_1}(z-z_2)^{-m_2}\ldots (z-z_n)^{-m_n}f(z)$$ is entire and does not vanish on $E_T.$ By the first part there exists $\varepsilon_0>0$ such that $g(z)$ does not vanish on $V_{\varepsilon_0}.$ Hence $$f(z)=(z-z_1)^{m_1}(z-z_2)^{m_2}\ldots (z-z_n)^{m_n}g(z)$$ does not vanish on the boundary of any subrectangle of $V_{\varepsilon_0},$ as the only roots in $V_{\varepsilon_0}$ are located on the line $x={1\over 2}.$