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The answer is no, consider $f_\epsilon(z) = \frac{e^{-\frac{\epsilon}{z}}-1}{100}$. Then for all $\epsilon < 1$ the function satisfies the required bound, the limit is $\frac{1}{100}$ regardless of $\ …

First of all both of these facts are wrong. For the first one we have to assume that $f$ is not a constant (since $f(z) = i$ works). For the second one it is actually a classical and very important re …

I am not sure if this is the sort of thing you want, but here it goes anyway. The idea is that if $f\in L^2(\mathbb{R}^n)$ and $A$ is very thin then by the Cauchy--Schwarz inequality $f$ is an entire …

The answer is yes. Your function definies a function of finite exponential type if and only if the counting function of $z_n$ is bounded by a linear function (that is, if the number of $z_n$ in the di …

No, of course not. Essentially, you want to bound a $2$-dimensional object (the complex Fourier transform. I talk about the complex dimension here) from the at most $1$-dimensional bound (that is, eve …

The answer is yes. I will prove it using the assumption only for $\xi > 0$ (almost the same proof works if we only assume it for $\xi < 0$). For $z\in U = \{z: Im (z) \ge 0, |z| \ge 1\}$, consider th …

Sure enough, the idea is that if $f(z_0) = 0$ then $g(z) = \frac{f(z)-f(z_0)}{z-z_0}$ is an entire function which is non-vanishing and for which we know the lower bound on the circle $|z-z_0| =\frac{\ …

Let me try to propose something: As I understand from your post you say you're fine with functions without zeroes since for them we have equality. So let us try to reduce to this case. For an analytic …

So, morally, the only real way to bound the decay of $\hat{f}$ is to obtain some bounds on the derivatives of $f$. Let me do a little dilation and consider $f(x) = \exp(-1/(1-x)^p - 1/(1+x)^p)$, it d …

So, we have a function $u_n(z) = \Re \sum_{j = 1}^n \frac{z^j}{j}$. As a real part of an analytic function, it is a harmonic function. We are interested in its behaviour on the circle $|z| = x = 1 + \ …

As I promised, here is the proof that for $A > 1$ such a series does not exists. Key idea is as follows: if $f(z) = \sum_{n=-\infty}^\infty c_nz^n$ has roots at $e^{2n}$ then $g(z) = \sum_{n = 0}^\in …

I believe that there is such a function and even more it is (almost) entire. Indeed, the most standard way to construct function with prescribed zeros is to consider Weierstrass product. Put $f(z) = …

I belive the following should work: basically the only thing we know is 2-dimensional estimate that you wrote (I'll always talk about real dimension to avoid ambiguity). Then to get estimate of $\log …

No, this analogue is false. For example function $$f(z) = \sum\limits_{n = 1}^\infty \frac{z^{2^n}}{n^2}$$ can not be analytically continued beyond $\mathbb{D}$ but $|f|$ is bounded by $10$ in $\mat …