The following conjecture about analytic functions arose as a way to show the asymptotic growth for certain PDE solutions. As I am unfamiliar with any results of this type, I thought I'd ask here.


In some sense, this is an analytic continuation result, as it says that if measure of points close to $0$ on which the function is very small is big enough, then $f(0)$ must be small.


**Conjecture:**


Let $f\_n : (-1,1) \to \mathbb{R}$ analytic functions such that
$$|f\_n^{(m)}(0)| \leq C^n m!$$
for some $0 < C < \infty$. Suppose also that $f\_n(0) = 1$ for all $n$. Does there exist $\delta >0$ depending on $C$, such that
$$\liminf\_n | \{x \in [0,\delta] : |f\_n(x)| \leq e^{-n} \}| = 0,$$
where by $|A|$ we mean the Lebesgue measure of $A$?



Quantitative analytic continuation estimate for a function small on a set of positive measure
Unfortunately, no, as requested:


Take any sequence $\delta\_j\in(0,1)$ decaying to $0$, choose small $\mu\_j>0$ such that $\prod\_j \delta\_j^{\mu\_j}=e^{-1}$ and put $f\_n(z)=e^n\prod\_j B\_{\delta\_j}(z)^{[\mu\_j n]}$ where $B\_\delta(z)=\frac{\delta-z}{1-\delta z}$ is the usual Blaschke factor. Then $|f\_n(0)|\ge 1$ and the set $\{|f\_n\|\le e^{-n}\}$ contains an interval of fixed length around each $\delta\_j$ (the interval where $|B\_{\delta\_j}|\le e^{-3\mu\_j^{-1}}$, say) for sufficiently large $n$.


