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In approximation theory, it is classical to use a result that can be considered a generalization of the Schwarz Lemma:

Let $f:[-1,1]\rightarrow\mathbb{C}$ be a function that is analytic in a domain $S$ containing all the points that are at a distance of $\leq \beta$ from $[-1,1]$, where $\beta>2$. Moreover, suppose that $|f|$ is bounded by $M$ in $S$. Then, if $f$ has $k$ zeros on $[-1,1]$ we have $$|f(x)|\leq M \left(\frac{\beta}{2}\right)^{-k}, \quad x\in[-1,1].$$

The proof is quick: If $f$ is analytic in $S$ and has $k$ zeros, then $f(z)/((z-x_1)\cdots(z-x_k))$ is analytic in $S$, where $x_1,\ldots,x_k$ are the zeros of $f$. Thus, by the maximum modulus principle we have, for any $x\in[-1,1]$, $$|f(x)|\leq |x-x_1|\cdots|x-x_k|\sup_{z\in\partial S}\frac{|f(z)|}{|z-x_1|\cdots|z-x_k|}\leq M \left(\frac{\beta}{2}\right)^{-k},$$ where the last equality follows because $|x-x_j|\leq 2$ and $|z-x_j|\geq \beta$.

I am interested in what can be said if I know $f$ only has $k$ near-zeros. That is instead of $f(x_1) = 0,\ldots,f(x_k)=0$, I have $|f(x_1)|\leq\epsilon,\ldots,|f(x_k)|\leq\epsilon$. The proof above seems to rely heavily on the hard zeros of $f$.

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Such estimates are standard tools in transcendental number theory. The following is a simplified version (usually one also require that derivatives are small): Suppose that $D$ is a domain with a nice boundary $\Gamma$ (piece-wise smooth is fine). Let $f$ be a holomorphic on $D$ and continuous on $\overline{D}$. Let $a_1,\dots,a_n\in D$. Let $z\in D$. Assume that numbers $r_k>0$ are so small that disks $|z-a_k|\leq r_k$ are in $D$ and don't contain $z$ and $a_l$ for $l\neq k$. Denote by $F(z)=\prod_{k=1}^n (z-a_k)$. Then $$ f(z)=\frac{1}{2\pi i}\int_\Gamma \frac{F(z)}{F(w)} \frac{f(w)dw}{w-z}-\frac{1}{2\pi i}\sum_{l=1}^n f(a_l)\int_{|w-a_l|=r_l} \frac{F(z)}{F(w)} \frac{dw}{w-z}.$$

If the values are small and points are spread, then the second term is small, and you get almost the same estimate.

To prove it, just use Cauchy formula for the first integral. You will get a sum of residues. One of them is $f(z)$. Other are in $z_k$ and that is the sum of integrals in the formula.

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  • $\begingroup$ Oh. This is perfect. Thank you very much! Great. $\endgroup$ – alext87 Apr 7 '15 at 22:45
  • $\begingroup$ This is totally obvious to me now. Not sure why I struggled. Thanks, Oleg. $\endgroup$ – alext87 Apr 7 '15 at 22:51
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I am somewhat confused as to what you are looking for. Surely as stated, any function that has at least one zero has infinitely many "near-zeros" in your sense, for every $\varepsilon$. Hence you cannot say anything.

If you assume that your function has no zeros, and perhaps (?) require your near-zeros to be local minima of $|f|$, then you could apply the original result to the derivative of $f$,restricted to a slightly smaller domain (note that, by the Schwarz lemma, this will give you a uniform bound on the derivative). It seems that this will give you a bound on the original function involving the minimum of $|f|$ and the number of critical points of $f$ on the interval in question.

Without more information, it is difficult to tell whether this is in the spirit of what you are looking for.

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