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 $f$. 



Generalized Schwarz Lemma for near-zeros
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.


