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 $f$.