Polynomials cannot be small in a large set Let $P$ be a degree $d$ polynomial where $d\geq 1$. Suppose $P$ has a root $x_0\in [0,1]$, and suppose that $\max_{x\in [0,1]}|P(x)|=1$. Is there a constant $C_d$ depending on $d$ only, such that for every $0<\epsilon<0.01$
$$
|\{x\in [0,1]:|P(x)|<\epsilon\}|\leq C_d\epsilon^{1/d}?
$$ 
Note that this is easy if $d=1$. Indeed, suppose $P(x)=a(x-x_0)$. Since $\max_{x\in [0,1]}|P(x)|=1$, there is $x_1\in [0,1]$ such that $|a(x_1-x_0)|=1$. Solving the inequality $|P(x)|<\epsilon$, we get $|x-x_0|<\epsilon/a$, and thus 
$$
|\{x\in [0,1]:|P(x)|<\epsilon\}|\leq 2\epsilon/|a|=2\epsilon |x_1-x_0|\leq 2\epsilon.
$$
But in higher dimensions, this seems much trickier.
 A: The answer is yes but, unfortunately, not in a satisfactory way (read the last
two lines just after the horizontal line).
We'll assume $P$ has only one zero of multiplicity $1$. 
Then $P(x)=(x-x_0)Q(x)$, where $x_0\in[0,1]$ and $Q$ is some polynomial which
doesn't vanish on $[0,1]$.
Let $\varepsilon>0$ and let $\alpha_0=\min_{x\in[0,1]}|Q(x)|>0$. If $x\in[0,1]$
satisfies $|P(x)|<\varepsilon$ then
$$
  |P(x)|=|x-x_0||Q(x)|<\varepsilon
  \ \Longrightarrow\ 
  |x-x_0|<\frac{\varepsilon}{|Q(x)|}\leq\frac{\varepsilon}{\alpha_0}
  .
$$
Therefore 
$\{x\in[0,1]\colon\:|P(x)|<\varepsilon\}
\subseteq\{x\in[0,1]\colon\:|x-x_0|\leq\varepsilon/\alpha_0\}$ and we
have the estimate
$$
  |\{x\in[0,1]\colon\:|P(x)|<\varepsilon\}|
  \leq 2\varepsilon/\alpha_0
  .
$$
If we take $\varepsilon<1$ then necessarilly $\varepsilon^{1/d}\geq\varepsilon$
for $d\geq 1$, so that
$$
  |\{x\in[0,1]\colon\:|P(x)|<\varepsilon\}|
  \leq 2\varepsilon/\alpha_0\leq\frac{2}{\alpha_0}\epsilon^{1/d}
$$
Lastly, pick $C_d=2/\alpha_0$ and notice that our constant $C_d$
only depends on our given polynomial $P$.
When our polynomial has more than one zero in $[0,1]$ we proceed
analogously. Even multiplicity won't be a problem

As you can see, this argument ignores completely the exact value of the
degree of $P$ (as long as it is $\geq 1$) and also the fact that $|P(x)|\leq 1$
for $x\in[0,1]$. (Also the fact that $\varepsilon<0.01$ is unimportant here
as long as $\varepsilon<1$.)
