Relations among different boolean approximations Essentially similar question to here Different boolean degrees polynomially related? (change being error condition $\epsilon\in(0,1)$).

Let $p$ be the minimum degree (of degree $d_f$) real polynomial that represents boolean function $f$ such that $f(x)=p(x)$. 

Let $p_{0,\epsilon}$ be the minimum degree (of degree $d_{0,f,\epsilon}$) real polynomial  that represents boolean function $f$ such that $$f(x)=0\implies p_{0,\epsilon}(x)=0$$$$f(x)=1\implies|p_{0,\epsilon}(x)-f(x)|\leq\epsilon.$$

Let $p_{1,\epsilon}$ be the minimum degree (of degree $d_{1,f,\epsilon}$) real polynomial  that represents boolean function $f$ such that $$f(x)=1\implies p_{1,\epsilon}(x)=1$$$$f(x)=0\implies|p_{1,\epsilon}(x)-f(x)|\leq\epsilon.$$

Is $d_{f}\leq d_{0,f,\epsilon}^{c_0}$ and $d_{f}\leq d_{1,f,\epsilon}^{c_1}$ for some  $c_0$ and $c_1$?

Above holds if $\epsilon\in(0,\frac{1}{2})$ as mentioned here in link Different boolean degrees polynomially related?.

However what happens if $\epsilon\in(0,1)$ instead of $(0,\frac{1}{2})$ (does polynomial relation still hold)?

That is we consider $0&lt;\epsilon&lt;\frac{1}{2}\leq\delta&lt;1$.

Note that defining $p_\delta$ makes little sense if $\delta\in[\frac{1}{2},1)$.

I am most interested in $\delta=1-\frac{1}{h(n)}$ with some function of $n$ (logarithmic/polynomial/exponential).
The new parameters are polynomially related to the old parameters. In fact, for every $0&lt;\delta,\epsilon&lt;1$ it is the case that $d_{0,f,\epsilon} = \Theta_{\delta,\epsilon}(d_{0,f,\delta})$. This is proved in the same way that the usual relation regarding $d_{f,\epsilon}$ is proved, as detailed below.

Suppose we are given $0&lt;\delta,\epsilon&lt;1$. Let $g$ be some continuous function satisfying $g(0) = 0$ and $g(x) = 1$ for $1-\delta \leq x \leq 1+\delta$. The Weierstrass approximation theorem shows that there is some polynomial $G$ such that $|G(0)| \leq \epsilon/2$ and $|G(x)-1| \leq \epsilon/2$ for all $1-\delta \leq x \leq 1+\delta$. Let $G_0(x) = G(x) - G(0)$, so that $G_0(0) = 0$ and $|G_0(x)-1| \leq \epsilon$ for all $1-\delta \leq x \leq 1+\delta$. The polynomial $P = G_0(p_{0,\delta})$ has degree $(\deg G_0)d_{0,f,\delta}$ and satisfies $P(x) = 0$ whenever $f(x) = 0$, and $|P(x)-1| \leq \epsilon$ whenever $f(x) = 1$. This shows that $d_{0,f,\epsilon} \leq (\deg G_0) d_{0,f,\delta}$. Since $G_0$ only depends on $\delta,\epsilon$, this proves that $d_{0,f,\epsilon} = O_{\delta,\epsilon}(d_{0,f,\delta})$.