Why do these inequalities of norms hold? Let $f: \Omega \to \mathbb{R}$ be a finite-dimensional polynomial defined on a convex domain $\Omega \subset \mathbb{R}^d$ with a characteristic length $h$, and $f_\epsilon$ be a canonical extension of $f$ to an $\epsilon$-neighborhood of $\Omega$, say, $\Omega_\epsilon := \{x \in \mathbb{R}^d ~|~ \operatorname{dist}(x,y) < \epsilon, ~\text{for any}~ y \in \Omega ~\text{and some}~ \epsilon > 0 \}$. 
A lecture note claims that the following inequalities of norms hold true
$$
\| \nabla f_\epsilon \|_{L^\infty(\Omega_\epsilon)} \le C_1 \| \nabla f \|_{L^\infty(\Omega)} \le C_2 \| f \|_{L^\infty(\Omega)} \le C_3 ~h^{-\frac{d}{2}}~ \| f \|_{L^2(\Omega)} 
$$ 
for some constants $C_1, C_2, C_3$ independent of $h$. But I don't see the proof explicitly. Can anyone explain the details for me? 
ps. I think the last inequality uses the norm equivalence on finite dimensional spaces, but I don't get the $h^{-\frac{d}{2}}$ term. 
 A: Let $\Omega = (0,1)$ and consider a sequence of polynomials $f_n = \frac{1}{n+1}x^{n+1}$. Then $\nabla f = x^n$ and $\|\nabla f_\epsilon\|_{L^\infty(\Omega_\epsilon)}= (1+\epsilon)^n$ while $\|\nabla f\|_{L^\infty(\Omega)}=1$. This makes it impossible for there to exist a constant $C_1$ so that
$$\|\nabla f_\epsilon\|_{L^\infty(\Omega_\epsilon)}≤\|\nabla f\|_{L^\infty(\Omega)}$$
for all polynomials $f$.
The second inequality is also false. Consider the power series expansion of $\frac1n \sin(n^2 x)$ and cut it off at a point where both the function and its derivative are "very closely" approximated in $L^\infty$ norm (say with error $1/n$). If $f_n$ is this polynomial, then:
$$\|\nabla f_n\|_{L^\infty(\Omega)}≥ n -\frac1n,\text{ while }\|f_n\|_{L^\infty(\Omega)}≤\frac1n+\frac1n$$
making it impossible for an inequality of the form $\|\nabla f \|_\infty ≤ C_2 \|f\|_\infty$ to hold for all polynomials simultaneously.
The final inequality is done in the same way. Choose some sequence of polynomials approximating $e^{- (nx)^2}$. The $L^2$ norm of this will go to $0$ while the $L^\infty$ norm remains at $1$.
