reverse poincare inequality for polynomials with vanishing boundary

Question

I know that for general function $f\in H_0^1(\Omega)$, it is not true that there is a constant $C(\Omega)$ such that $\|\nabla f\|_{L^2}\leq C \|f\|_{L^2}$ (See this question).

So I wonder if the reverse Poincare inequality is true for polynomials with vanishing boundary: Given a bounded domain $\Omega$, and a polynomial $p , p\mid_{\partial \Omega}=0$ , then is it true that there is a constant $C(\Omega)$ such that $\|\nabla p\|_{L^2}\leq C \|p\|_{L^2}$ ? If it is right, how to derive the constant $C$ ? (Maybe answer the 1D case is enough )

Any help is appreciated.

Answer 1

Not true. Choose $p_n(x) := (x-1)x^n$ that vanishes on the boundary of $[0, 1]$ for all $n \in \mathbb{N}$. Then $$ \lVert p_n \rVert_{L^2([0, 1])}^2 = \dfrac{1}{4n^3+12n^2+11n+3}, \quad \lVert p_n' \rVert_{L^2([0, 1])}^2 = \dfrac{n}{4n^2-1}. $$ Observe: $$ \frac{\lVert p_n' \rVert_{L^2([0, 1])}^2}{\lVert p_n \rVert_{L^2([0, 1])}^2} \in O(n^2) $$

So this quotient goes to infinity as $n \rightarrow \infty$ which means that such constant $C$ cannot exist.

But note: If you restrict the degree of the polynomials to some number $N \in \mathbb{N}$, then such constant $C$ exists. This is because the space $P$ of polynomials with degree $\leq N$ that vanishes on the boundary is a finite dimensional space - which makes $\nabla : P\rightarrow P$ a linear continuous map.