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Let $H^1(\Omega)=\{v\in L^2(\Omega):\nabla v\in L^2(\Omega)^2\}$, where $L^2(\Omega)$ is the usual space of square-integrable functions.

There exists an operator $$P:H^1(\Omega)\longrightarrow P_k(\Omega)$$ such that $\|Pv\|{\color{red}\geq} C\|v\|$, with $C>0$ a constant independent of $v$? (note the red sign).

Here, $P_k(\Omega)$ is the set of polynomials defined on $\Omega$ less than $k\geq 0$, and the norm is:

$$\|v\|^2=\|v\|_0^2+\|\nabla v\|_0^2$$

for all $v\in H^1(\Omega)$, where $\|\cdot \|_0$ is the norm induced by the $L^2$ inner product:

$$\|v\|_0^2=\displaystyle\int_\Omega v^2\,dx$$

Note that $P$ is not an projector and then not requires to hold any projector propertie.

yemino
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  • When you say projection-like, is $P^2=P$? – Calvin Khor Feb 27 '18 at 22:29
  • No, I wrote "projector-like" just to say that is an operator from the continuous space $H^1(\Omega)$ to its subespace $P_k(\Omega)$, but is not an projector because the red sign. – yemino Feb 27 '18 at 22:38

1 Answers1

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Let $p \in P_k(\Omega)$ be fixed with $\|p\| = 1$. Then, you can define $P(v) := \|v\| \, p$. This yields $\|P v \| = \|v\| \, \|p\| = \|v\|$. That is, your desired inequality holds with $C = 1$ (and is even an equality).

gerw
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