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Let $\Omega$ be some domain. Let $V $ be a function space over this domain. Let the $L^2(\Omega)$ projection $Pf$ of an $f \in L^2(\Omega)$ be given by the function such that

$$ (f-Pf,g)=0, $$ for every $g \in V$.

Then, my question! I can't assume any regularity on $f$. I see often inequalities of the type $$ \|f-Pf\|_{L^2} \leq C \|D^2 f\|_{L^2(\Omega)}, $$ where this is proved by using a good enough interpolant, but, since I don't have any differentiability on $f$, I can't do this. I want to prove something like

$$ \|f-Pf\| \leq C \|f\|_{L^2(\Omega)}, $$ (where if I were to guess, $C$ would depend on the dimension of $V$, somehow.)

Any ideas for how I would prove such a thing? Anyone know an interpolation or projection or something which allows me to do something similar?

Thanks!

EDIT: I can show that $Pf$ is the best approximation, i.e. that for any other $v \in V$, it holds that

$$ \|f-Pf\|\leq \|f-v\|... $$

If the zero function were to be in $V$, then I would be done right? Or am I missing something...

edo
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1 Answers1

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The method in your edit works fine. $V$ is a vector space, so $0 \in V$ by definition, and by setting $v = 0$, you're done.

Alternatively, you can show directly from the definition that the Pythagorean theorem holds: $$\begin{align*}\|f\|^2 &= \|(f-Pf)+Pf\|^2 \\ &= \|f-Pf\|^2 + \|Pf\|^2 + (f-Pf,Pf) + (Pf, f-Pf) \\ &= \|f-Pf\|^2 + \|Pf\|^2 \\ &\geq \|f-Pf\|^2\end{align*}$$

Brian Moehring
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