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Let $C^\omega(\Bbb R)$ denote the set of all real-analytic functions on $\Bbb R$. I was trying the following question:

$(\mathscr{Q1})$ Let $f \in C^\omega(\Bbb R) \cap L^2(\Bbb R)$ such that $f^{(2k)} \in L^2(\Bbb R) \:, \forall k \ge 1$. Then given $\varepsilon >0$ does there exist $f_{\varepsilon} \in C^\infty(\Bbb R) \setminus C^\omega(\Bbb R)$ such that $\|f^{(2k)}- f^{(2k)}_{\varepsilon}\|_{L^2(\Bbb R)} < \varepsilon ,\: \forall k = 0,1,2,\dots \:?$

My most naive approach was to obtain $f_{\varepsilon}$ by modifying $f$ on some set of measure $0$, while keeping its smoothness intact. But of course, it does not work, because then $f-f_{\varepsilon}$ would be a $C^\infty$ function which is non-zero only on a set of zero measure, which is not possible!

Then I thought about considering a convolution with non-analytic mollifier or approximate identity kind of argument, but I don't know how to preserve the norm closeness at the derivatives level.

Is there some Sobolev density result which would be useful here?

$(\mathscr{Q2})$ The higher dimensional analogue of $(\mathscr{Q1})$ with the even derivatives replaced by the iterates of the Laplacian?

Thanks in advance for help!

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  • $\begingroup$ Hello. For question 1, it has to be the same function $f_{\varepsilon}$ for all values of $k$? $\endgroup$ Oct 26, 2022 at 13:23
  • $\begingroup$ @Jean-ArmandMoroni Yes. $\endgroup$
    – Brozovic
    Oct 27, 2022 at 11:31

1 Answer 1

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Off hand I would try to argue that this is impossible for the following reasons.

If the error function $g(x)= f(x)- f_{\epsilon}(x)$ exists then the integrals $I_k=\int |\hat g(\xi)|^2 |\xi|^{2k} d\xi $ are bounded uniformly for all $k$, which should imply that

(i) $\hat g(\xi)$ can only be supported in $\xi\in[-1,1]$

(because for larger values of $\xi$ the powers $|\xi|^{2k}$ explode to infinity.

Then from (i) it would follow from one version of the Paley-Wiener theorem that since $\hat g$ has compact support, $g(x )$ extends analytically to an entire function of exponential type.

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