# Can we approximate a real-analytic function and its derivatives by non-analytic smooth functions in the following way?

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?

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

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.