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I read in Kolmgorov-Fomin's Элементы теории функций и функционального анализа (p. 408 here) that the set of continuously differentiable functions are dense everywhere in space $L^1[a,b]$ of Lebesgue integrable functions on $[a,b]$ endowed with distance $d(f,g)=\int_{[a,b]}|f-g|d\mu$.

I know that the set of continuous functions is dense everywhere in $L^1[a,b]$, but how to prove that the set of continuously differentiable functions is dense everywhere too? I thank you all very much!!!

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It is a direct consequence of Stone-Weierstrass' theorem. Let $f\in L_1[a,b]$, and $\epsilon>0$. You know that there is some $g\in C[a,b]$ such that $\int_a^b|f-g|d\mu<\epsilon/2$. Now, let $p$ a polynomial such that $\|p-g\|_\infty<\epsilon/[2(b-a)]$.

Then $$\int_a^b|f-p|d\mu\le\int_a^b|f-g|d\mu+\int_a^b|g-p|d\mu<\epsilon$$

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  • $\begingroup$ A polynomial! I know Weierstrass theorem from my studies in numerical analysis with a proof based on Bernstein polynomials, fortunately, because Kolmogorov-Fomin's uses the above result precisely to prove Weierstrass theorem, which would be a circular argument. Your very reasoning shows that the sets of functions differentiable $k$ times are everywhere dense. Thank you so much!!! $\endgroup$ – Self-teaching worker Nov 16 '14 at 11:10

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