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Suppose that $X \subset L^2[0,1]$ is the set $$ X = \{f \in C^\infty[0,1] : f^{(n)}(0)=f^{(n)}(1)=0 \ \forall n\geq0 \}. $$ Is this a dense subspace of $L^2[0,1]$? I know that the space of continuous functions is dense in $L^2[0,1]$. My approach would be to approximate a continuous function with a function in $X$ with respect to the $L^2[0,1]$ norm but couldn't write down a formal proof so far.

Muzi
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    Combine two facts: 1) by "mollification", every continuous function can be approximated arbitrarily closely in L2 by smooth functions, so smooth function are dense in L2 too. Then 2) by multiplying a sequence of smooth functions by a sequence of smooth bump functions which vanish on an increasingly tiny nbhd of the endpoints we get the desired approximating sequence. – Keeley Hoek Nov 08 '21 at 23:41
  • It contains $C^\infty_c((0,1))$, which is dense in $L^2$. – daw Nov 09 '21 at 11:04

1 Answers1

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Yes. We need a lemma which can be found in any introduction to smooth manifold:

Lemma: For (sufficiently small) $\epsilon>0$, there exists a smooth function $\rho_{\epsilon}(x)$ on $[0, 1]$ such that

(1) $0\le \rho_{\epsilon} \le 1$

(2) $\rho_{\epsilon}(x)=\begin{cases} 0 & x\in [0, \epsilon/2) \cup (1-\epsilon/2, 1] \\ 1 & x\in (\epsilon, 1-\epsilon) \end{cases}$

As $\{e_n(x) = e^{2\pi i n x}: n\in\mathbb Z\}$ form an orthonormal basis of $L^2[0, 1]$, and $e_n\cdot \rho_{\epsilon}\rightarrow e_n$ as $\epsilon\rightarrow 0$, the claim is established.

Just a user
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