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I was asked to prove that $\{{t^{3n}}\}_{n=0}^{\infty}$ is complete in $L_2[0,1]$ (complete system). And the solution says that it is sufficient to show completeness in $C[0,1]$. Why is that? Is it simply because continuity on a closed interval implies integrability on that interval? I am not even sure how correct it is, but what I ultimately want to know is the simple claim the enables this equivalence.

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    Recall that $C[0,1]$ is dense in $L^2[0,1]$. – 2017-02-27
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    If $A$ is dense in $B$ and $B$ is dense in $C$ then $A$ is dense in $C$; this is a restatement of $\overline{\overline{A}}=\overline{A}$. – 2017-02-27

2 Answers 2

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Let $\mathcal{M}$ be the linear span of $\{ t^{3n} \}_{n=0}^{\infty}$. The closure of $\mathcal{M}$ in $C[0,1]$ is a subspace of $L^2[0,1]$, and $$ \overline{\mathcal{M}}^{C} \subseteq \overline{\mathcal{M}}^{L^2} $$ To see why this is true, note that if $f \in C[0,1]$ is in the $C[0,1]$ closure of $\mathcal{M}$, then $f$ is in the $L^2[0,1]$ closure of $\mathcal{M}$ because there is a sequence $\{ m_n \}\subset\mathcal{M}$ such that $\|f-m_n\|_{C}\rightarrow 0$, which forces $$ \|f-m_n\|_{L^2} \le \|f-m_n\|_{C} \rightarrow 0. $$ Once you can show that $\overline{\mathcal{M}}^{C} = C[0,1]$, then you know that $$ C[0,1]=\overline{\mathcal{M}}^{C} \subseteq\overline{\mathcal{M}}^{L^2} \\ \implies L^2=\overline{C[0,1]}^{L^2}\subseteq \overline{\mathcal{M}}^{L^2} \\ \implies L^2 = \overline{\mathcal{M}}^{L^2}. $$

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By Muntz theorem the subspace of $C^0[0,1]$ generated by $x^{\lambda_1},x^{\lambda_2},\ldots$ is a dense subspace of $C^0[0,1]$ iff $\sum_{n\geq 1}\frac{1}{\lambda_n}$ is diverging. That is clearly the case if $\lambda_n=3n$. At last, it is enough to exploit the density of $C^0[0,1]$ in $L^2[0,1]$.

On the other hand, we do not need the full power of Muntz theorem to prove the first part. If $f(x)$ is a continuous function on $[0,1]$, so it is $g(x)=f(\sqrt[3]{x})$, and by Weierstrass approximation theorem there is a sequence of polynomials $\{p_n(x)\}_{n\geq 1}$ uniformly convergent to $g(x)$ on $[0,1]$. It follows that $\left\{p_n(x^3)\right\}_{n\geq 1}$ is a sequence of polynomials in the subspace generated by $1,x^3,x^6,\ldots$ uniformly convergent to $f(x)$ on $[0,1]$.

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    This looks really good! Unfortunately, I am not familiar with this theorem. – 2017-02-27