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The set $\{f_n : n \in \mathbb{Z}\}$ with $f_n(x) = e^{2πinx}$ forms an orthonormal basis of the complex space $L_2([0,1])$.

I understand why its ON but not why its a basis?

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Assume that you can find an $f$ in $L^2$ that is orthogonal to all $\sin (nx)$ and $\cos (nx)$. Then show that $f$ has to be zero almost everywhere. – Rudy the Reindeer Apr 20 '12 at 12:32
up vote 0 down vote accepted

It is known that orthonormal system $\{f_n:n\in\mathbb{Z}\}$ is a basis if $$ \operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=L_2([0,1]) $$ where $\operatorname{cl}_{L_2}$ means the closure in the $L_2$ norm.

Denote by $C_0([0,1])$ the space of continuous functions on $[0,1]$ which equals $0$ at points $0$ and $1$. It is known that for each $f\in C_0([0,1])$ the Feier sums of $f$ uniformly converges to $f$. This means that $$ \operatorname{cl}_{C}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=C_0([0,1]) $$ where $\operatorname{cl}_{C}$ means the closure in the uniform norm.

Since we always have inequality $\|f\|_{L_2([0,1])}\leq\|f\|_{C([0,1])}$, then $$ \operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=C_0([0,1]) $$

It is remains to say that $C_0([0,1])$ is dence subspace of $L_2([0,1])$, i.e. $$ \operatorname{cl}_{L_2}(C_0([0,1]))=L_2([0,1]) $$ then we obtain $$ \operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))= \operatorname{cl}_{L_2}(C_0([0,1]))=L_2([0,1]) $$

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Thanks A lot! Very informative. – rk101 Apr 20 '12 at 12:37

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