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It follows from the definition that a Schauder basis must be linear independent, i.e. every finite subset of the Schauder basis is linear independent.

I wonder if the following "converse" of this is also true:

Let $X$ be a Banach space and $\{e_n:n \ge 1 \}$ be a linear independent subset of $X$. If for any element $x \in X$, there exists a sequence $(a_n)_{n \ge 1}$ such that $x=\sum_{n=1}^\infty a_n e_n$, then such a sequence must be unique, i.e. $\{e_n:n \ge 1 \}$ is a Schauder basis.

Clearly it is true for finite dimensional $X$, but how about the infinite dimensional case? Does $X$ have to be complete? If it is not true, are there any counterexamples?

Thanks in advance!

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The answer is no. Take $e_{2n} = (t \mapsto t^n) $ and $e_{2n+1} = t \mapsto e^{in\pi t}$

If your function is an entiere serie and has also its Fourier serie that converge to itself, you have (at least !) two sequences $(a_n)$ that converge to $f$ :

  • $a_{2n} = b_n$, the coefficient of $z^n$ in the entiere serie developpement of $f$ and $a_{2n+1} = 0$
  • $a_{2n} = 0$, and $a_{2n+1}=c_n$ the coefficient of the Fourier serie.
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  • $\begingroup$ Oups, I meant $e^{in\pi t }$ $\endgroup$ – Tryss Dec 4 '15 at 17:13
  • $\begingroup$ Thank you! But how to show that $(e^{in\pi t})_{n \ge 1}$ are linear independent? $\endgroup$ – No One Dec 4 '15 at 17:17
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    $\begingroup$ @TiWen : Suppose that $$\forall t, \sum_{k=0}^n \lambda_k e^{itk\pi} = 0$$. Consider the polynom $$\sum_{k=0}^n \lambda_k (e^{it\pi} )^k $$, it has at most $n$ roots if it's not $0$, but $e^{it\pi}$ is a root for every $t$ by hypothesis : the polynom is zero and every $\lambda_k=0$ $\endgroup$ – Tryss Dec 4 '15 at 17:26
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Another example to think of: Consider the space $C_\infty[-1,1]$. The seqeunce $\{x^n\}_{n\geq0}$ is a complete set, and also the sequence $\{x^{5n}\}_{n\geq 1}$ is a complete set. (This does not automatically imply the representation is not unique, but i think that with some adjustments it can be understood that way).

I also wanted to add the following theorem: If $X$ is a Hilbert space, and you also assume that your sequence is orthonormal, then indeed it forms a basis (meaning; every vector has a unique representation).

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  • $\begingroup$ Thank you! But I don't quite understand what you mean by "complete set". And what "adjustment" should we make? $\endgroup$ – No One Dec 4 '15 at 18:43

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