Linear independence and Schauder basis

Question

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!

Answer 1

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.

Answer 2

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).