Markushevich basis

In functional analysis, a Markushevich basis (sometimes M-basis^[1]) is a biorthogonal system that is both complete and total.^[2]

Definition

Let ${\displaystyle X}$ be Banach space. A biorthogonal system ${\displaystyle \{x_{\alpha };f_{\alpha }\}_{x\in \alpha }}$ in ${\displaystyle X}$ is a Markushevich basis if $${\displaystyle {\overline {\text{span}}}\{x_{\alpha }\}=X}$$ and $${\displaystyle \{f_{\alpha }\}_{x\in \alpha }}$$ separates the points of ${\displaystyle X}$.

In a separable space, biorthogonality is not a substantial obstruction to a Markushevich basis; any spanning set and separating functionals can be made biorthogonal. But it is an open problem whether every separable Banach space admits a Markushevich basis with ${\displaystyle \|x_{\alpha }\|=\|f_{\alpha }\|=1}$ for all ${\displaystyle \alpha }$.^[3]

Examples

Every Schauder basis of a Banach space is also a Markushevich basis; the converse is not true in general. An example of a Markushevich basis that is not a Schauder basis is the sequence $${\displaystyle \{e^{2i\pi nt}\}_{n\in \mathbb {Z} }\quad \quad \quad ({\text{ordered }}n=0,\pm 1,\pm 2,\dots )}$$ in the subspace ${\displaystyle {\tilde {C}}[0,1]}$ of continuous functions from ${\displaystyle [0,1]}$ to the complex numbers that have equal values on the boundary, under the supremum norm. The computation of a Fourier coefficient is continuous and the span dense in ${\displaystyle {\tilde {C}}[0,1]}$; thus for any ${\displaystyle f\in {\tilde {C}}[0,1]}$, there exists a sequence $${\displaystyle \sum _{|n|<N}{\alpha _{N,n}e^{2\pi int}}\to f{\text{.}}}$$But if ${\displaystyle f=\sum _{n\in \mathbb {Z} }{\alpha _{n}e^{2\pi nit}}}$, then for a fixed ${\displaystyle n}$ the coefficients ${\displaystyle \{\alpha _{N,n}\}_{N}}$ must converge, and there are functions for which they do not.^[3][4]

The sequence space ${\displaystyle l^{\infty }}$ admits no Markushevich basis, because it is both Grothendieck and irreflexive. But any separable space (such as ${\displaystyle l^{1}}$) has dual (resp. ${\displaystyle l^{\infty }}$) complemented in a space admitting a Markushevich basis.^[3]