Past exam of the functional analysis course of the University of Oxford / 2011 / 3 / ii

Fix $i$ and let $M_i=\overline{\mathrm{span}}\{x_j:j\ne i\}$. The hypothesis gives $M_i\ne X$. Moreover $x_i\notin M_i$, since otherwise the closed span of all the vectors would be contained in $M_i$. Hence $d_i=\mathrm{dist}(x_i,M_i)>0$.

On $M_i+\mathbb{R}{x}_i$, define ${\varphi }_i(m+t{x}_i)=t$. It is well-defined because $x_i\notin M_i$. For $t\ne 0$,so $|{\varphi }_i(m+t{x}_i)|\le \Vert m+t{x}_i\Vert /d_i$; the inequality also holds for $t=0$. The Hahn-Banach theorem extends ${\varphi }_i$ without increasing its norm to $x_i^{\ast }\in X^{\ast }$. This gives $x_i^{\ast }(x_j)={\delta }_{ij}$.

If two continuous functionals have these values, their difference vanishes on the linear span of every $x_j$, which is dense in $X$. Continuity then makes the difference zero everywhere. Thus each functional is unique, proving the required unique biorthogonal system. In fact its norm is $\Vert x_i^{\ast }\Vert =1/d_i$, by taking vectors in $M_i$ approaching the distance infimum. This is the coordinate-functional construction for a minimal total family in a normed space.