Let $(e_n)$ be a normalized basis for a Banach space $X$ and suppose there exists $x^{*} \in X^{*}$ with $x^{*}(e_n)=1$ for all n. Show that the sequence $(e_n-e_{n-1})_{n=1}^{\infty}$ is also a basis for $X$ (we let $e_0=0$ in this definition).

I was trying to prove it is a Schauder Basis and was trying to define bi-orthogonal functionals on $X$. But I am not being able to come up with anything. Any help is greatly appreciated!


Let $v \in X$, $v = \sum_{k = 1}^\infty a_k e_k$ in the (Schauder) sense that $$ \lim_{N \to \infty} \left\lVert v - \sum_{k = 1}^N a_k e_k\right\rVert = 0. $$

Note that the condition $x^*(e_n) = 1$ implies $\sum_{k = 1}^N a_k \to x^*(v)$ as $N \to \infty$, and define $$ b_n = x^*(v) - \sum_{k = 1}^{n-1} a_k. $$

Then, $$ \sum_{n = 1}^N b_n(e_n-e_{n-1}) = x^*(v)e_N - \sum_{n = 1}^N \sum_{k = 1}^{n-1} a_k(e_n-e_{n-1}) = x^*(v)e_N - \sum_{k = 1}^{N-1} \sum_{n = k+1}^{N} a_k(e_n-e_{n-1}) = x^*(v)e_N - \sum_{k = 1}^{N-1} a_k(e_N-e_k) = e_N\left( x^*(v) - \sum_{k = 1}^{N-1} a_k \right)+ \sum_{k = 1}^{N-1} a_ke_k, $$ hence $$ \left\lVert v - \sum_{k = 1}^N b_n (e_n - e_{n-1})\right\rVert \leq \left\lVert v - \sum_{k = 1}^{N-1} a_k e_k\right\rVert + \left\lvert x^*(v) - \sum_{k = 1}^{N-1} a_k \right\rvert, $$ which goes to zero. To show uniqueness of the $b_n$'s, simply note that one can recover $a_k = b_{k+1}-b_k$, and the $a_k$'s are unique by hypothesis.

  • $\begingroup$ What is $b_1$ according to your definition? $\endgroup$ – Topology Oct 2 '18 at 5:45
  • $\begingroup$ $b_1 = x^*(v)$, since the empty sum is zero. $\endgroup$ – Hugo Oct 2 '18 at 11:44

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