# To prove $(e_n-e_{n-1})$ is basis for $X$

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