# Non-orthogonal basis satisfying the Bessel's inequality

As we know that

If $H$ is a Hilbe$\{e_n\}$ be an orthonormal sequence then for all $x\in H$ $$\|x\|^2\ge \sum_{k=1}^\infty|\langle x,e_k\rangle|^2$$ and the infinite sum converges to $$\sum_{k=1}^\infty\langle x,e_k \rangle e_k.$$

I was thinking for that if we don't have an orthonormal basis then can we give an example of such sequence which satisfies $\|x\|^2=\sum_{k=1}^\infty| \langle x,e_k\rangle|^2\ \$ for all $x\in H$.

• What do you mean by 'don't have an orthonormal basis'? Do you mean any sequence or a Schauder basis that is not orthornomal? – copper.hat Dec 21 '16 at 19:43
• Any sequence or a Schauder basis anything. – Sachchidanand Prasad Dec 21 '16 at 19:47

If you have the Parseval equality, then the $e_n$ are orthogonal because $$\|e_n\|^2 = \sum_{k=1}^{\infty}|\langle e_n,e_k\rangle|^2=\|e_n\|^2+\sum_{k\ne n}|\langle e_n,e_k\rangle|^2.$$