Hilbert space is orthornormality needed for representation?

Question

In a Hilbert space $H$ with countable basis, if I know there is a countable basis $\{h_n\}$ of $H$ then can I express every element $h\in H$ therein as: \begin{equation} h = \sum_n \langle h,h_n\rangle h_n? \end{equation}

Answer 1

For a simple example to demonstrate the importance of the basis being orthonormal, consider $\mathbb{R}^2$ with the standard inner product and the basis $h_1=(1,0)$ and $h_2=(1,1)$. If $h=(0,1)$ then $$ h=-h_1+h_2 $$ but $\langle h,h_1\rangle=0$.

By the way, a Hilbert space basis of $H$ is different from a basis of $H$ as an abstract vector space (called a Hamel basis). The latter is generally much larger, as only finite linear combinations of the elements are permitted.

Answer 2

Re : The proposer's comment to carmichael561's answer: Let $\{e_n:n\in N\}$ be an orthonormal Hilbert-space basis for $H.$ Let $h_1=e_1$ and $h_2=e_1-e_2. $ For $n\geq 3$ let $h_n=e_1+2e_{n-1}-e_n.$ Then $e_1$ has two representations with respect to $\{h_n: n\in N\},$ namely $$(i) \quad e_1=h_1.$$ $$(ii) \quad e_1=\sum_{n=1}^{\infty} h_n 2^{-n}.$$