Considering $f_k \in l^2 $ defined as
$$\begin{align} f_{2n-1}&=\frac{e_{2n-1}-e_{2n}}{\sqrt{2}} \\
f_{2n}&= \frac{e_{2n-1}+e_{2n}}{\sqrt{2}},
\end{align}$$ where $e_k=(0,0,0,\dots, 0,1,0,\dots)$ with $1$ in $k$-th place.How to show that $\{f_k\}$ is an orthonormal set?
I tried to find the integral of inner product however ended up with something different.
Best Answer
First we show that $\left<e_n,e_m\right>=0$ for $n\neq m$. We do this by considering $$\left<e_n,e_m\right> = \sum_{k=1}^\infty e_n^ke_m^k.$$ Remark that $e_n^k$ is only nonzero if $k=n$, and likewise $e_m^k$ is only nonzero if $k=m$. It follows that in every term, at least one of the factors is zero, since $n\neq m$. By consequence, the sum evaluates to zero.
The following proof for orthonormality illustrates how we apply this.
For $f_{2n}$ and $f_{2m}$ with $n < m$ we get $$ \begin{align*} \left<f_{2n},f_{2m}\right> = {} &\frac12\left<e_{2n-1} + e_{2n},e_{2m-1} + e_{2m}\right> \\ = {} & \frac12\left(\left<e_{2n-1},e_{2m-1}\right> + \left<e_{2n},e_{2m-1}\right> + \left<e_{2n-1},e_{2m}\right> + \left<e_{2n},e_{2m}\right>\right). \end{align*} $$ Because $n<m$, the operands are orthogonal in every scalar product, which means every scalar product is zero. I will do the calculations for the third product, since it's slightly less trivial. Since $n<m$ we have $n\leq m-1$, which means $2n\leq 2m-2 < 2m-1$. Therefore, $e_{2n}$ is not equal (and hence is orthogonal) to $e_{2m-1}$. For the other cases, we make the same inequality argument, but finding the needed inequality is easier.
For $f_{2n}$ and $f_{2m-1}$ with $n < m$, we get the same four scalar products, but the last two will be prepended by a minus sign, which doesn't invalidate the above argument.
For $f_{2n-1}$ and $f_{2n}$ with $n < m$ we get
$$ \begin{align*} \left<f_{2n-1},f_{2n}\right> = {} &\frac12\left<e_{2n-1} - e_{2n},e_{2n-1} + e_{2n}\right> \\ = {} & \frac12\left(\left<e_{2n-1},e_{2n-1}\right> - \left<e_{2n},e_{2n-1}\right> + \left<e_{2n-1},e_{2n}\right> - \left<e_{2n},e_{2n}\right>\right) \\ = {} & \frac12(1+0+0-1) = 0. \end{align*} $$
We have now proven that the $f_k$ are orthogonal. To complete the argument, we consider $$ \begin{align*} \left<f_{n},f_{n}\right> = {} &\frac12\left<e_{n-1} \pm e_{n},e_{n-1} \pm e_{n}\right> \\ = {} & \frac12\left(\left<e_{n-1},e_{n-1}\right> \pm \left<e_{n-1},e_{n}\right> \pm \left<e_{n},e_{n-1}\right> + \left<e_{n},e_{n}\right>\right) \\ = {} & \frac12(1 \pm 0 \pm 0 + 1) = 1. \end{align*} $$