Orthonormal basis in inner product space

Question

Given that $V$ is an inner product space

and ${e_n} \in V ,n=1,2,3...$ is an Orthonormal basis

and $\lambda_n$ is a scalar series, How did they get from the first step to the second step?

Answer 1

$$\big\|\sum_{n=1}^N (\lambda_n-\langle u,e_n\rangle) e_n\big\|^2=\left\langle\sum_{n=1}^N (\lambda_n-\langle u,e_n\rangle )e_n,\sum_{m=1}^N (\lambda_m-\langle u,e_m\rangle )e_m \right\rangle\\=\sum_{n=1}^N \sum_{m=1}^N (\lambda_n-\langle u,e_n\rangle )\cdot(\lambda_m-\langle u,e_m\rangle ) \langle e_n,e_m \rangle=\sum_{n=1}^N \sum_{m=1}^N (\lambda_n-\langle u,e_n\rangle )\cdot(\lambda_m-\langle u,e_m\rangle ) \delta_{n,m}\\=\sum_{n=1}^N \left |\lambda_n-\langle u,e_n\rangle \right|^2$$

That checks the first equality.

The second equality follows from finite sums: $$\sum_{n=1}^N (\lambda_n-\langle u,e_n\rangle)e_n=\sum_{n=1}^N \lambda_n e_n-\langle u,e_n\rangle e_n=\sum_{n=1}^N \lambda_n e_n-\sum_{n=1}^N\langle u,e_n\rangle e_n$$

The last part is the triangle inequality: $$\|\sum_{n=1}^N \lambda_n e_n-\sum_{n=1}^N\langle u,e_n\rangle e_n\|=\|\left(-v+\sum_{n=1}^N \lambda_n e_n\right)+\left(v-\sum_{n=1}^N\langle u,e_n\rangle e_n\right)\|\\ \leq \|\left(-v+\sum_{n=1}^N \lambda_n e_n\right)\|+\|\left(-v+\sum_{n=1}^N\langle u,e_n\rangle e_n\right)\|$$

Though I imagine if they want this to tend to zero, when $N\to\infty$ they would want $v=u$.