Show max $\lVert proj_{w^\perp} u_i \rVert^2 = 1, \quad \lVert u_i \rVert = 1$

Question

given an orthonormal basis $w_1,...,w_k$ for $w^\perp$ and a unit vector $u_i$, how can I show that $$0 \le\lVert proj_{w^\perp} u_i \rVert^2 = \lVert \sum_{j=1}^k \langle u_i, w_j \rangle w_j \rVert ^2 \le 1 $$ ?

What I have tried is this $$ \lVert \sum_{j=1}^k \langle u_i, w_j \rangle w_j \rVert \le \sum_{j=1}^k \lVert \langle u_i, w_j \rangle w_j \rVert = \sum_{j=1}^m \lvert \langle u_i ,w_j \rangle \rvert \lVert w_j \rVert = \sum_{j=1}^m \lvert \langle u_i ,w_j \rangle \rvert$$ How can that last sum be shown to be less than or equal to 1? Thanks!

Answer 1

Hint: Observe \begin{align} \Big\| \sum^k_{j=1}\langle u_i, w_j\rangle w_j\Big\|^2 = \sum^k_{j=1}|\langle u_i, w_j\rangle|^2 \leq \|u_i\|^2 = 1. \end{align}