Let $X$ be a Hilbert space an $(e_k)_{k\geq 1}$ be a complete orthonormal set (i.e. a basis). Then define $$F_n:X\rightarrow X;~~~x\mapsto \sum_{k=1}^n \langle x,e_k\rangle e_k$$I want to prove that $||F_n-F_m||=1$ if $n\neq m$.
W.l.o.g. we can assume $m<n$, then $$\begin{align}||F_n-F_m||&=\sup_{||x||=1}||F_n(x)-F_m(x)||\\&=\sup_{||x||=1} \left|\left|\sum_{k=m+1}^n \langle x,e_k\rangle e_k \right|\right|\\&=\sup_{||x||=1} \left(\sum_{k=m+1}^n |\langle x, e_k\rangle|^2\right)^{1/2}\end{align}$$
Now somehow I don't see why the last equality should be true. I read something that it has to do with the fact that $(e_k)$ is orthonormal, but I don't see why.
Can maybe someone explain me the last equality?
Let $\alpha_j:= \langle x,e_j\rangle$. By definition of the norm $$ \left\lVert\sum_{k=m+1}^n \alpha_k e_k\right\rVert^2 =\left\langle \sum_{j=m+1}^n \alpha_j e_j,\sum_{k=m+1}^n \alpha_k e_k\right\rangle $$ and by bilinearity of inner product, $$ \left\lVert\sum_{k=m+1}^n \alpha_k e_k\right\rVert^2 =\sum_{j=m+1}^n\sum_{k=m+1}^n \alpha_j\overline{\alpha_k } \left\langle e_j, e_k\right\rangle $$ and by orthonormality, $\left\langle e_j, e_k\right\rangle=1$ if $j=k$ and $0$ otherwise.