Triangle inequality for the finite NUFFT Euclidean norm

First prove Cauchy–Schwarz for these arrays. If $y=0$, its inner product with $x$ is zero. If $y\mathrm{\ne }0$, choose $e_0$ with $y_{e_0}\mathrm{\ne }0$; [F2]–[F3] give $\mathrm{\parallel }y{\mathrm{\parallel }}_2^2={\sum }_e\mathrm{\mid }{y}_e{\mathrm{\mid }}^2>0$. Put $a=⟨x,y{⟩}_E$ and $c=a\mathrm{/}\mathrm{\parallel }y{\mathrm{\parallel }}_2^2$. Expanding the finite sum using [F1] and conjugation laws gives $0\le \mathrm{\parallel }x-cy{\mathrm{\parallel }}_2^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2-\frac{\mathrm{\mid }a{\mathrm{\mid }}^2}{\mathrm{\parallel }y{\mathrm{\parallel }}_2^2}\mathrm{.}$ Multiplying by the positive denominator yields $\mathrm{\mid }⟨x,y{⟩}_E{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2\mathrm{\parallel }y{\mathrm{\parallel }}_2^2$; all quantities are nonnegative, so [F4] gives $\mathrm{\mid }⟨x,y{⟩}_E\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }y{\mathrm{\parallel }}_2$.

Expand the squared norm: $\mathrm{\parallel }x+y{\mathrm{\parallel }}_2^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2+2\mathrm{Re}⟨x,y{⟩}_E+\mathrm{\parallel }y{\mathrm{\parallel }}_2^2\le (\mathrm{\parallel }x{\mathrm{\parallel }}_2+\mathrm{\parallel }y{\mathrm{\parallel }}_2{)}^2,$ where the inequality follows from step 1.1 and $\mathrm{Re}z\le \mathrm{\mid }z\mathrm{\mid }$, itself immediate from $\mathrm{\mid }a+ib{\mathrm{\mid }}^2=a^2+b^2\ge a^2$ and [F4]. Both sides before squaring are nonnegative, so [F4] gives the stated triangle inequality.