Dimostrazione.
Prendiamo $\mathbf{w} \neq 0, \,\, \mathbf{v} \in \mathbf{H}, \quad \alpha, \beta \in \mathbf{C}$, allora
$0 \leq \ll \alpha \mathbf{v}+ \beta \mathbf{w},\alpha \mathbf{v}+\beta \mathbf{w}\gg=$
$= \ll \alpha \mathbf{v},\alpha \mathbf{v}\gg+\ll\alpha \mathbf{v},\beta \mathbf...
...beta \mathbf{w},\alpha \mathbf{v}\gg+\ll \beta \mathbf{w},\beta \mathbf{w} \gg=$
$=\alpha \overline{\alpha} \ll \mathbf{v},\mathbf{v}\gg+\alpha \overline{\beta} ...
...l \mathbf{w},\mathbf{v} \gg +\beta \overline{\beta}\ll \mathbf{w},\mathbf{w}\gg$.
Sostituendo i valori $\alpha=\ll \mathbf{w},\mathbf{w}\gg$ e $\beta=- \ll \mathbf{v},\mathbf{w} \gg$ otteniamo (svolgendo i calcoli):
$0 \leq \vert\vert\mathbf{w}\vert\vert^4\vert\vert\mathbf{v}\vert\vert^2-\vert\v...
...\vert^2 \ll \mathbf{v},\mathbf{w} \gg \overline{\ll \mathbf{v},\mathbf{w} \gg} $; e poiché
$ \ll \mathbf{v},\mathbf{w} \gg \overline{\ll \mathbf{v},\mathbf{w} \gg}=\vert\ll \mathbf{v},\mathbf{w}\gg\vert^2$
si ha $\vert\vert\mathbf{w}\vert\vert^2 \cdot \vert\ll \mathbf{v},\mathbf{w}\gg\vert^2 \leq \vert\vert\mathbf{w}\vert\vert^4 \vert\vert\mathbf{v}\vert\vert^2$;
quindi dividendo per $\vert\vert\mathbf{w}\vert\vert^2$ otteniamo il risultato voluto sotto la condizione che $\alpha \mathbf{v}+\beta \mathbf{w}=0$, quindi solo se i due vettori sono l.d., ovvero paralleli.
c.v.d.