Statement :- We assume the orthagonal sequence [tex]{{\{\phi\}}_{1}^{\infty}}[/tex] in Hilbert space, now [tex]{\forall \sf \:v\in \mathbb{V}}[/tex], the Fourier coefficients are given by:
[tex]{\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}[/tex]
Then Bessel's inequality give us:
[tex]{\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}[/tex]
Proof :- We assume the following equation is true
[tex]{\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}[/tex]
So that, [tex]{\bf v_n}[/tex] is projection of [tex]{\bf v}[/tex] onto the surface by the first [tex]{\bf n}[/tex] of the [tex]{\bf \phi_{i}}[/tex] . For any event, [tex]{\sf (v-v_{n})\perp v_{n}}[/tex]
Now, by Pythagoras theorem:
[tex]{:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}[/tex]
[tex]{:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}[/tex]
Now, we can deduce that from the above equation that;
[tex]{:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}[/tex]
For [tex]{\sf n\to \infty}[/tex], we have
[tex]{:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}[/tex]
Hence, Proved
