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State and prove bessel inequality​

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Answer:

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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

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