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lozanna [386]
2 years ago
5

State and prove bessel inequality​

Physics
2 answers:
maria [59]2 years ago
5 0

Statement :- We assume the orthagonal sequence {{\{\phi\}}_{1}^{\infty}} in Hilbert space, now {\forall \sf \:v\in \mathbb{V}}, the Fourier coefficients are given by:

{\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}

Then Bessel's inequality give us:

{\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}

Proof :- We assume the following equation is true

{\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}

So that, {\bf v_n} is projection of {\bf v} onto the surface by the first {\bf n} of the {\bf \phi_{i}} . For any event, {\sf (v-v_{n})\perp v_{n}}

Now, by Pythagoras theorem:

{:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}

{:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}

Now, we can deduce that from the above equation that;

{:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i}  \vert^{2}\leqslant \Vert v\Vert^{2}}

For {\sf n\to \infty}, we have

{:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}

Hence, Proved

tensa zangetsu [6.8K]2 years ago
3 0

Answer:

hope this helps and if you have any questions please feel free to ask me any time

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<em><u>I hope this helps!</u></em>

8 0
2 years ago
Read 2 more answers
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