Bessels Inequality

Often used to find a lower bound for a function. Heavily used in optimization.

Corrolary

• let $\{{\alpha }_1,\dots ,{\alpha }_n\}$ be an orthogonal set of non-zero vectors in an Inner Product Space
• If $\beta$ is any vector in $V$
• Then, ${\sum }_{i=1}^n\frac{|⟨\beta |{\alpha }_k⟩{|}^2}{||{\alpha }_k|{|}^2}\le ||\beta |{|}^2$
  • With equality $⟺\beta ={\sum }_{i=1}^n\frac{|⟨\beta |{\alpha }_k⟩{|}^2}{||{\alpha }_k|{|}^2}{\alpha }_k$

Proof

1. Let $g={\sum }_{i=1}^n\frac{|⟨\beta |{\alpha }_k⟩{|}^2}{||{\alpha }_k|{|}^2}{\alpha }_k$ and $\delta =\beta -\gamma$
2. Then, $\beta =\alpha +\delta$ by definition of $\gamma$, $⟨\delta |\gamma ⟩=0$
3. Then, by Generalized Pythagorean Theorem $||\beta |{|}^2=||\gamma |{|}^2+||\delta |{|}^2$
4. By definition of $\gamma$, $||\gamma |{|}^2={\sum }_{i=1}^n\frac{|⟨\beta |{\alpha }_2{⟩}^2|}{||{\alpha }_k|{|}^2}\le ||\gamma |{|}^2+||\delta |{|}^2=||\beta |{|}^2$, with equality $⟺||\delta |{|}^{2=0}$
5. It is now sufficient to show $||\delta ||={\sum }_{i=1}^n\frac{|⟨\beta |{\alpha }_k⟩{|}^2}{||{\alpha }_k|{|}^2}$