cauchy-schwarz theorem

[[concept]]

Topics

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Cauchy-Schwarz Theorem

Let $V,⟨,⟩$ be an inner product space over $K$. Then for all $x,y\in V$ we have $|⟨x,y⟩{|}^2\le ⟨x,x⟩⟨y,y⟩$ Or equivalently, if working with a norm induced by an inner product, $|⟨x,y⟩|\le ||x||\cdot ||y||$

$K=\mathbb{C}$) For any $t,\theta \in \mathbb{R}$ it is true that and

(here we prove for

0 &\leq \langle te^{i\theta} + y, te^{i\theta}x,y\rangle \\ &= t^2 e^{i\theta}\overline{e^{i\theta}} \langle x,x \rangle + t e^{i\theta} \langle x,y \rangle + te^{i\theta} \langle y, x \rangle + \langle y, y \rangle \\ &= t^2 \langle x,x \rangle + 2 t Re[e^{i\theta} \langle x,y \rangle] + \langle y, y \rangle \\ \text{choose }\theta \text{ s.t. } Re[e^{i\theta} \langle x,y \rangle] &= \lvert \langle x, y \rangle \rvert \\ &= t^2 \langle x,x \rangle + 2 t \lvert \langle x, y \rangle \rvert + \langle y, y \rangle \\ & \geq 0 \;\;\;\forall t \in \mathbb{R} \end{aligned}$$ And this is a quadratic in $\mathbb{R}$! So the discriminant must be negative, that is $$[2\lvert \langle x, y \rangle \rvert]^2 - 4[\langle x,x \rangle \langle y,y \rangle] \leq 0$$ $$\implies \lvert \langle x, y \rangle \rvert^2 \leq \langle x,x \rangle \langle y,y \rangle$$

References

References

See Also

Mentions

Mentions

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