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Consider a vector space ${\cal S}$ with inner product $(\cdot, \cdot)$. The Cauchy-Schwarz Inequality reads $$ (y_1, y_1) (y_2, y_2) \geq \left| (y_1, y_2) \right|^2~~\forall y_1, y_2 \in {\cal S} $$ This inequality is saturated when $y_1 = \lambda y_2$. In particular, this implies $$ \boxed{ (y_1, y_1) (y_2, y_2) \geq \left| \text{Im} (y_1, y_2) \right|^2 } $$ My question is the following. Given a fixed $y_1 \in {\cal S}$. Is it always possible to find $y_2 \in {\cal S}$ such that the boxed inequality is saturated?

If not in general, under what conditions is this possible?

PS - I am a physicist, and not that well-versed with math jargon.

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You have $|y_1|^2|y_2|^2\geq|(y_1,y_2)|^2\geq|\text{Im}(y_1,y_2)|$. If the right hand side is equal to the left-most hand side then, in particular, you have 'saturation' of Cauchy's inequality. Then $y_2=\lambda y_1$ because the saturation of Cauchy's inequality happens if and only if the vectors are proportional. You then only need to have $|\lambda|^2|y_1|^4=|(y_1,\lambda y_1)|^2=|\text{Im}(\overline{\lambda}(y_1,y_1))|^2=|\text{Im}(\overline{\lambda})|y_1|^2|^2=|\text{Im}(\lambda)|^2|y_1|^4$. This is ok for $\lambda$ imaginary.

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  • $\begingroup$ Ah! I see. Of course, that was stupid of me. Thank you! $\endgroup$ – Prahar Jul 16 '13 at 4:03

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