Use the quadratic formula to derive the Cauchy-Schwarz Inequality.Let f and g belong to L2(E). From the linearity of integration, for any numberλ:λ2∫Ef2+2λ∫Ef⋅g+∫Eg2=∫E(λf+g)2≥0From this and the quadratic formula directly derives the Cauchy-Schwarz Inequality.

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chabinka61jx · Accepted Answer

Step 1You made your mistake in assuming λ is real, in fact you want just the opposite (hence the reversal of your inequality).Since f, g are real valued, for any real λ,(λf+g)2≥0, hence ∫(λf+g)2≥0. Suppose ∫(λf+g)2, then λf+g=0 almost everywhere, so f=−gλIn this case, you can prove the Cauchy-Schwartz inequality directly:∫|fg|=∫g2λ=∫(gλ)2∫g2=∫f2∫g2.Otherwise ∫(λf+g)2&amp;gt;0 for all real λ, so we must have a complex root, so the discriminant is &amp;lt;0, and the same argument you made but with the signs reversed gives you the correct answer.Note that you even get the full version of Cauchy Schwartz, with gives equality iff f,g are linearly dependent.

Use the quadratic formula to derive the Cauchy-Schwarz

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