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The Cauchy-Schwarz integral inequality is as follows:

$$ \displaystyle \left({\int_a^b f \left({t}\right) g \left({t}\right) \ \mathrm d t}\right)^2 \le \int_a^b \left({f \left({t}\right)}\right)^2 \mathrm d t \int_a^b \left({g \left({t}\right)}\right)^2 \mathrm d t $$

How do I prove this using multivariable calculus methods, preferably with double integrals?

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    $\begingroup$ This isn't what you asked for, but to me it seems most clear to prove Cauchy-Schwarz in an abstract inner product space (guided by our intuition from $\mathbb R^n$), then get this inequality as a special case. $\endgroup$ – littleO Jan 3 '15 at 3:14
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If by double-integral you mean the identity: $$\frac{1}{2}\int_a^b\int_a^b (f(x)g(y) - g(x)f(y))^2\,dx\,dy \\= \int_a^b f^2(x)\,dx\int_a^b g^2(x)\,dx - \left(\int_a^b f(x)g(x)\,dx\right)^2$$

Then note that the integrand $\displaystyle (f(x)g(y) - g(x)f(y))^2 \ge 0$, hence the inequality follows.

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HINT: Use the polynomial function $$P(x) = \int_a^b \left(f(t) + xg(t)\right) ^2dt$$

And you will have two cases to prove : either $g = \Theta_{[a, b] \to \mathbb{R}}$ or not (the first case is pretty easy).

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So we take $P(\lambda)=\int_a^b (|f|+\lambda|g|)^2$

Then $$ P(\lambda)=\int_a^b |f|^2 +2\lambda\int_a^b |fg| +\lambda^2 \int_a^b |g|^2 $$

P is a 2nd degre polynome and beacause he never cancel his discriminant is negative then we have the inegality ! If I don't make a mistake sure...

Shadock

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  • $\begingroup$ It's actually not always negative tho, that's why you need $\leq$ and not $<$ $\endgroup$ – servabat Jan 3 '15 at 3:06
  • $\begingroup$ @servabat If I write "his discriminant is negative" is it right? $\endgroup$ – Hexacoordinate-C Jan 3 '15 at 3:08
  • $\begingroup$ Well I guess yes. I usually say negative or zero to be more clear for $\leq 0$, but that's just some habit. $\endgroup$ – servabat Jan 3 '15 at 3:09
  • $\begingroup$ @servabat I agree but by definition ... :) $\endgroup$ – Hexacoordinate-C Jan 3 '15 at 3:11
  • $\begingroup$ Yes, you are right. $\endgroup$ – servabat Jan 3 '15 at 3:11

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