Use the quadratic formula to derive the Cauchy-Schwarz Inequality.

Question

This question comes from Royden & Fitzpatrick Real Analysis. Please help me find my mistake.

Let $f$ and $g$ belong to $L^2(E)$. From the linearity of integration, for any number $\lambda: \lambda^2 \int_E f^2 + 2 \lambda \int_E f \cdot g + \int_E g^2 = \int_E (\lambda f + g)^2 \ge 0$. From this and the quadratic formula directly derives the Cauchy-Schwarz Inequality.

The Cauchy Schwarz Inequality: Let $E$ be a measurable set and $f$ and $g$ measurable functions on $E$ for which $f^2$ and $g^2$ are integrable over $E$. Then their product $f \cdot g$ also is integrable over $E$ and $\int_E |fg| \le \sqrt{ \int_E f^2} \cdot \sqrt{ \int_E g^2}$.

I clearly need a nap because I'm arriving at the opposite result. Here is my approach: I am using the quadratic formula to determine $\lambda$ when the expression =0. Meaning, $\lambda = \frac{-b +/- \sqrt{ b^2 - 4ac}}{2a}$ for $a = \int_E f^2$, $b= 2 \int_E fg$, and $c= \int_E g^2$. The fact that $\lambda$ is a real number indicates $b^2 - 4ac \ge 0$.

By substitution then $(2 \int_E fg)^2 - 4 \cdot \int_E f^2 \cdot \int_E g^2 \ge 0$ iff $4(\int_E fg)^2 - 4 \cdot \int_E f^2 \cdot \int_E g^2 \ge 0$ iff $(\int_E fg)^2 - \int_E f^2 \cdot \int_E g^2 \ge 0$ iff $(\int_E fg)^2 \ge \int_E f^2 \cdot \int_E g^2$ - which is the opposite of the result I want.

Am I supposed to have a different approach? Have I made some ridiculously silly mistake? Thank you.

Answer 1

Because the polynomial $ \lambda^2 \int_E f^2 + 2 \lambda \int_E f \cdot g + \int_E g^2 \ge 0$ (in $\lambda)$ is nonnegative, it can have at most one real root (think of a parabola---to have two roots it needs to dip below the x-axis). Thus the discrimant is actually less than or equal to $0$, so you have the inequality backwards.

Answer 2

Hint:

You've forgotten that a quadratic polynomial with real coefficients has a constant sign or is zero if and only its (reduced) discriminant is non positive. The reduced discriminant of $ax^2+2b'x+c$ is $$\Delta'=b'^2-ac=\tfrac14\Delta.$$

Answer 3

You made your mistake in assuming $\lambda$ is real, in fact you want just the opposite (hence the reversal of your inequality).

Since $f,g$ are real valued, for any real $\lambda$, $(\lambda f + g)^2 \geq 0$, hence $\int (\lambda f + g)^2 \geq 0$. Suppose $\int (\lambda f + g)^2$, then $\lambda f + g = 0 $ almost everywhere, so $f = -g/\lambda$. In this case, you can prove the Cauchy-Schwartz inequality directly: $$\int |fg| = \int g^2/\lambda = \sqrt{\int (g/\lambda)^2} \sqrt{\int g^2} = \sqrt{\int f^2} \sqrt{\int g^2}.$$

Otherwise $\int (\lambda f + g)^2 > 0$ for all real $\lambda$, so we must have a complex root, so the discriminant is $< 0$, 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.