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.
Best Answer
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 $$\Delta'=b'^2-ac=\tfrac14\Delta.$$