Linear Algebra Proof Exercise Involving Cauchy-Schwarz Inequality

Question

So, for continuous $f: [1,\infty) \rightarrow [0,\infty)$, I must show that $$\left(\int_1^{\infty} f(x) dx\right)^2 \le \left(\int_1^{\infty} x^2(f(x))^2 dx\right).$$

I know I that I should define the inner product to be $\langle f,g\rangle = \int_1^{\infty} fg)$ and that the Cauchy-Schwarz inequality states that $|\langle f,g\rangle| \le ||f||||g||$.

The left side of the problem's inequality can be written as $\langle f, 1\rangle^2$ which is less than or equal to $||f||^2||1||^2$. My thoughts were that since $x$ is between $1$ and $\infty$, $||f||^2||1||^2 \le ||f||^2 ||x||^2$ but this evaluates to $(\int_1^{\infty} (f(x))^2 dx)(\int_1^{\infty} x^2 dx)$, which is not what I need. I've tried some other approaches, but this seems to be the closest, but I am not sure where to go from here.

Edit: The norm is $||f|| = \sqrt{\langle f,f\rangle}$ here.

Answer 1

You want to apply the Cauchy-Schwarz inequality $$ \langle F, G \rangle \le \Vert F \Vert \cdot \Vert G \Vert $$ with the inner product $\langle F, G \rangle = \int_1^\infty F(x) G(x) \, dx$ and the corresponding norm $\Vert F \Vert = \sqrt{\langle F, F \rangle}$.

In order to get the expression $\int_1^\infty x^2 f(x)^2 \, dx$ on the right-hand side of the inequality we can choose $F(x) = x f(x)$, and to get $\int_1^\infty f(x) \, dx$ on the left-hand side we must then choose $G(x) = 1/x$.

This choice does indeed lead to the desired result: $$ \begin{align} \int_1^\infty f(x) \, dx &= \int_1^\infty xf(x) \cdot \frac 1x \, dx \\ &= \langle F, G \rangle \le \Vert F \Vert \cdot \Vert G \Vert \\ &= \left( \int_1^\infty x^2 f(x)^2 \, dx\right)^{1/2} \cdot \underbrace{\left( \int_1^\infty \frac{dx}{x^2}\right)^{1/2}}_{= 1} \, . \end{align} $$

Applying the Cauchy-Schwarz inequality to $\langle f, 1 \rangle$ does not help because $\int_1^\infty 1 \, dx$ is not finite.