# Linear Algebra Proof Exercise Involving Cauchy-Schwarz Inequality

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.

• Hint: $f = \frac 1x \cdot (xf)$. Commented Mar 24 at 20:25
• Thank you. Would it be too much to ask if my thoughts so far are on the right track? If that is too revealing, no need to answer. Commented Mar 24 at 20:26
• I got it! Thanks! Commented Mar 24 at 20:47
• I see how this works, but may I also ask what the intuition behind it is? I see why this works, but how does one come to it? Commented Mar 24 at 21:01

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.