# Proving the Cauchy-Schwarz integral inequality in a different way

Suppose $\alpha$ is monotonically increasing on $[a,b]$. Also suppose $f,g: [a,b] \rightarrow \mathbf{R}$, and both functions are Riemann-Stieltjes integrable with respect to $\alpha$. That is, $\forall \epsilon > 0$ there is a partition $P_\epsilon$ such that for all upper and lower sums: \begin{equation*} U(P_\epsilon,f,\alpha) - L(P_\epsilon,f,\alpha) < \epsilon \end{equation*}

Use the Cauchy-Schwarz inequality

To prove the following: \begin{equation*} \left|\int_a^b f(x)g(x)d\alpha (x)\right|^2 \leq \left(\int_a^b [f(x)]^2 d\alpha(x)\right) \left(\int_a^b [g(x)]^2 d\alpha(x)\right) \end{equation*}

I've seen this proof using done by looking at $0 \leq \int_a^b (\lambda f(x) + g(x))^2 dx$, and then looking at the resulting quadratic equation. This problem, however, seems to be a more general case. How could I approach this?

• Strange that the LHS of the inequality to be proven doesn't contain $\alpha$. – mathcounterexamples.net Mar 18 '17 at 20:33
• It does. That was my mistake - I just editted it. – user20354139 Mar 18 '17 at 20:37

I know two beautiful direct proofs of this fact. I like a lot the second one!

First proof. If $\alpha$ is monotonically increasing on the set $[a,b]\subsetneq \mathbb R$ , let $\mathcal{R}_\alpha[a, b]$ denote the algebra of Riemann-Stieltjes integrable functions with respect to $\alpha$.
First, if $f, g\in \mathcal{R}_\alpha[a, b]$, then $f^2, g^2\in \mathcal{R}_\alpha[a, b]$.

Let us denote $\lambda:= \displaystyle\sqrt{\int\limits_{a}^{b}f(x)^2\,d\alpha}$ and $\mu:= \displaystyle\sqrt{\int\limits_{a}^{b}g(x)^2\,d\alpha }$ . Define $h:[a,b]\to \mathbb{R}$ by $h(x)=\lambda g(x)- \mu f(x)$. Clearly $\lambda,\mu\geq0$ and $h\in \mathcal{R}_\alpha[a, b]$, so
\begin{align}&\\ 0&\leq \int\limits_{a}^{b} h(x)^2\,d\alpha=\int\limits_{a}^{b}\big(\lambda g(x)- \mu f(x)\big)^2\,d\alpha\\ &=\lambda^2\int\limits_{a}^{b}g(x)^2\,d\alpha-2\lambda \mu \int\limits_{a}^{b}f(x) g(x) \,d\alpha+\mu^2 \int\limits_{a}^{b}f(x)^2\,d\alpha \\&=\lambda^2 \mu^2 -2\lambda\mu \int\limits_{a}^{b}f(x)g(x)\,d\alpha+\mu^2 \lambda^2 =2\lambda \mu \left( \lambda \mu- \int\limits_{a}^{b}f(x)g(x)\,d\alpha \right). \end{align}
Since $f, g\in \mathcal{R}_\alpha[a, b]$ are assumed to be both non-zero (otherwise, the desired inequality is obvious), then $2\lambda \mu >0$, so it has to be that $\displaystyle \lambda \mu- \int\limits_{a}^{b}f(x)g(x)\,d\alpha \geq 0$, that is $$\int\limits_{a}^{b}f(x)g(x)\,d\alpha \leq\lambda \mu \qquad\qquad (*)$$ On the other hand, $\displaystyle -\int\limits_{a}^{b}f(x)g(x)\,d\alpha=\int\limits_{a}^{b}\big( (-f)(x) \big) g(x)\,d\alpha$, so, by means of $(*)$, \begin{align} -\int\limits_{a}^{b}f(x)g(x)\,d\alpha&=\int\limits_{a}^{b}\big( (-f)(x) \big) g(x)\,d\alpha\leq \sqrt{\, \int\limits_{a}^{b} \big(-f(x)\big)^2 \,d\alpha}\,\cdot \, \sqrt{\, \int\limits_{a}^{b} g(x)^2\,d\alpha }\\&= \sqrt{\, \int\limits_{a}^{b} f(x)^2 \,d\alpha} \, \cdot \, \sqrt{\, \int\limits_{a}^{b} g(x)^2\,d\alpha }. \end{align} That is, we have proved that $\displaystyle\left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|\leq \lambda \mu$. Equivalently, $$\left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|^2\leq \lambda^2 \mu^2= \left(\int\limits_a^b f(x) ^2 \,d\alpha \right) \left(\int\limits_a^b g(x) ^2 \,d\alpha \right),$$ as it was to be shown.

Second proof. We take $\lambda,\mu$ as before, and since $\displaystyle\left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|\leq \int\limits_{a}^{b}\big|f(x)g(x)\big| \,d\alpha$, the Cauchy-Schwarz inequality can be proved just by showing that $$\int\limits_{a}^{b}\big|f(x)g(x)\big| \,d\alpha\leq \lambda\mu =\sqrt{\int\limits_{a}^{b}f(x)^2\,d\alpha }\:\cdot \sqrt{\int\limits_{a}^{b}g(x)^2\,d\alpha } .$$ Equivalently (if we asumme that $f\neq 0 \neq g$), $$\int\limits_{a}^{b}\frac{\big|f(x)g(x)\big|}{\lambda \mu }\,d\alpha=\int\limits_{a}^{b}\frac{\big|f(x) \big|}{\lambda } \cdot\frac{\big| g(x)\big|}{ \mu }\,d\alpha \leq 1.$$ Indeed, since for every pair of real numbers $\zeta, \eta\in\mathbb{R}$ we have that $|\zeta\eta|\leq \frac{\zeta^2+\eta^2}{2}$, it follows that \begin{align*} \int\limits_{a}^{b}\frac{\big|f(x) \big|}{\lambda } \cdot\frac{\big| g(x)\big|}{ \mu }\,d\alpha&\leq \int\limits_{a}^{b}\frac{\frac{f(x)^2}{\lambda^2}+ \frac{g(x)^2}{\mu^2} }{2}\,d\alpha=\frac{1}{2}\left(\, \int\limits_{a}^{b}\frac{f(x)^2}{\lambda^2}\,d\alpha +\int\limits_{a}^{b} \frac{g(x)^2}{\mu^2}\,d\alpha \right)=\frac{1}{2}\left(\frac{\lambda^2}{\lambda^2 }+\frac{ \mu^2}{ \mu^2}\right)=1. \end{align*}

Hint: Taking $[0,1]$ for simplicity and letting $d\alpha = dx,$, note

$$\sum_{k=1}^{n}f\left (\frac{k}{n}\right )g\left ( \frac{k}{n}\right )\frac{1}{n} = \sum_{k=1}^{n}\left [f\left (\frac{k}{n}\right )\frac{1}{\sqrt n}\right ]\left [g\left ( \frac{k}{n}\right )\frac{1}{\sqrt n}\right ] .$$

Take $a_k,b_k$ to be the terms inside the brackets.

• Then as $n \to \infty$ the sums on both sides of the Cauchy inequality approach integrals, right? Doesn't this assume the partition is evenly spaced? I'm unsure how you would be able to generalize this for any interval when each interval length is not uniformly $1/n$ but rather $\alpha(x_k) - \alpha(x_{k-1})$. – user20354139 Mar 18 '17 at 21:16
• @antsatsui: If the finite integration exists, then you can choose whatever partition you'd like, and you'll still arrive at the one, and only result; so choosing evenly spaced partition is the simplest way to go. – user49685 Mar 19 '17 at 2:01
• @antsatsui Replace $1/n$ with $\Delta \alpha_k.$ – zhw. Mar 20 '17 at 1:46