Can Cauchy Schwarz inequality be proven using Jensen's inequality?

Question

After reading a comment on If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also exists, I wonder if Cauchy Schwarz inequality can be proven using Jensen's inequality?

Answer 1

Here are two such proofs.

1. One standard way of proving Hölder's inequality derives it from Young's inequality, which follows from AM/GM, which can be proven from Jensen's inequality applied to the convexity of $x\mapsto e^x$. If you take that proof of Hölder and specialize it to the case $p=q=2$, you'll have a proof of Cauchy-Schwarz which is at root from convexity.

2. I think this one is slightly less well-known. Consider two functions $f$ and $g$ (on $[0,1]$, say, or $\{1,\dotsc,n\}$ for the sequence case). Define $$ F = \frac{f}{\|f\|+\|g\|} \quad\text{and}\quad G = \frac{g}{\|f\|+\|g\|} $$ where $\|\cdot\|$ is the $L_2$ norm. (We do not assume that it is a norm; in fact we will prove the triangle inequality.) Note that $\|F\|+\|G\|=1$, so we can use them as coefficients in a convex combination. By Jensen's inequality and the convexity of $x\mapsto x^2$, $$ (f+g)^2 = \left(\|F\|\frac{f}{\|F\|} + \|G\|\frac{g}{\|G\|}\right)^2 \le \|F\|\left(\frac{f}{\|F\|}\right)^2 + \|G\|\left(\frac{g}{\|G\|}\right)^2 = \frac{f^2}{\|F\|} + \frac{g^2}{\|G\|} $$ Integrating yields $$ \|f+g\|^2 \le \frac{\|f\|^2}{\|F\|} + \frac{\|g\|^2}{\|G\|} = \|f\|\big(\|f\|+\|g\|\big) + \|g\|\big(\|f\|+\|g\|\big) = \big(\|f\|+\|g\|\big)^2 $$ Expanding the leftmost and rightmost expressions, we get $$ \|f\|^2 + 2\langle f,g\rangle + \|g\|^2 \le \|f\|^2 + 2\|f\|\|g\| + \|g\|^2 $$ and simplifying yields Cauchy-Schwarz. (This proof readily generalizes to give Minkowski's inequality without going through Hölder; I learned it from Garling's Inequalities: A Journey into Linear Analysis (Cambridge UP, 2007).)

Answer 2

We can prove Hölder's inequality for $p\in(1,\infty)$ (Cauchy-Schwarz is the case $p=2$ of Hölder) as follows.

Let $q(1-1/p)=1$ and assume that $f,g\ge0,\|g\|_q=1$. Let $E_g=\{x\in E:g(x)\gt0\}$. Note that on $E_g$, we also have $\|g\|_q=1$. $$ \begin{align} \int_Efg\,\mathrm{d}x &=\int_{E_g}fg\,\mathrm{d}x\tag{1a}\\[3pt] &=\int_{E_g}fg^{-q/p}g^q\,\mathrm{d}x\tag{1b}\\ &\le\left(\int_{E_g}f^pg^{-q}g^q\,\mathrm{d}x\right)^{1/p}\tag{1c}\\ &\le\left(\int_Ef^p\,\mathrm{d}x\right)^{1/p}\tag{1d}\\[9pt] &=\|f\|_p\|g\|_q\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: $g=0$ on $E\setminus E_g$
$\text{(1b)}$: $q(1-1/p)=1$
$\text{(1c)}$: apply Jensen on $E_g$ with the unit measure $g^q\mathrm{d}x$
$\text{(1d)}$: $E_g\subset E$
$\text{(1e)}$: $\|g\|_q=1$

Inequality $(1)$ scales linearly in $\|g\|_q$, so we can lift the requirement that $\|g\|_q=1$.