The Cauchy-Schwarz Inequality in Lebesgue Spaces 📂Lebesgue Spaces

The Cauchy-Schwarz Inequality in Lebesgue Spaces

Theorem¹

If $f,g \in L^{2} (E)$, then $fg \in L^{1}(E)$ and the following holds.

$$ \left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2} $$

Here, $\| \cdot \|_{2}$ is the norm of the $L^{2}$ space, and $\| \cdot \|_{1}$ is the norm of the $L^{1}$ space.

Explanation

If you are studying something like functional analysis, you should immediately sense why this inequality carries the name Cauchy-Schwarz. In fact, whenever an inner product is defined, the Cauchy-Schwarz inequality can be found everywhere. It can be generalized to the Hölder inequality.

Proof

$$ \int_{E} fg dm \le \int_{E} |fg| dm \le \int_{E} | f + g |^2 dm < \infty $$

therefore $fg \in L^{1}$. Meanwhile, from $\displaystyle (x - y)^2 \ge 0$, we obtain the following.

$$ xy \le \dfrac{1}{2} \left( x^2 + y^2 \right) $$

Case 1. $\left\| f \right\|_{2} = 0$ or $\left\| g \right\|_{2} = 0$
Since $f = 0$ almost everywhere or $g = 0$ almost everywhere, we have $f\overline{g} = 0$ almost everywhere. Therefore $\displaystyle \left| \int_{E} f \overline{g} dm \right| = \left\| fg \right\|_{1} = 0$ and $\left\| f \right\|_{2} \left\| g \right\|_{2} = 0$, so the inequality is satisfied.
Case 2. $\left\| f \right\|_{2} = \left\| g \right\|_{2} = 1$
$$ \left| \int_{E} f \overline{g} dm \right| \le \int_{E} \left| f \overline{g} \right| dm = \left\| fg \right\|_{1} \le {{1} \over {2}} (1 + 1) = 1 = \left\| f \right\|_{2} \left\| g \right\|_{2} $$
therefore the inequality is satisfied.
Case 3. Otherwise
Let us newly define the normalized functions $\displaystyle \hat{ f } : = {{f} \over {\left\| f \right\|_{2}}}$ and $\displaystyle \hat{ g } : = {{g} \over {\left\| g \right\|_{2}}}$. Then, by Case 2,
$$ \left| \int_{E} \hat{f} \overline{\hat{g} } dm \right| \le \left\| \hat{f} \hat{g} \right\|_{1} \le \left\| \hat{f} \right\|_{2} \left\| \hat{g} \right\|_{2} $$
Written out,
$$ \left| \int_{E} {{f} \over {\left\| f \right\|_{2}}} \overline{{{g} \over {\left\| g \right\|_{2}}} } dm \right| \le \left\| {{f} \over { \left\| f \right\|_{2}}} {{g} \over {\left\| g \right\|_{2}}} \right\|_{1} \le \left\| {{f} \over {\left\| f \right\|_{2}}} \right\|_{2} \left\| {{g} \over {\left\| g \right\|_{2}}} \right\|_{2} $$
Organizing the scalars $\left\| f \right\|_{2} , \left\| g \right\|_{2} \in (0, \infty)$, we obtain
$$ \left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2} $$
.

■