Cauchy-Schwarz Inequality in Lebesgue Spaces 📂Lebesgue Spaces

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

Explanation

If one has studied functional analysis, the reason this inequality is named after Cauchy-Schwarz should be immediately apparent. In fact, the Cauchy-Schwarz inequality can be found anywhere an inner product is defined. It can be generalized as the Hölder inequality.

Proof

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

Thus, it is $fg \in L^{1}$. Meanwhile, from $\displaystyle (x - y)^2 \ge 0$, the following is obtained.

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

Case 1. Either $\left\| f \right\|_{2} = 0$ or $\left\| g \right\|_{2} = 0$
Almost everywhere it’s either $f = 0$ or $g = 0$, thus, almost everywhere it’s $f\overline{g} = 0$. Therefore, it is $\displaystyle \left| \int_{E} f \overline{g} dm \right| = \left\| fg \right\|_{1} = 0$ and since $\left\| f \right\|_{2} \left\| g \right\|_{2} = 0$, the inequality is satisfied.
Case 2. $\left\| f \right\|_{2} = \left\| g \right\|_{2} = 1$
$$ \left| \int_{E} f \overline{g} dm \right| \le \int_{E} \left| f \overline{g} \right| dm = \left\| fg \right\|_{1} \le {{1} \over {2}} (1 + 1) = 1 = \left\| f \right\|_{2} \left\| g \right\|_{2} $$
Thus, the inequality is satisfied.
Case 3. All other cases
Define the normalized functions $\displaystyle \hat{ f } : = {{f} \over {\left\| f \right\|_{2}}}$ and $\displaystyle \hat{ g } : = {{g} \over {\left\| g \right\|_{2}}}$ anew. Then, by Case 2
$$ \left| \int_{E} \hat{f} \overline{\hat{g} } dm \right| \le \left\| \hat{f} \hat{g} \right\|_{1} \le \left\| \hat{f} \right\|_{2} \left\| \hat{g} \right\|_{2} $$
Explained,
$$ \left| \int_{E} {{f} \over {\left\| f \right\|_{2}}} \overline{{{g} \over {\left\| g \right\|_{2}}} } dm \right| \le \left\| {{f} \over { \left\| f \right\|_{2}}} {{g} \over {\left\| g \right\|_{2}}} \right\|_{1} \le \left\| {{f} \over {\left\| f \right\|_{2}}} \right\|_{2} \left\| {{g} \over {\left\| g \right\|_{2}}} \right\|_{2} $$
Sorting the scalar $\left\| f \right\|_{2} , \left\| g \right\|_{2} \in (0, \infty)$,
$$ \left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2} $$
Is obtained.

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