logo

Cauchy-Schwarz Inequality in Lebesgue Spaces 📂Lebesgue Spaces

Cauchy-Schwarz Inequality in Lebesgue Spaces

Theorem1

If f,gL2(E)f,g \in L^{2} (E) then fgL1(E)fg \in L^{1}(E) and the following holds:

Efgdmfg1f2g2 \left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2}

Here, 2\| \cdot \|_{2} is the norm of the L2L^{2} space, 1\| \cdot \|_{1} is the norm of the L1L^{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

EfgdmEfgdmEf+g2dm< \int_{E} fg dm \le \int_{E} |fg| dm \le \int_{E} | f + g |^2 dm < \infty

Thus, it is fgL1fg \in L^{1}. Meanwhile, from (xy)20\displaystyle (x - y)^2 \ge 0, the following is obtained.

xy12(x2+y2) xy \le \dfrac{1}{2} \left( x^2 + y^2 \right)

  • Case 1. Either f2=0\left\| f \right\|_{2} = 0 or g2=0\left\| g \right\|_{2} = 0

    Almost everywhere it’s either f=0f = 0 or g=0g = 0, thus, almost everywhere it’s fg=0f\overline{g} = 0. Therefore, it is Efgdm=fg1=0\displaystyle \left| \int_{E} f \overline{g} dm \right| = \left\| fg \right\|_{1} = 0 and since f2g2=0\left\| f \right\|_{2} \left\| g \right\|_{2} = 0, the inequality is satisfied.

  • Case 2. f2=g2=1\left\| f \right\|_{2} = \left\| g \right\|_{2} = 1

    EfgdmEfgdm=fg112(1+1)=1=f2g2 \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 f^:=ff2\displaystyle \hat{ f } : = {{f} \over {\left\| f \right\|_{2}}} and g^:=gg2\displaystyle \hat{ g } : = {{g} \over {\left\| g \right\|_{2}}} anew. Then, by Case 2

    Ef^g^dmf^g^1f^2g^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,

    Eff2gg2dmff2gg21ff22gg22 \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 f2,g2(0,)\left\| f \right\|_{2} , \left\| g \right\|_{2} \in (0, \infty),

    Efgdmfg1f2g2 \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.

See Also


  1. Capinski. (1999). Measure, Integral and Probability: p132. ↩︎