Cauchy-Schwarz Inequality in Lebesgue Spaces
📂Lebesgue Spaces Cauchy-Schwarz Inequality in Lebesgue Spaces Theorem If f , g ∈ L 2 ( E ) f,g \in L^{2} (E) f , g ∈ L 2 ( E ) then f g ∈ L 1 ( E ) fg \in L^{1}(E) f g ∈ L 1 ( E ) and the following holds:
∣ ∫ E f g ‾ d m ∣ ≤ ∥ f g ∥ 1 ≤ ∥ f ∥ 2 ∥ g ∥ 2
\left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2}
∫ E f g d m ≤ ∥ f g ∥ 1 ≤ ∥ f ∥ 2 ∥ g ∥ 2
Here, ∥ ⋅ ∥ 2 \| \cdot \|_{2} ∥ ⋅ ∥ 2 is the norm of the L 2 L^{2} L 2 space , ∥ ⋅ ∥ 1 \| \cdot \|_{1} ∥ ⋅ ∥ 1 is the norm of the L 1 L^{1} 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 ∫ E f g d m ≤ ∫ E ∣ f g ∣ d m ≤ ∫ E ∣ f + g ∣ 2 d m < ∞
\int_{E} fg dm \le \int_{E} |fg| dm \le \int_{E} | f + g |^2 dm < \infty
∫ E f g d m ≤ ∫ E ∣ f g ∣ d m ≤ ∫ E ∣ f + g ∣ 2 d m < ∞
Thus, it is f g ∈ L 1 fg \in L^{1} f g ∈ L 1 . Meanwhile, from ( x − y ) 2 ≥ 0 \displaystyle (x - y)^2 \ge 0 ( x − y ) 2 ≥ 0 , the following is obtained.
x y ≤ 1 2 ( x 2 + y 2 )
xy \le \dfrac{1}{2} \left( x^2 + y^2 \right)
x y ≤ 2 1 ( x 2 + y 2 )
Case 1. Either ∥ f ∥ 2 = 0 \left\| f \right\|_{2} = 0 ∥ f ∥ 2 = 0 or ∥ g ∥ 2 = 0 \left\| g \right\|_{2} = 0 ∥ g ∥ 2 = 0
Almost everywhere it’s either f = 0 f = 0 f = 0 or g = 0 g = 0 g = 0 , thus, almost everywhere it’s f g ‾ = 0 f\overline{g} = 0 f g = 0 . Therefore, it is ∣ ∫ E f g ‾ d m ∣ = ∥ f g ∥ 1 = 0 \displaystyle \left| \int_{E} f \overline{g} dm \right| = \left\| fg \right\|_{1} = 0 ∫ E f g d m = ∥ f g ∥ 1 = 0 and since ∥ f ∥ 2 ∥ g ∥ 2 = 0 \left\| f \right\|_{2} \left\| g \right\|_{2} = 0 ∥ f ∥ 2 ∥ g ∥ 2 = 0 , the inequality is satisfied.
Case 2. ∥ f ∥ 2 = ∥ g ∥ 2 = 1 \left\| f \right\|_{2} = \left\| g \right\|_{2} = 1 ∥ f ∥ 2 = ∥ g ∥ 2 = 1
∣ ∫ E f g ‾ d m ∣ ≤ ∫ E ∣ f g ‾ ∣ d m = ∥ f g ∥ 1 ≤ 1 2 ( 1 + 1 ) = 1 = ∥ f ∥ 2 ∥ g ∥ 2
\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}
∫ E f g d m ≤ ∫ E ∣ f g ∣ d m = ∥ f g ∥ 1 ≤ 2 1 ( 1 + 1 ) = 1 = ∥ f ∥ 2 ∥ g ∥ 2
Thus, the inequality is satisfied.
Case 3. All other cases
Define the normalized functions f ^ : = f ∥ f ∥ 2 \displaystyle \hat{ f } : = {{f} \over {\left\| f \right\|_{2}}} f ^ := ∥ f ∥ 2 f and g ^ : = g ∥ g ∥ 2 \displaystyle \hat{ g } : = {{g} \over {\left\| g \right\|_{2}}} g ^ := ∥ g ∥ 2 g anew. Then, by Case 2
∣ ∫ E f ^ g ^ ‾ d m ∣ ≤ ∥ f ^ g ^ ∥ 1 ≤ ∥ f ^ ∥ 2 ∥ g ^ ∥ 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}
∫ E f ^ g ^ d m ≤ f ^ g ^ 1 ≤ f ^ 2 ∥ g ^ ∥ 2
Explained,
∣ ∫ E f ∥ f ∥ 2 g ∥ g ∥ 2 ‾ d m ∣ ≤ ∥ f ∥ f ∥ 2 g ∥ g ∥ 2 ∥ 1 ≤ ∥ f ∥ f ∥ 2 ∥ 2 ∥ g ∥ g ∥ 2 ∥ 2
\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}
∫ E ∥ f ∥ 2 f ∥ g ∥ 2 g d m ≤ ∥ f ∥ 2 f ∥ g ∥ 2 g 1 ≤ ∥ f ∥ 2 f 2 ∥ g ∥ 2 g 2
Sorting the scalar ∥ f ∥ 2 , ∥ g ∥ 2 ∈ ( 0 , ∞ ) \left\| f \right\|_{2} , \left\| g \right\|_{2} \in (0, \infty) ∥ f ∥ 2 , ∥ g ∥ 2 ∈ ( 0 , ∞ ) ,
∣ ∫ E f g ‾ d m ∣ ≤ ∥ f g ∥ 1 ≤ ∥ f ∥ 2 ∥ g ∥ 2
\left| \int_{E} f \overline{g} dm \right| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2}
∫ E f g d m ≤ ∥ f g ∥ 1 ≤ ∥ f ∥ 2 ∥ g ∥ 2
Is obtained.
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See Also