Notice that by (∗ * ∗ λ > 0 \lambda > 0 λ > 0
m ( { ∣ f ∣ > λ } ) ≤ 1 λ ∫ { ∣ f ∣ > λ } ∣ f ( x ) ∣ d x ≤ m ( { ∣ f ∣ > λ } ) λ ⟹ m ( { ∣ f ∣ > λ } ) ≤ 1 λ 2 . \begin{gathered}
m\p{\set{\abs{f} > \lambda}}
\leq \frac{1}{\lambda} \int_{\set{\abs{f} > \lambda}} \abs{f\p{x}} \,\diff{x}
\leq \frac{\sqrt{m\p{\set{\abs{f} > \lambda}}}}{\lambda} \\
\implies
m\p{\set{\abs{f} > \lambda}} \leq \frac{1}{\lambda^2}.
\end{gathered} m ( { ∣ f ∣ > λ } ) ≤ λ 1 ∫ { ∣ f ∣ > λ } ∣ f ( x ) ∣ d x ≤ λ m ( { ∣ f ∣ > λ } ) ⟹ m ( { ∣ f ∣ > λ } ) ≤ λ 2 1 . Hence, because m ( [ 0 , 1 ] ) = 1 m\p{\br{0,1}} = 1 m ( [ 0 , 1 ] ) = 1
∥ f ∥ L p p = ∫ 0 ∞ p λ p m ( { ∣ f ∣ > λ } ) d λ = ∫ 0 1 p λ p m ( { ∣ f ∣ > λ } ) d λ + ∫ 1 ∞ p λ p m ( { ∣ f ∣ > λ } ) d λ ≤ ∫ 0 ∞ p λ p − 2 d λ + ∫ 1 ∞ p λ p d λ < ∞ , \begin{aligned}
\norm{f}_{L^p}^p
&= \int_0^\infty p\lambda^p m\p{\set{\abs{f} > \lambda}} \,\diff\lambda \\
&= \int_0^1 p\lambda^p m\p{\set{\abs{f} > \lambda}} \,\diff\lambda + \int_1^\infty p\lambda^p m\p{\set{\abs{f} > \lambda}} \,\diff\lambda \\
&\leq \int_0^\infty p\lambda^{p-2} \,\diff\lambda + \int_1^\infty p\lambda^p \,\diff\lambda \\
&< \infty,
\end{aligned} ∥ f ∥ L p p = ∫ 0 ∞ p λ p m ( { ∣ f ∣ > λ } ) d λ = ∫ 0 1 p λ p m ( { ∣ f ∣ > λ } ) d λ + ∫ 1 ∞ p λ p m ( { ∣ f ∣ > λ } ) d λ ≤ ∫ 0 ∞ p λ p − 2 d λ + ∫ 1 ∞ p λ p d λ < ∞ , since p − 2 > − 1 p - 2 > -1 p − 2 > − 1 p > 1 p > 1 p > 1
First, for 0 < a < b ≤ 1 0 < a < b \leq 1 0 < a < b ≤ 1 0 ≤ h ≤ a 0 \leq h \leq a 0 ≤ h ≤ a
b − a ≤ b − h − a − h . \sqrt{b} - \sqrt{a}
\leq \sqrt{b - h} - \sqrt{a - h}. b − a ≤ b − h − a − h . Indeed, let φ ( h ) = b − h − a − h \phi\p{h} = \sqrt{b - h} - \sqrt{a - h} φ ( h ) = b − h − a − h
φ ′ ( h ) = 1 2 a − h − 1 2 b − h ≥ 0 , \phi'\p{h}
= \frac{1}{2\sqrt{a - h}} -\frac{1}{2\sqrt{b - h}}
\geq 0, φ ′ ( h ) = 2 a − h 1 − 2 b − h 1 ≥ 0 , so φ \phi φ h = 0 h = 0 h = 0 0 ≤ a 1 < b 1 ≤ a 2 < b 2 ≤ ⋯ ≤ a n < b n ≤ 1 0 \leq a_1 < b_1 \leq a_2 < b_2 \leq \cdots \leq a_n < b_n \leq 1 0 ≤ a 1 < b 1 ≤ a 2 < b 2 ≤ ⋯ ≤ a n < b n ≤ 1 c 0 = 0 c_0 = 0 c 0 = 0 c k = c k − 1 + ( b k − a k ) c_k = c_{k-1} + \p{b_k - a_k} c k = c k − 1 + ( b k − a k ) k ≥ 1 k \geq 1 k ≥ 1 h = a k − c k − 1 h = a_k - c_{k-1} h = a k − c k − 1
∫ ⋃ k = 1 n ( a k , b k ) 1 2 x d x = ∑ k = 1 n ( b k − a k ) ≤ ∑ k = 1 n ( b k − h − a k − h ) ≤ ∑ k = 1 n ( c k − c k − 1 ) = c n − c 0 = ∑ k = 1 n ( b k − a k ) . \begin{aligned}
\int_{\bigcup_{k=1}^n \p{a_k,b_k}} \frac{1}{2\sqrt{x}} \,\diff{x}
&= \sum_{k=1}^n \p{\sqrt{b_k} - \sqrt{a_k}} \\
&\leq \sum_{k=1}^n \p{\sqrt{b_k - h} - \sqrt{a_k - h}} \\
&\leq \sum_{k=1}^n \p{\sqrt{c_k} - \sqrt{c_{k-1}}} \\
&= \sqrt{c_n} - \sqrt{c_0} \\
&= \sqrt{\sum_{k=1}^n \p{b_k - a_k}}.
\end{aligned} ∫ ⋃ k = 1 n ( a k , b k ) 2 x 1 d x = k = 1 ∑ n ( b k − a k ) ≤ k = 1 ∑ n ( b k − h − a k − h ) ≤ k = 1 ∑ n ( c k − c k − 1 ) = c n − c 0 = k = 1 ∑ n ( b k − a k ) . By regularity of the Lebesgue measure, continuity of t \sqrt{t} t E ⊆ [ 0 , 1 ] E \subseteq \br{0, 1} E ⊆ [ 0 , 1 ]
∫ E 1 2 x d x ≤ m ( E ) , \int_E \frac{1}{2\sqrt{x}} \,\diff{x}
\leq \sqrt{m\p{E}}, ∫ E 2 x 1 d x ≤ m ( E ) , so 1 2 x \frac{1}{2\sqrt{x}} 2 x 1 ∗ * ∗ L 2 ( [ 0 , 1 ] ) L^2\p{\br{0,1}} L 2 ( [ 0 , 1 ] )