Given f : [ 0 , 1 ] → R \func{f}{\br{0,1}}{\R} f : [ 0 , 1 ] → R L 1 ( d x ) L^1\p{\diff{x}} L 1 ( d x ) n ∈ { 1 , 2 , 3 , … } n \in \set{1, 2, 3, \ldots} n ∈ { 1 , 2 , 3 , … }
f n ( x ) = n ∫ k / n ( k + 1 ) / n f ( y ) d y for x ∈ [ k n , k + 1 n ) and k = 0 , … , n − 1. f_n\p{x} = n \int_{k/n}^{\p{k+1}/n} f\p{y} \,\diff{y}
\quad\text{for } x \in \pco{\frac{k}{n}, \frac{k+1}{n}}
\quad\text{and}\quad k = 0, \ldots, n-1. f n ( x ) = n ∫ k / n ( k + 1 ) / n f ( y ) d y for x ∈ [ n k , n k + 1 ) and k = 0 , … , n − 1. Prove f n → f f_n \to f f n → f L 1 L^1 L 1
Solution.
Notice that
n ∫ k / n ( k + 1 ) / n d y = n ⋅ 1 n = 1 n \int_{k/n}^{\p{k+1}/n} \,\diff{y}
= n \cdot \frac{1}{n}
= 1 n ∫ k / n ( k + 1 ) / n d y = n ⋅ n 1 = 1 for any k = 0 , … , n − 1 k = 0, \ldots, n-1 k = 0 , … , n − 1
Let ε > 0 \epsilon > 0 ε > 0 C ( [ 0 , 1 ] ) C\p{\br{0, 1}} C ( [ 0 , 1 ] ) L 1 ( [ 0 , 1 ] ) L^1\p{\br{0, 1}} L 1 ( [ 0 , 1 ] ) g g g ∥ f − g ∥ L 1 < ε \norm{f - g}_{L^1} < \epsilon ∥ f − g ∥ L 1 < ε [ 0 , 1 ] \br{0, 1} [ 0 , 1 ] g g g δ > 0 \delta > 0 δ > 0 ∣ x − y ∣ < δ ⟹ ∣ g ( x ) − g ( y ) ∣ < ε \abs{x - y} < \delta \implies \abs{g\p{x} - g\p{y}} < \epsilon ∣ x − y ∣ < δ ⟹ ∣ g ( x ) − g ( y ) ∣ < ε
∥ f n − f ∥ L 1 = ∫ ∣ f n ( x ) − f ( x ) ∣ d x = ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∣ ∫ k / n ( k + 1 ) / n f ( y ) − f ( x ) d y ∣ d x ≤ ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − f ( x ) ∣ d y d x ≤ ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ + ∣ g ( y ) − g ( x ) ∣ + ∣ g ( x ) − f ( x ) ∣ d y d x . \begin{aligned}
\norm{f_n - f}_{L^1}
&= \int \abs{f_n\p{x} - f\p{x}} \,\diff{x} \\
&= \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\abs{\int_{k/n}^{\p{k+1}/n} f\p{y} - f\p{x} \,\diff{y}} \,\diff{x} \\
&\leq \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{f\p{y} - f\p{x}} \,\diff{y} \,\diff{x} \\
&\leq \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{f\p{y} - g\p{y}} + \abs{g\p{y} - g\p{x}} + \abs{g\p{x} - f\p{x}} \,\diff{y} \,\diff{x}.
\end{aligned} ∥ f n − f ∥ L 1 = ∫ ∣ f n ( x ) − f ( x ) ∣ d x = k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∣ ∣ ∫ k / n ( k + 1 ) / n f ( y ) − f ( x ) d y ∣ ∣ d x ≤ k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − f ( x ) ∣ d y d x ≤ k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ + ∣ g ( y ) − g ( x ) ∣ + ∣ g ( x ) − f ( x ) ∣ d y d x . We will look at each term separately: In the following, everything is non-negative, so by Fubini-Tonelli,
∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d y d x = ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d x d y = ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d y = ∥ f − g ∥ L 1 ≤ ε . \begin{aligned}
\sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{f\p{y} - g\p{y}} \,\diff{y} \,\diff{x}
&= \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{f\p{y} - g\p{y}} \,\diff{x} \,\diff{y} \\
&= \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} \abs{f\p{y} - g\p{y}} \,\diff{y} \\
&= \norm{f - g}_{L^1} \\
&\leq \epsilon.
\end{aligned} k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d y d x = k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d x d y = k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n ∣ f ( y ) − g ( y ) ∣ d y = ∥ f − g ∥ L 1 ≤ ε . Similarly,
∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( x ) − g ( x ) ∣ d y d x = ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n ∣ f ( x ) − g ( x ) ∣ d x = ∥ f − g ∥ L 1 ≤ ε \begin{aligned}
\sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{f\p{x} - g\p{x}} \,\diff{y} \,\diff{x}
&= \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} \abs{f\p{x} - g\p{x}} \,\diff{x} \\
&= \norm{f - g}_{L^1} \\
&\leq \epsilon
\end{aligned} k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ f ( x ) − g ( x ) ∣ d y d x = k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n ∣ f ( x ) − g ( x ) ∣ d x = ∥ f − g ∥ L 1 ≤ ε Finally, if n n n
∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ g ( y ) − g ( x ) ∣ d y d x ≤ ∑ k = 0 n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ε d y d x = ε . \begin{aligned}
\sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \abs{g\p{y} - g\p{x}} \,\diff{y} \,\diff{x}
&\leq \sum_{k=0}^{n-1} \int_{k/n}^{\p{k+1}/n} n\int_{k/n}^{\p{k+1}/n} \epsilon \,\diff{y} \,\diff{x} \\
&= \epsilon.
\end{aligned} k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ∣ g ( y ) − g ( x ) ∣ d y d x ≤ k = 0 ∑ n − 1 ∫ k / n ( k + 1 ) / n n ∫ k / n ( k + 1 ) / n ε d y d x = ε . Overall, we get ∥ f n − f ∥ L 1 ≤ 3 ε \norm{f_n - f}_{L^1} \leq 3\epsilon ∥ f n − f ∥ L 1 ≤ 3 ε ε → 0 \epsilon \to 0 ε → 0