Infi-2 14

Equivalent definitions of uniform convergence (oscillation) #theorem

\begin{matrix} f_{n} ⇉ f ⟺ d_{n} \to 0 \\ Proof: \\ ⟹ Let f_{n} ⇉ f \\ Let ε > 0 \\ \forall x \in A : \exists N_{ε} : \forall n > N_{ε} : | f_{n} (x) - f (x) | < \frac{ε}{2} \\ ⟹ \exists N_{ε} : \forall n > N_{ε} : sup_{x \in A} | f_{n} (x) - f (x) | \leq \frac{ε}{2} \\ ⟹ \exists N_{ε} : \forall n > N_{ε} : | d_{n} - 0 | \leq \frac{ε}{2} < ε \\ ⟹ d_{n} \to 0 \\ ⟸ Let d_{n} \to 0 \\ Let ε > 0 \\ \exists N_{ε} : \forall n > N_{ε} : | d_{n} - 0 | < ε \\ ⟹ \exists N_{ε} : \forall n > N_{ε} : sup_{x \in A} | f_{n} (x) - f (x) | < ε \\ ⟹ \forall x \in A : \exists N_{ε} : \forall n > N_{ε} : | f_{n} (x) - f (x) | < ε \\ ⟹ f_{n} ⇉ f \end{matrix}

\begin{matrix} Let f_{n} be integrable on [a, b] \\ Then f is integrable on [a, b] and \forall x \in [a, b] : \int_{a}^{x} f_{n} (t) d t ⇉ \int_{a}^{x} f (t) d t \\ Proof: \\ Let x \in [a, b] \\ \int_{a}^{x} f_{n} (t) d t \to \int_{a}^{x} f (t) d t ⟺ \int_{a}^{x} f_{n} (t) d t - \int_{a}^{x} f (t) d t \to 0 \\ ⟺ \int_{a}^{x} f_{n} (t) - f (t) d t \to 0 \\ ⟺ | \int_{a}^{x} f_{n} (t) - f (t) d t | \to 0 \\ \overset{| \int f | \leq \int | f |}{⟺} \int_{a}^{x} | f_{n} (t) - f (t) | d t \to 0 \\ \overset{| f_{n} (t) - f (t) | \leq d_{n}}{⟺} \int_{a}^{x} d_{n} d t \to 0 = \underset{\to 0}{\underset{⏟}{d_{n}}} \cdot (x - a) \to 0 \\ | \int_{a}^{x} f_{n} (t) - f (t) d t | \leq d_{n} (x - a) \\ ⟹ sup_{x \in [a, b]} | \int_{a}^{x} f_{n} (t) - f (t) d t | \leq d_{n} (x - a) \\ ⟹ sup_{x \in [a, b]} | \int_{a}^{x} f_{n} (t) - f (t) d t | \to 0 \\ ⟹ \int_{a}^{x} f_{n} (t) d t ⇉ \int_{a}^{x} f (t) d t \end{matrix}

\begin{matrix} \sum_{n = 1}^{\infty} f_{n} (x) \\ For example: geometric series \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}, x \in (- 1, 1) \end{matrix}

\begin{matrix} S_{N} (x) = \sum_{n = 1}^{N} f_{n} (x) \\ lim_{N \to \infty} S_{N} (x) = S (x) ⟺ \sum_{n = 1}^{\infty} f_{n} (x) = S (x) \end{matrix}

\begin{matrix} f_{n} is continuous/differentiable/integrable ⟹ S_{N} is too \\ And even more: \int S_{N} = \int \sum_{n = 0}^{N} f_{n} = \sum_{n = 0}^{N} \int f_{n} = {(\sum_{n = 0}^{N} f_{n})}^{'} = \sum_{n = 0}^{N} f_{n}^{'} \end{matrix}

\begin{matrix} 1. Let f_{n} (x) be continuous \\ Then S (x) is also continuous \\ 2. Let f_{n} (x) be integrable \\ Then S (x) is also integrable and \\ \int S_{N} (x) \to \int S (x) \\ \int \sum_{n = 0}^{N} f_{n} (x) \to \int \sum_{n = 0}^{\infty} f_{n} (x) \\ \int_{a}^{x} \sum_{n = 0}^{\infty} f_{n} (t) d t = \sum_{n = 0}^{\infty} \int_{a}^{x} f_{n} (t) d t \\ 3. Let S_{N}^{'} ⇉ g (x) \\ Let \exists x_{0} \in A : \sum_{n = 0}^{\infty} f_{n} (x) \to M \\ Then S_{N}^{'} (x) ⇉ S^{'} (x) \\ Or \sum_{n = 0}^{\infty} f_{n}^{'} (x) ⇉ {(\sum_{n = 0}^{\infty} f_{n} (x))}^{'} \end{matrix}