Infi-2 7

\begin{matrix} a_{n} : N \to R \\ f_{n} : N \to R^{D} \end{matrix}

Pointwise convergence

\begin{matrix} f_{n} \to f \\ \forall x \in D : \forall ε > 0 : \exists N_{x, ε} : \forall n \geq N_{x, ε} : | f_{n} (x) - f (x) | < ε \end{matrix}

\begin{matrix} (0, \infty) \\ f_{n} (x) = n (\sqrt[n]{x} - 1) \\ lim_{n \to \infty} f_{n} (x) = lim_{n \to \infty} \frac{\sqrt[n]{x} - 1}{\frac{1}{n}} \overset{t = \frac{1}{n}}{=} lim_{t \to 0} \frac{x^{t} - 1}{t} = \ln (x) \end{matrix}

\begin{matrix} f_{n} (x) = \frac{n^{2}}{1 + n^{2} x^{2}} \\ lim_{n \to \infty} f_{n} (x) = lim_{n \to \infty} \frac{1}{\frac{1}{n^{2}} + x^{2}} - = \frac{1}{x^{2}} \end{matrix}

\begin{matrix} Let {f_{n} (x)} \to f (x) on [a, b] \\ Let \forall n, x \in [a, b] : f_{n} (x) is bounded \\ Then not necessarily f is bounded \end{matrix}

\begin{matrix} f_{n} ⇉ f \\ \forall ε > 0 : \exists N_{ε} : \forall n \geq N_{ε} : \forall x \in D : | f_{n} (x) - f (x) | < ε \\ Let d_{n} = sup_{x \in D} | f_{n} (x) - f (x) | \\ f_{n} ⇉ f ⟺ d_{n} \to 0 \end{matrix}

\begin{matrix} f_{n} ⇉ f and f_{n} is continuous ⟹ f is continuous \end{matrix}

\begin{matrix} Let f_{n} ⇉ f and f_{n} (x) be bounded \\ ⟹ f is bounded \\ And exists M that bounds f_{n} for every n \\ \forall n \geq N : | f_{n} | \leq | f_{n} - f | + | f - f_{N} | + | f_{N} | < 2 + M_{N} \\ ⟹ \forall n \in N : | f_{n} | \leq max {M_{1}, M_{2}, \dots, M_{N - 1}, 2 + M_{N}} \end{matrix}

\begin{matrix} [0, 1] \\ f_{n} (x) = \frac{x}{1 + n^{2} x^{2}} \\ f_{n} \to 0 \\ d_{n} = sup_{x \in [0, 1]} \frac{x}{1 + n^{2} x^{2}} = max_{x \in [0, 1]} \frac{x}{1 + n^{2} x^{2}} \\ x = 0 ⟹ d_{n} = 0 \\ x = 1 ⟹ d_{n} = \frac{1}{1 + n^{2}} \\ 0 = {(\frac{x}{1 + n^{2} x^{2}})}^{'} = \frac{(1 + n^{2} x^{2}) - 2 n^{2} x^{2}}{(1 + n^{2} x^{2})^{2}} = \frac{1 - n^{2} x^{2}}{(1 + n^{2} x^{2})^{2}} \\ ⟹ x^{2} = \frac{1}{n^{2}} ⟹ x = \frac{1}{n} \\ f_{n} (\frac{1}{n}) = \frac{1}{2 n} \\ ⟹ d_{n} = \frac{1}{2 n} \to 0 \\ ⟹ f_{n} ⇉ 0 \end{matrix}