00:00 Intro

00:14 Uniform convergence for sequence of functions

01:09 Theorem for uniform limit of continuous functions

02:18 Proof of the Theorem

07:41 Credits

Q1: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: I \rightarrow \mathbb{R}$. What is the correct definition for the uniform convergence to a function $f: I \rightarrow \mathbb{R}$?

A1: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \leq N ~ : ~ |f_n(x)-f(x)|<\varepsilon $

A2: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~~ \forall x \in I ~ : ~ |f_n(x)-f(x)|<\varepsilon $

A3: $ \displaystyle \forall x \in I ~~ \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \geq N ~ : ~ |f_n(x)-f(x)|<\varepsilon $

A4: $ \displaystyle \forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall x \in I ~~ \forall n \leq N ~ : ~ |f_n(x)-f(x)|<\varepsilon $

Q2: Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be continuous for all $n \in \mathbb{N}$. Which of the following cases can not occur?

A1: $(f_1, f_2, f_3, \ldots)$ is pointwisely convergent to $f$ and $f$ is not continuous.

A2: $(f_1, f_2, f_3, \ldots)$ is pointwisely convergent to $f$ and $f$ is continuous.

A3: $(f_1, f_2, f_3, \ldots)$ is uniformly convergent to $f$ and $f$ is continuous.

A4: $(f_1, f_2, f_3, \ldots)$ is uniformly convergent to $f$ and $f$ is not continuous.

Q3: Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be continuous for all $n \in \mathbb{N}$ and $(f_n)_{n \in \mathbb{N} }$ be pointwisely convergent to $f$ given by: $$ \displaystyle f(x) = \begin{cases} 1 &, ~~ x = 0\ 0 &, ~~x \neq 0\end{cases} $$ Which statement is true?

A1: $| f - f_n |_\infty \xrightarrow{n \rightarrow \infty} 0 $

A2: The $(f_n)_{n \in \mathbb{N} }$ is not uniformly convergent to $f$.

A3: The $(f_n)_{n \in \mathbb{N} }$ is uniformly convergent to $f$.