Uniform convergence - definition / notation clarification

Question

My professor gave the following definition in class for uniform convergence:

$(f_{n}: A \subset \mathbb{R}^{k} \rightarrow \mathbb{R}^{l})_{n=1}^{\infty}$ converges uniformly to $f$ on $A$ if and only if $\forall \epsilon >0 \exists N \in \mathbb{N}\backslash \{0 \}$ such that $n \geq N \Rightarrow \lvert f_{n}(x) - f(x) \rvert < \epsilon \forall x \in A$.

I don't understand what the $f(x)$ is in this definition. That is, say I have $f_{n}(x) = \frac{sin(x)}{n}$. My understanding is I need to show $\lvert \frac{sin(x)}{n} - f(x) \rvert < \epsilon$. What is the $n$ value for $f(x)$? Am I using $f_{1}(x)= sin(x)$?

Answer 1

The important part of the definition is that $N$ is independent of $x$. This is what makes the convergence uniform.

The limit function is $f(x)=0$, as mentioned above. So for any $\epsilon>0$, $\vert f_n(x) -f(x)\vert = \vert \sin x / n \vert \le 1/n < \epsilon \Rightarrow N\ge 1/\epsilon$. And since $N$ is independent of $x$, $f_n$ converges uniformly.

Answer 2

Note that $|\sin x|\leq 1$ for all $x\in \mathbb{R}$. So, for each $x\in\mathbb R,$ we have $(\sin x)/n\to 0$ as $n\to\infty$. That is $f_n$, tends pointwise to the constant zero function as $n\to\infty$. This gives a good candidate for $f$. Now show that $f_n\to f$ uniformly as $n\to\infty$. Given $\varepsilon>0$, find $N\in\mathbb N$ large enough so that $|f_n(x)|<\varepsilon$ for all $n\geq N$ and all $x\in X$.