A precise definition for a sequence of functions that "Converge Pointwise, but not Uniformly"

Question

I am trying to lay down a precise deifinition of "Pointwise Convergent but NOT Uniformly Convergent" sequence of functions $(f_n)$ with domain $D \subset \mathbb{R}$ and codomain $\mathbb{R}$.

We denote $D \setminus S(\epsilon, k) = T $, and $f(x)$ be the limit function of the sequence. $S( \epsilon, k)$ denotes a set dependent on $\epsilon$ and $k$.

Here is the definition:

$\forall \epsilon>0$, $\ \exists$ a $k\in \mathbb{N}$ and a maximal set $S( \epsilon, k)$, such that $|f_n (x) -f(x)| < \epsilon$, $\ \forall n \geq k$, $\ \forall x \in S$, with $T \neq \emptyset$ for any finite $k $ but $T \to \emptyset$ as $k \to \infty$.

Is this definition rigorous? Or does it have any flaws which I am missing?

Answer 1

The definition of pointwise convergence is:

For every $x \in D$, and for every $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|f_n(x)-f(x)|<\epsilon$.

The definition of uniform convergence is:

For every $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for every $n \geq N$ and every $x \in D$, $|f_n(x)-f(x)|<\epsilon$.

Therefore, the definition of not converging uniformly is:

There exists $\epsilon>0$ such that for all $N \in \mathbb{N}$, there exist $n \geq N$ and $x \in D$ such that $|f_n(x)-f(x)| \geq \epsilon$.

So finally, the definition of pointwise convergence without uniform convergence is:

For every $x \in D$, and every $\epsilon >0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|f_n(x)-f(x)|<\epsilon$. However, there exists $\epsilon>0$ such that for all $N \in\mathbb{N}$, there exist $n \geq N$ and $x \in D$ such that $|f_n(x)-f(x)| \geq \epsilon$.

Now what you are trying to do is say for a given $\epsilon$, some $x$'s will have the same value of $N$ "work". So define $$S(N,\epsilon) = \{x \in D : \text{for all } n \geq N, |f_n(x)-f(x)|<\epsilon\}.$$ We know by the definition of pointwise convergence that for every $x \in D$ and every $\epsilon >0$, there exists $N$ such that $x \in S(N,\epsilon)$. Moreover, it's clear that $S(N_1,\epsilon) \subseteq S(N_2,\epsilon)$ if $N_1<N_2$. So we can say the $S(N,\epsilon)$'s are increasing to $D$, meaning $$S(1,\epsilon) \subseteq S(2,\epsilon) \subseteq S(3,\epsilon) \subseteq \cdots \qquad \text{ and } \qquad \bigcup_{N=1}^\infty S(N,\epsilon)=D.$$ This is equivalent to saying the sets $T(N,\epsilon):=D\setminus S(N,\epsilon)$ are decreasing to $\emptyset$, meaning $$T(1,\epsilon) \supseteq T(2,\epsilon) \supseteq T(3,\epsilon) \supseteq \cdots \qquad \text{ and } \qquad \bigcap_{N=1}^\infty T(N,\epsilon) = \emptyset.$$

Now the definition of pointwise convergence can be phrased as:

For every $\epsilon>0$, the sets $S(N,\epsilon)$ defined above are increasing to $D$ as $N \to \infty$.

However, this says nothing about not converging uniformly. To phrase this in terms of the $S(N,\epsilon)$'s we can write:

There exists $\epsilon>0$ such that for all $N \in \mathbb{N}$, there exists $x \notin S(N,\epsilon)$ (i.e., $S(N,\epsilon) \neq D$).

So finally, the definition of pointwise convergence without uniform convergence, in terms of the $S(N,\epsilon)$'s, is:

For every $\epsilon >0$, the sets $S(N,\epsilon)$ increase to $D$ as $N \to \infty$, but there exists $\epsilon >0$ such that none of the sets $S(N,\epsilon)$ are equal to $D$.

Note that we can't say "for all $\epsilon >0$, and all $N \in \mathbb{N}$, $S(N,\epsilon) \neq D$." That is too strong, since it is completely possible that for some $\epsilon$ (say $\epsilon = 1000000$), all of the $S(N,\epsilon)$ are equal to $D$. This would happen if, for instance, all of the functions have codomain $[0,1000000]$. All that is required for uniform convergence to fail is that there exists some epsilon so that the same $N$ will not work for every $x$.