📂Analysis

The Difference between Pointwise Convergence and Uniform Convergence of Functions

Let’s define the function $f : E \to \mathbb{R}$ and the sequence of functions $\left\{ f_{n} : E \to \mathbb{R} \right\}_{n=1}^{\infty}$ for the subset $E \ne \emptyset$ of $\mathbb{R}$.

Pointwise Convergence

It is said that $f_{n}$ converges pointwise to $f$ in $E$ if, for every $\varepsilon > 0$ and $x \in E$, there exists $N \in \mathbb{N}$ satisfying $n \ge N \implies | f_{n} (x) - f(x) | < \varepsilon$, and it is denoted as follows:

$$f_n \rightarrow f$$

Uniform Convergence

It is said that $f_{n}$ converges uniformly to $f$ in $E$ if, for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ satisfying $n \ge N \implies | f_{n} (x) - f(x) | < \varepsilon$, and it is denoted as follows:

$$f_n \rightrightarrows f$$

Explanation

The difference between Pointwise Convergence and Uniform Convergence can be seen as the difference between the convergence of function values and the convergence of the function itself. When comparing both types of convergence, the only difference is whether there is mention of $x \in E$. From the perspective of a mathematician, it could be described with the following nuance.

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Pointwise Convergence: Looking at each $x \in E$, for every $\varepsilon > 0$ given, each respective $N_{x} \in \mathbb{N}$ that satisfies $n \ge N \implies | f_{n} (x) - f(x) | < \varepsilon$ indeed exists.

Uniform Convergence: Not only every $\varepsilon > 0$ but also regardless of what $x \in E$ is, there exists $N \in \mathbb{N}$ that solely satisfies $n \ge N \implies | f_{n} (x) - f(x) | < \varepsilon$.

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To speak more extremely with a focus on functions, Pointwise Convergence could be considered “fake convergence”, whereas Uniform Convergence might be seen as “real convergence”. The distinction between Pointwise Convergence and Uniform Convergence comes from a kind of “speed of convergence”. Though both eventually converge, in the case of Pointwise Convergence, because each point is viewed separately, it’s permissible for each $x$ to have its own $N_{x}$. Conversely, if it Uniformly Converges, without discriminating any $x_{1}$ or ignoring any $x_{2}$, the entire $E$ forces a simultaneous, uniform condition $| f_{N} (x) - f(x) | < \varepsilon$ to be met by the existence of - a single $N$.

Example

Consider the following $f , f_{n}$ defined in $E := [0,1)$.

$$f(x) := 0$$

$$f_{n} (x) := x^{n}$$

For $0 \le x<1$, regardless of what $\varepsilon$ is, there exists $N \in \mathbb{N}$ that satisfies $n \ge N \implies x^{n} < \varepsilon$. Suppose $\varepsilon = 0.5$ is given and let’s only check three points, $x_{7} = 0.7$, $x_{8} = 0.8$, and $x_{9} = 0.9$.

$$N_{7} = 2 \implies 0.7^{2} = 0.49 < 0.5$$

$$N_{8} = 4 \implies 0.8^{4} = 0.4096 < 0.5$$

$$N_{9} = 7 \implies 0.9^{7} = 0.4782969 < 0.5$$

As such, as there exists respective $N_{r}$ for each $x_{r}$, it’s correct to say that $f_{n}$ converges pointwise to $f$ in Pointwise Convergence. However, in this way, as $x_{r}$ gets closer to $1$, $N_{r}$ must also increase significantly. No matter how large $N \in \mathbb{N}$ is suggested, there must exist $(0.9…9 )$ that satisfies $(0.9…9)^{N} > 0.5$. Therefore, it’s not Uniform Convergence.

Note that while $N_{8}=4$ might cover up to $N_{7}=2$, it couldn’t cover up to $N_{9}=7$. If there’s a race where each $x$ converges to $0$, at $N=6$, since $x_{9} = 0.9$ is $0.9^{6}=0.531441$, the convergence speed is slow and falls behind the criteria of being smaller than $\varepsilon = 0.5$ - this is considered “dropped” in the context. Pointwise Convergence allows this, but Uniform Convergence does not. In discussing the convergence of functions, the leisurely premise of ’eventually converging at some point’ is insufficient; genuine convergence calls for unconditional, collective convergence.