KindiakMath

The Limit of Functions

Let’s talk about summing up functions, finitely and infinitely. Throughout this post, let K be a set and \mathbb K be an ordered field.

Definition 1. For any i \in \mathbb N, let f_i : K \to \mathbb K be a \mathbb K-valued function. For any n, define the sum function \displaystyle F_n := \sum_{i=1}^n f_i : K \to \mathbb K by

\displaystyle F_n(x) := \sum_{i=1}^n f_i(x),\quad x \in K.

Finite sums aren’t the problem—infinite sums are. Given the sequence \{f_n\} of functions, we can create the sequence of sums \{F_n\}. When does the function series \{F_n\} converge? Finally, what does the series converge to?

When we studied series of real numbers, we first analysed sequences of real numbers. The same idea holds for series of functions—let’s first analyse sequences of functions. There turns out to be some subtleties worth expanding on.

Definition 2. Let \{f_n\} be a sequence of functions, i.e. for each n, f_n : K \to \mathbb K. Suppose for each x \in K, the sequence \{f_n(x) \} converges. Define the pointwise limit \displaystyle f := \lim_{n \to \infty} f_n : K \to \mathbb K by

\displaystyle f(x) := \lim_{n \to \infty} f_n(x).

The definition seems rather straightforward. For each x \in K, take the limit of the sequence \{f_n(x) \}. Unfortunately, many desirable properties like continuity and differentiability fail spectacularly. We will illustrate continuity for example.

Example 1. For each n \in \mathbb N, define f_n : [0, 1] \to \mathbb R by f_n(x) = x^n. Each f_n is continuous. However, \displaystyle f := \lim_{n \to \infty} f_n is not continuous. Why? Observe that for x \in [0, 1),

\displaystyle f(x) = \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} x^n = 0,

while f(1) = 1. Hence,

\displaystyle f(x) := \begin{cases} 0, & \text{if}\ x \in [0, 1), \\ 1, & \text{if}\ x = 1. \end{cases}

Thus, f is clearly discontinuous at 1, since

\displaystyle \lim_{x \to 1^-} f(x) = 0 \neq 1 = f(1).

Examples hold for differentiability as well.

The problem here is that our notion of convergence is point-wise. That is, we write f_n \to f to mean that for any x \in K, f_n(x) \to f(x). This dependency on x is a local property—it has impacts near x, but not so much beyond.

However, we need to control global properties, that is, we need a handle of the whole function, not just near some point. To achieve this goal, we will adopt a stronger notion of convergence, known as uniform convergence. This subtlety is analogous to the difference between (point-wise) continuity and uniform continuity.

Definition 3. Let \{f_n\} be a sequence of functions f_n : K \to \mathbb K and f : K \to \mathbb K be a function.

We say that f_n \to f point-wise if

\forall \epsilon > 0\quad \forall x \in K\quad \exists N \in \mathbb N: \quad (n > N \quad \Rightarrow \quad |f_n(x) - f(x)| < \epsilon).

We say that f_n \to f uniformly if

\forall \epsilon > 0\quad \exists N \in \mathbb N: \quad (n > N \quad \Rightarrow \quad (\forall x \in K \quad |f_n(x) - f(x)| < \epsilon)).

For point-wise convergence, given any fixed x \in K, we require an N \in \mathbb N that yields |f_n(x) - f(x)| < \epsilon. However, in the uniform convergence case, our choice of N should yield the same estimate |f_n(x) - f(x)| < \epsilon regardless of x \in K.

Let’s first check that uniform convergence is a stronger property than point-wise convergence in the following sense:

Theorem 1. Let \{f_n\} be a sequence of functions f_n : K \to \mathbb K and f : K \to \mathbb K be a function. If f_n \to f uniformly, then f_n \to f point-wise.

Proof. Fix \epsilon > 0 and x \in K. Since f_n \to f uniformly, there exists N \in \mathbb N such that for any n > N and for any u \in K, |f_n(u) - f(u)| < \epsilon. Particularising to u = x yields the estimate |f_n(x) - f(x)| < \epsilon. Thus, f_n \to f point-wise.

In fact, uniform convergence is so strong it carries on into continuity in the following sense.

Theorem 2. Suppose K \subseteq \mathbb R. Let \{f_n\} be a sequence of continuous functions f_n : K \to \mathbb R and f : K \to \mathbb R be a function. Suppose f_n \to f uniformly on K. Then f is continuous on K.

Proof. Fix c \in K. We need to prove that f is continuous at c. To that end, fix \epsilon > 0. Since f_n \to f uniformly, for any k_1 > 0, there exists N \in \mathbb N such that for any u \in K,

n > N \quad \Rightarrow \quad |f_n(u) - f(u)| < k_1 \cdot \epsilon.

Since f_{N+1} is continuous on K, and thus continuous at x, for any k_2 > 0, there exists \delta > 0 such that

|x - c| < \delta \quad \Rightarrow \quad |f_{N+1}(x) - f_{N+1}(c)| < k_2 \cdot \epsilon.

Fix x \in K such that |x-c| < \delta. Particularising the uniform convergence condition for x, c,

|f_{N+1}(x) - f(x)| < k_1 \cdot \epsilon,\quad |f_{N+1}(c) - f(c)| < k_1 \cdot \epsilon.

Applying the triangle inequality,

\begin{aligned}|f(x) - f(c)| &\leq |f(x) - f_{N+1}(x)| + |f_{N+1}(x) - f_{N+1}(c)| + |f_{N+1}(c) - f(c)| \\ & < k_1 \cdot \epsilon + k_2 \cdot \epsilon + k_1 \cdot \epsilon \\ &= (2 \cdot k_1 + k_2) \cdot \epsilon. \end{aligned}

To conclude the proof, set k_1=1/4, k_2 = 1/2. In fact, any choice of k_1 > 0,k_2 > 0 that yields 2\cdot k_1 + k_2 \leq 1 works. The common trick is called the \epsilon/3-trick, where we set k_1 = k_2 = 1/3.

Using this analysis, we observe how the uniform convergence condition is used twice in order to derive continuity.

So we see the power of uniform convergence in helping us “transfer” properties of functions in a sequence to the point-wise limit of those functions. But when do we guarantee uniform convergence?

—Joel Kindiak, 15 Jan, 2219H

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