Uniform convergence of sequence of functions

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Consider a sequence of functions \(\{f_n(x)\}\) on interval \(I\):

\[f_1(x), f_2(x), f_3(x), \cdots.\]

This is an indexed collection of functions on \(I\). We can consider different types of convergence of such a sequence of functions: \(f_n(x) \to f(x)\). If the convergence is ``uniform,'' then some properties of the functions \(f_n(x)\), such as continuity, integrability, and differentiability, are inherited to the limiting function \(f(x)\).




Given a sequence of functions \(\{f_n(x)\}\) on \(I\), we can define a sequence of real numbers \(\{f_n(a)\}\) for each \(a \in I\). If the sequence \(\{f_n(a)\}\) converges for each \(a\in I\), we can define its limit as \(f(a)\) thereby defining a function \(f(x)\) on \(I\). In this case, we say the sequence of functions \(\{f_n(x)\}\) converges to the function \(f(x)\) and write

\[\lim_{n\to\infty}f_n(x) = f(x).\]

We refer to this type of convergence as point-wise convergence. 

Remark. In a logical form, the point-wise convergence is denoted as

\[\forall a \in I ~ \forall \varepsilon > 0 ~ \exists N \in \mathbb{N} ~ \forall n \in \mathbb{N} ~ (n \geq N \implies |f(a) - f_n(a)| < \varepsilon).\]

Note in particular that the value of \(N\) may depend on \(a\) in this expression. □

On the contrary, we say the sequence of functions \(\{f_n(x)\}\) uniformly converges to the function \(f(x)\) if the following holds:

\[\forall \varepsilon > 0 ~ \exists N \in \mathbb{N} ~ \forall a \in I ~ \forall n \in \mathbb{N} ~ (n \geq N \implies |f(a) - f_n(a)| < \varepsilon).\]

Note the difference of the position of the quantifier \(\forall a \in I\). In this expression, the value of \(N\) does not depend on \(a\) (but does depend on \(\varepsilon\))  That's why this is called "uniform" convergence. This corresponds to the condition (*) in the proof of the Theorem (Continuous power series).

See also: Calculus of power series

Thus, we have the following theorem.

Theorem (Uniform convergence and continuity)

Suppose that the sequence \(\{f_n(x)\}\) of functions on interval \(I\) uniformly converges to the function \(f(x)\) on \(I\). If each \(f_n(x)\) is continuous on \(I\), then so is \(f(x)\).

Proof. Exercise. (See Step 2 of the proof of the Theorem (Continuous power series) in Calculus of power series.) ■

Example. For each \(n\in \mathbb{N}\), let us define the function \(f_n(x)\) on \(\mathbb{R}\) as follows:

\[f_n(x) = \left\{ \begin{array}{cc} 0 & (x \leq 0),\\ nx & \left(0 < x < \frac{1}{n}\right),\\ 1 & \left(x \geq \frac{1}{n}\right). \end{array} \right.\]

The sequence of function \(\{f_n(x)\}\) converges to the following function \(f(x)\):

\[f(x) = \left\{ \begin{array}{cc} 0 & (x \leq 0),\\ 1 & (x > 0). \end{array} \right.\]

For each \(n\), \(f_n(x)\) is a continuous function. However, \(f(x)\) is not continuous. In fact, \(\lim_{n\to\infty}f_n(x) = f(x)\) is point-wise convergence, not uniform convergence. □

Exercise. (Try to) show directly that the sequence \(\{f_n(x)\}\) in the above example does not converge uniformly to \(f(x)\). □

Theorem (Uniform convergence and definite integral)

Suppose that the sequence \(\{f_n(x)\}\) of functions on a bounded interval \(I\) uniformly converges to the function \(f(x)\) on \(I\). For each \(a \in I\), let \(F_n(x) = \int_a^xf_n(t)dt\) and \(F(x) = \int_a^xf(t)dt\). Then the sequence of functions \(\{F_n(x)\}\) uniformly converges to \(F(x)\) on \(I\).

Proof. Suppose \(I \subset [\alpha, \beta]\). For any \(\varepsilon > 0\), we can find a natural number \(N\) such that, for all \(n\geq N\) and for any \(x \in I\), \(|f(x) - f_n(x)| < \frac{\varepsilon}{\beta - \alpha}\) (uniform convergence). Then, for any \(x \in I\), we have

\[|F(x) - F_n(x)| \leq \int_a^x|f(t) - f_n(t)|dt < \varepsilon\cdot\frac{|x-a|}{\beta - \alpha} \leq \varepsilon.\]

This means that \(\{F_n(x)\}\) uniformly converges to \(F(x)\) on \(I\). ■

Theorem (Uniform convergence and derivative)

Let \(\{f_n(x)\}\) be a sequence of \(C^1\) functions that converges to \(f(x)\) on \(I\) (point-wise). Suppose that the sequence of derivatives \(\{f'_n(x)\}\) uniformly converges to the function \(g(x)\) on \(I\). Then, \(f(x)\) is also of class \(C^1\) and \(f'(x) = g(x)\).

Proof. On the one hand, by the above Theorem (Uniform convergence and definite integral), \(\int_a^xf'_n(t)dt\) uniformly converges to \(\int_a^xg(t)dt\) on \(I\) where \(a \in I\) is arbitrary. On the other hand,

\[\int_a^xf'_n(t)dt = f_n(x) - f_n(a) \to f(x) - f(a)\]

as \(n \to \infty\). Thus, the convergence of \(\{f_n(x)\}\) to \(f(x)\) is uniform and

\[f(x) - f(a) = \int_a^xg(t)dt.\]

By differentiating both sides with respect to \(x\), we have \(f'(x) = g(x)\). ■


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