Uniform convergence of sequence of functions

-

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 \(f'(x) = g(x)\). ■


Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Newton's method

-
Image
Newton's method (the Newton-Raphson method) is a very powerful numerical method for solving nonlinear equations.  Suppose we'd like to solve a non-linear equation \(f(x) = 0\), where \(f(x)\) is a (twice) differentiable non-linear function. Newton's method generates a sequence of numbers \(c_1, c_2, c_3, \cdots\) that converges to a solution of the equation. That is, if \(\alpha\) is a solution (i.e., \(f(\alpha) = 0\)) then,  \[\lim_{n\to\infty}c_n = \alpha,\] and this sequence \(\{c_n\}\) is generated by a series of linear approximations of the function \(f(x)\). Theorem (Newton's method) Let \(f(x)\) be a function that is twice differentiable on an open interval \(I\) that contains the closed interval \([a, b]\) (i.e., \([a,b]\subset I\)) and satisfy the following conditions: \(f(a) < 0\) and \(f(b) > 0\); For all \(x\in [a, b]\), \(f'(x) > 0\) and \(f''(x) > 0\). Let us define the sequence \(\{c_n\}\) by \[ \begin{eqnarray} c_1 &
Read more