Uniform Continuity
A function \( f: S \to \mathbb{R} \) is uniformly continuous on \( S \) if for every \( \epsilon \gt 0 \), there exists a \( \delta \gt 0 \) such that for all \( x, y \in S \): \[ |x - y| \lt \delta \implies |f(x) - f(y)| \lt \epsilon \]
Key Difference: In pointwise continuity, \( \delta \) can depend on both \( \epsilon \) and the point \( x \). In uniform continuity, \( \delta \) must work for all points \( x, y \) in the domain simultaneously. It depends only on \( \epsilon \).
Visualizing Uniform Continuity
Imagine a rectangle of width \( 2\delta \) and height \( 2\epsilon \). If a function is uniformly continuous, you can slide this rectangle along the graph, and the graph will never exit through the top or bottom edges of the rectangle while passing through the center.
Controls
Move mouse to slide the box.
Important Theorems
Theorem 19.2
If \( f \) is continuous on a closed interval \( [a, b] \), then \( f \) is uniformly continuous on \( [a, b] \).
Theorem 19.4
If \( f \) is uniformly continuous on a bounded open interval \( (a, b) \), then \( f \) can be extended to a continuous function on \( [a, b] \).
(Equivalently, \(\lim_{x \to a^+} f(x)\) and \(\lim_{x \to b^-} f(x)\) must exist).
Theorem 19.5
If \( f \) is uniformly continuous on \( S \) and \( (s_n) \) is a Cauchy sequence in \( S \), then \( (f(s_n)) \) is a Cauchy sequence.
Practice Problems
True/False
1. If \( f \) is uniformly continuous on a set \( S \), then \( f \) is continuous on \( S \).
True. Uniform continuity implies pointwise continuity.
2. If \( f \) is continuous on \( (0, 1) \), then \( f \) is uniformly continuous on \( (0, 1) \).
False. Consider \( f(x) = 1/x \). It is continuous but not uniformly continuous on \( (0, 1) \).
3. If \( f \) is continuous on a closed and bounded interval \( [a, b] \), then \( f \) is uniformly continuous on \( [a, b] \).
True. This is a standard theorem (Heine-Cantor).
4. The function \( f(x) = x^2 \) is uniformly continuous on \( \mathbb{R} \).
False. The slope gets arbitrarily large, so for a fixed \( \epsilon \), no single \( \delta \) works for all \( x \).
Fill in the Blank
1. In the definition of uniform continuity, the value of \( \delta \) depends only on , whereas in pointwise continuity it may also depend on the point \( x \).
Answer: \( \epsilon \)
2. A continuous function on a set is always uniformly continuous.
Answer: compact (or closed and bounded interval)
3. To show \( f \) is NOT uniformly continuous, we can find an \( \epsilon_0 \gt 0 \) and two sequences \( (x_n) \) and \( (y_n) \) such that \( |x_n - y_n| \to 0 \) but \( |f(x_n) - f(y_n)| \ge \) .
Answer: \( \epsilon_0 \) (or some positive constant)
Full Problems
1. Prove that \( f(x) = 3x + 1 \) is uniformly continuous on \( \mathbb{R} \).
Let \( \epsilon \gt 0 \). We want to find \( \delta \gt 0 \) such that \( |x - y| \lt \delta \implies |f(x) - f(y)| \lt \epsilon \).
Consider \( |f(x) - f(y)| = |(3x+1) - (3y+1)| = |3x - 3y| = 3|x - y| \).
We want \( 3|x - y| \lt \epsilon \), which means \( |x - y| \lt \epsilon/3 \). So, choose \( \delta = \epsilon/3 \). This \( \delta \) works for all \( x, y \).
2. Show that \( f(x) = x^2 \) is not uniformly continuous on \( \mathbb{R} \).
Let \( x_n = n \) and \( y_n = n + \frac{1}{n} \). Then \( |x_n - y_n| = \frac{1}{n} \to 0 \).
However, \( |f(x_n) - f(y_n)| = |n^2 - (n + \frac{1}{n})^2| = |n^2 - (n^2 + 2 + \frac{1}{n^2})| = 2 + \frac{1}{n^2} \gt 2 \).
Since the difference in function values stays above 2 even as the points get arbitrarily close, \( f \) is not uniformly continuous.
3. Show that \( f(x) = \sqrt{x} \) is uniformly continuous on \( [0, \infty) \).
This is tricky. We can split the domain into \( [0, 1] \) and \( [1, \infty) \).
- On \( [0, 1] \), \( f \) is continuous on a closed bounded interval, so it is uniformly continuous.
- On \( [1, \infty) \), the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) is bounded by \( 1/2 \). By the Mean Value Theorem, \( |f(x) - f(y)| \le \frac{1}{2}|x - y| \), so it is Lipschitz and thus uniformly continuous.
Since it is uniformly continuous on both pieces (and they overlap/connect), it is uniformly continuous on the union.
4. Let \( f \) be uniformly continuous on a bounded interval \( (a, b) \). Show that \( f \) is bounded on \( (a, b) \).
Since \( f \) is uniformly continuous, we can choose \( \epsilon = 1 \) and find a corresponding \( \delta \).
Partition \( (a, b) \) into a finite number of subintervals of length less than \( \delta \). Let these points be \( x_0, x_1, \dots, x_n \).
On each subinterval, \( |f(x) - f(x_i)| \lt 1 \), so \( f(x) \) is bounded by \( |f(x_i)| + 1 \). Since there are finitely many subintervals, \( f \) is bounded on the whole interval.