Example · Undergraduate
Terms
| Term | This example | Why |
|---|
| continuous |
✓ |
|
| continuously differentiable |
✓ |
$x \mapsto \sin(1/x)$ is continuously differentiable, and so continuous, on $(0, 1]$.
|
| locally Lipschitz continuous |
✓ |
$x \mapsto \sin(1/x)$ on $(0, 1]$ is locally Lipschitz continuous, because each point $x_0 \in (0, 1]$ has a neighborhood $[x_0 / 2, 1]$ on which the derivative $-\cos(1/x) / x^2$ is bounded by $4 / x_0^2$.
|
| uniformly continuous |
✗ |
$x \mapsto \sin(1/x)$ on $(0, 1]$ is not uniformly continuous, because points that are arbitrarily close together have values that differ by $2$.
For each natural number $n$, let
\[x_n = \frac{1}{\frac{\pi}{2} + 2 \pi n} \quad \text{and} \quad y_n = \frac{1}{\frac{3\pi}{2} + 2 \pi n}.\]
Then $\sin(1/x_n) = 1$ and $\sin(1/y_n) = -1$, so $\abs{\sin(1/x_n) - \sin(1/y_n)} = 2$ for every $n$, but $\abs{x_n - y_n} \to 0$ as $n \to \infty$.
Therefore, for $\varepsilon = 1$, no $\delta > 0$ satisfies the definition of uniform continuity: some pair $x_n, y_n$ is closer than $\delta$, but the values of the pair differ by more than $\varepsilon$.
|
| Lipschitz continuous |
✗ |
$x \mapsto \sin(1/x)$ on $(0, 1]$ is not Lipschitz continuous, because the slopes between pairs of points on its graph are unbounded.
With the points $x_n$ and $y_n$ from the uniform continuity explanation, the slope between $(x_n, 1)$ and $(y_n, -1)$ has absolute value
\[\frac{2}{\abs{x_n - y_n}} = \frac{2 \left(\frac{\pi}{2} + 2 \pi n\right) \left(\frac{3\pi}{2} + 2 \pi n\right)}{\pi},\]
which grows without bound as $n \to \infty$, so no Lipschitz constant $L$ works.
|
Distinctions using this example
All examples · All distinctions · All definitions