Leture 10
Definition 5.3.1. Let S⊆R, and let f:S→R be a function. Suppose for every ε>0 there exists a δ>0 such that whenever x,c∈S and |x−c|<δ, then |f(x)−f(c)|<ε. Then we say f is uniformly continuous.
Uniform vs Nonuniform continuity
Uniform vs Nonuniform continuity
f:[0,1]→R defined by f(x):=x2 is uniformly continuous.
Proof. Note that 0≤x,c≤1. Then
|x2−c2|
=|x+c||x−c|
≤(|x|+|c|)|x−c|
≤(1+1)|x−c|.
Therefore, given ε>0, let δ:=ε2. If |x−c|<δ, then |x2−c2|<ε.
On the other hand, g:R→R defined by g(x):=x2 is not uniformly continuous.
Proof. Suppose it is uniformly continuous, then for every ε>0, there would exist a δ>0 such that if |x−c|<δ, then |x2−c2|<ε. So consider ε=1. If such δ existed and c=x+δ/2 ⇒|x2−(x+δ2)2|<1.
However |xδ+δ24|<1 ❗
which is a contradiction, since we can choose x large. ◼
Comparison between continuity and uniform continuity
Let S⊆R and let f:S→R be a function.
Continuous function |
Uniform continuous |
---|
The function f:(0,1)→R defined by f(x):=1x is not uniformly continuous.
Proof. Choose ε=2 and 0<δ. Set δ0=min, x= \delta_0, and y = 2 \delta_0. Then x,y\in(0,1) and \abs{x-y}= \delta_0 \lt \delta but
\ds\abs{\frac{1}{x}-\frac{1}{y}} =\ds\abs{\frac{y-x}{xy}} \ds= \abs{\frac{\delta_0}{2\delta_0^2} } =\ds\abs{ \frac{1}{2\delta_0}} \geq 2 = \epsilon.
Hence f is not uniformly continuous. \; \bs
Nonuniform continuity criterion
Theorem 5.3.1. Let S\subseteq \R and let f \colon S \to \R. Then the following statements are equivalent:
Nonuniform continuity criterion
Example 5.3.3. We can apply the previous result to show that f(x)=\dfrac{1}{x} is not uniformly continuous on (0,\infty).
Consider x_n=\dfrac{1}{n} and y_n=\dfrac{1}{n+1} in (0,\infty). Then
\ds\lim_{n\ra \infty} (x_n-y_n) =\ds\lim_{n\ra \infty} \left(\frac{1}{n}-\frac{1}{n+1}\right) =0,
but \,\abs{\,f(x_n)- f(y_n)}=1\, for all \,n\in \N.
Theorem 5.3.2. Let f \colon [a,b] \to \R be a continuous function. Then f is uniformly continuous.
👀 Complementary reading 📖
Definition 5.3.2. A function f \colon S\subseteq \R \to \R is Lipschitz continuous, if there exists a K > 0, such that \begin{equation*} \abs{\,f(x)-f(y)} \leq K \abs{x-y} \;\;\text{for all } x \text{ and } y \text{ in } S. \end{equation*}
Theorem 5.3.3. A Lipschitz continuous function is uniformly continuous.
Theorem 5.3.3. A Lipschitz continuous function is uniformly continuous.
Proof. Let f \colon S \to \R be a function and let K be a constant such that \abs{\,f(x)-f(y)} \leq K \abs{x-y} for all x, y in S. For \epsilon > 0 be given. Take \delta := \dfrac{\epsilon}{K}. For all x and y in S such that \abs{x-y} \lt \delta,
\ds\abs{\,f(x)-f(y)} \ds \leq K \abs{x-y} \ds\lt K \delta \ds = K \frac{\epsilon}{K} \ds = \epsilon .
Hence f is uniformly continuous. \; \bs
Not every uniformly continuous function is Lipschitz continuous.
For example f(x):=\sqrt{x} defined on I=[0,2].
f is uniformly continuous on I but there is no number K>0 such that |\,f(x)-f(0)|\leq K|x-0| for all x\in I.
Definition 5.4.1.
Let S \subseteq \R.
We say f \colon S \to \R is increasing
(resp. strictly increasing) if x,y \in S with
x \lt y implies f(x) \leq f(y) (resp. f(x) \lt f(y)).
We define
decreasing and
strictly decreasing in the same way by switching the
inequalities for f.
If a function is either increasing or decreasing, we say it is monotone. If it is strictly increasing or strictly decreasing, we say it is strictly monotone.
One-sided limits for monotone functions are computed by computing infima and suprema.
Let S \subseteq \R, c \in \R, f \colon S \to \R be increasing, and g \colon S \to \R be decreasing.
If c is a cluster point of S \cap (-\infty,c), then
\ds \lim_{x \to c^-} f(x) = \sup \{ \,f(x) : x \lt c, x \in S \}
\ds \lim_{x \to c^-} g(x) = \inf \{ g(x) : x \lt c, x \in S \} .
One-sided limits for monotone functions are computed by computing infima and suprema.
Let S \subseteq \R, c \in \R, f \colon S \to \R be increasing, and g \colon S \to \R be decreasing.
If c is a cluster point of S \cap (c, \infty), then
\ds \lim_{x \to c^+} f(x) = \inf \{ \,f(x) : x \gt c, x \in S \}
\ds \lim_{x \to c^+} g(x) = \sup \{ g(x) : x \gt c, x \in S \} .
\ds \lim_{x \to c^+} f(x) = \inf \{ \,f(x) : x \gt c, x \in S \}
\ds \lim_{x \to c^-} f(x) = \sup \{ \,f(x) : x \lt c, x \in S \}
Theorem 5.4.1. If I \subseteq \R is an interval and f \colon I \to \R is monotone and not constant, then f(I) is an interval if and only if f is continuous.
Theorem 5.4.2. Let I \subseteq \R be an interval and f \colon I \to \R be monotone. Then f has at most countably many discontinuities.
Theorem 5.4.3. If I \subseteq \R is an interval and f \colon I \to \R is strictly monotone, then the inverse f^{-1} \colon f(I) \to I is continuous.
Theorem 5.4.3 does not require \,f\, itself to be continuous.
Let
f \colon \R \to \R be defined by
\ds
f(x):=
\begin{cases}
x & \text{if } x \lt 0, \\
x+1 & \text{if } x \geq 0. \\
\end{cases}
The function f is not continuous at 0.
The image of I = \R is the set
(-\infty,0)\cup [1,\infty), not an interval.
Then f^{-1} \colon (-\infty,0)\cup [1,\infty)
\to \R can be written as
\begin{equation*}
f^{-1}(y) =
\begin{cases}
y & \text{if } y \lt 0, \\
y-1 & \text{if } y \geq 1.
\end{cases}
\end{equation*}
It is not difficult to see that f^{-1} is a continuous function.
\ds
f(x):=
\begin{cases}
x & \text{if } x \lt 0, \\
x+1 & \text{if } x \geq 0. \\
\end{cases}
|
\ds
f^{-1}(y) =
\begin{cases}
y & \text{if } y \lt 0, \\
y-1 & \text{if } y \geq 1.
\end{cases}
|
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