44 Uniform Continuity Practice

#todo/school : Read §4.4 📅 2024-11-09 ✅ 2024-11-09
#todo/school : Do problems 4.4.2, 3, 4, 6, 7, 8, 13 📅 2024-11-10 ✅ 2024-11-11
Hint: for 4.4.7 note that $| \sqrt{x} - \sqrt{y} | \leq | \sqrt{x} + \sqrt{y} |$ for $x, y \in [0, \infty)$ . Then try and bound $| \sqrt{x} - \sqrt{y} |^{2}$ .

4.4.2

Question

a. Is $f (x) = \frac{1}{x}$ uniformly continuous on $(0, 1)$ ?
b. Is $g (x) = \sqrt{x^{2} + 1}$ uniformly continuous on $(0, 1)$ ?
c. Is $h (x) = x \sin (\frac{1}{x})$ uniformly continuous on $(0, 1)$ ?

Proof

a. No. Intuitively the slope becomes unbounded near 0. For the proof, consider using the Sequential Criterion for Absence of Uniform Continuity. Choose $ε_{0} = \frac{1}{2} > 0$ . Consider the sequences $(x_{n}) = \frac{2}{n}$ and $(y_{n}) = \frac{1}{n}$ . We have $| x_{n} - y_{n} | = | \frac{1}{n} | \to 0$ and $| f (x_{n}) - f (y_{n}) | = | \frac{- n}{2} | \geq ε_{0} = \frac{1}{2}$ . Thus it fails to be uniformly continuous here.

b. The slopes don't get too crazy so we intuitively expect this to be uniformly continuous. To prove it, let $ε > 0$ . We want to choose some $δ$ such that any $| x - y | < δ$ gives:

\begin{aligned} | g (x) - g (y) | & = | \sqrt{x^{2} + 1} - \sqrt{y^{2} + 1} | \\ = \frac{| x^{2} + 1 - y^{2} - 1 |}{\sqrt{x^{2} + 1} + \sqrt{y^{2} + 1}} \\ = \frac{| (x - y) (x + y) |}{\sqrt{x^{2} + 1} + \sqrt{y^{2} + 1}} \\ < \frac{| (x - y) (x + y) |}{\sqrt{x^{2}} + \sqrt{y^{2}}} \\ = \frac{| (x - y) (x + y) |}{| x | + | y |} \\ = | x - y | & (x, y > 0) \\ < δ \\ = ε \end{aligned}

Thus using $δ = ε$ here should work.

c. The slopes near $0$ again become nearly vertical so we expect this to not be uniformly continuous, especially near 0. However, the value 0 can be added to $h (0)$ to consider an extended version of $h$ on $[0, 1]$ (in order to use Uniformly Continuous iff Compact and Continuous). Namely, consider the extension of the function $h$ , denoted $h_{0}$ , such that $h_{0} (0) = 0$ . Thus notice that since $[0, 1]$ is compact and $h_{0}$ is continuous on that interval (consider all of the subfunctions like $\frac{1}{x}, x, \sin (x)$ that are continuous and make up $h$ ), then $h_{0}$ is uniformly continuous. Thus it shows that since $h = h_{0}$ on the interval $(0, 1)$ then $h$ itself is uniformly continuous on that interval.

☐

4.4.3

Question

Show that $f (x) = \frac{1}{x^{2}}$ is uniformly continuous on the set $[1, \infty)$ but not on the set $(0, 1]$ .

Proof

Notice that since $f$ is continuous over $[1, \infty)$ and that domain is compact, then since Uniformly Continuous iff Compact and Continuous implies $f$ is uniformly continuous over the set.

We can show this explicitly too. Let $ε > 0$ . Then choose $δ = \frac{ε}{2}$ . Then for $| x - y | < δ$ :

\begin{aligned} | f (x) - f (y) | & = | \frac{1}{x^{2}} - \frac{1}{y^{2}} | \\ = | \frac{y^{2} - x^{2}}{x^{2} y^{2}} | \\ = \frac{| x - y | | x + y |}{x^{2} y^{2}} \\ = | x - y | \cdot | \frac{1}{x y^{2}} + \frac{1}{x^{2} y} | \\ \leq | x - y | \cdot 2 \\ < δ \cdot 2 \\ = \frac{ε}{2} \cdot 2 \\ = ε \end{aligned}

Now it is not uniformly continuous over $(0, 1]$ , as this set is definitely not compact (and thus the argument above doesn't work). Since $f$ is continuous on the set but not compact, then it cannot be uniformly continuous (use the negative of that statement).

To explicitly show this we'll again use Sequential Criterion for Absence of Uniform Continuity. Construct $(x_{n}) = \frac{1}{n}$ and $(y_{n}) = \frac{1}{2 n}$ . Notice $| x_{n} - y_{n} | = \frac{1}{2 n} \to 0$ but:

\begin{aligned} | f (x_{n}) - f (y_{n}) | & = | \frac{1}{({\frac{1}{n}}^{2})} - \frac{1}{({\frac{1}{2 n}}^{2})} | \\ = | n - 4 n^{2} | \\ = 3 n^{2} \end{aligned}

Thus using $ε_{0} = 3$ here then $| f (x_{n}) - f (y_{n}) | = 3 n^{2} \geq 3 = ε_{0}$ .

☐

4.4.4

Question

Decide whether each of the following statements is true or false, justifying each conclusion.
a. If $f$ is continuous on $[a, b]$ with $f (x) > 0$ for all $a \leq x \leq b$ , then $\frac{1}{f}$ is bounded on $[a, b]$ (so $\frac{1}{f}$ has bounded range).
b. If $f$ is uniformly continuous on a bounded set $A$ then $f (A)$ is bounded.
c. If $f$ is defined on $R$ and $f (K)$ is compact where $K$ is compact, then $f$ is continuous on $R$ .

Proof

a. True. Since $f$ is continuous and $\frac{1}{x}$ is continuous on all values positive (which $f (x) > 0$ ) then their composition $\frac{1}{f}$ must be continuous over $f ([a, b])$ . Since the Compactness Is Preserved By Continuity, then $f ([a, b])$ must also be compact. As a result, then it must be bounded.

b. True. Similar to (a) since $f$ is uniformly continuous over $A$ then $A$ is bounded (given) and closed (thus it's compact), while $f$ is continuous. Since the Compactness Is Preserved By Continuity, then $f (A)$ must also be compact, and thus bounded.

c. False. Consider $K = [1, 2]$ (a compact set) and $f (x) = \frac{1}{x}$ . Clearly $f$ is continuous on $[1, 2]$ so then $f (K)$ must be compact. However, $f$ is certainly not continuous over all of $R$ .

☐

4.4.6

Question

Give an example of each of the following, or state that such a request is impossible. For any that are impossible, supply a short explanation for why this is the case.
a. A continuous function $f : (0, 1) \to R$ and a Cauchy Sequence $(x_{n})$ such that $f (x_{n})$ is not Cauchy.
b. A uniformly continuous function $f : (0, 1) \to R$ and a Cauchy sequence $(x_{n})$ such that $f (x_{n})$ is not a Cauchy sequence.
c. A continuous function $f : [0, \infty) \to R$ and a Cauchy sequence $(x_{n})$ such that $f (x_{n})$ is not a Cauchy sequence.

This exercise shows why both conditions are needed in Uniformly Continuous iff Compact and Continuous.

Proof

a. Possible. Use $f (x) = \frac{1}{x}$ over $(0, 1)$ is continuous. But the Cauchy sequence $(x_{n}) = \frac{1}{n}$ has $x_{n} \to 0$ but $f (x_{n}) = n$ diverges (and thus isn't Cauchy).

b. Impossible. If $f$ is uniformly continuous then if $(x_{n}) \to x$ is Cauchy, then by the definition of Convergence of a Sequence then letting $ε > 0$ gives some $N \in N$ where $\forall n, m \geq N$ then $| x_{n} - x_{m} | < ε$ . Since $f$ is uniformly continuous then this implies that $| f (x_{n}) - f (x_{m}) | < ε$ as we needed to show that $f (x_{n})$ is Cauchy.

c. Impossible. Since $[0, \infty)$ is compact then since $f$ is continuous over it and Compactness Is Preserved By Continuity then $f ([0, \infty))$ must be compact. Thus if such a $(x_{n}) \to x \in [0, \infty)$ is proposed, then $f (x_{n}) \subseteq f ([0, \infty))$ is both bounded (by $f ([0, \infty))$ being compact) and convergent (since $(x_{n})$ is convergent then $f (x_{n})$ is convergent to $f (x) \in f ([0, \infty))$ by the Characterization of Continuity (3)), thus being Cauchy.

☐

4.4.7

Question

Prove that $f (x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$ .

Proof

Notice that $[0, \infty)$ is compact, and since $\sqrt{x}$ is continuous on that interval then it must be Uniformly Continuous iff Compact and Continuous. See Lecture 25 - Continuity#Continuous Examples (4) for the proof of this.

To do the proof more directly, let $ε > 0$ . We want to produce some $δ$ where for any $| x - y | < δ$ :

\begin{aligned} | \sqrt{x} - \sqrt{y} | & \leq | \sqrt{x} + \sqrt{y} | \\ = \sqrt{x} + \sqrt{y} \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

Where we used the fact that $\sqrt{x}$ was continuous on the second to last lime. Using the $δ$ from $\sqrt{x}$ being continuous should work here. Namely $δ = min {δ_{x}, δ_{y}}$ for each $δ$ we get from $\sqrt{x}, \sqrt{y}$ being continuous.

☐

4.4.8

Question

Give an example of each of the following, or provide a short argument for why the request is impossible.
a. A continuous function defined on $[0, 1]$ with range $(0, 1)$ .
b. A continuous function defined on $(0, 1)$ with range $[0, 1]$ .
c. A continuous function defined on $(0, 1]$ with range $(0, 1)$ .

Proof

a. Impossible. Since $[0, 1]$ is compact and the function $f$ is continuous, then it's range $f ([0, 1]) = (0, 1)$ must be continuous via Compactness Is Preserved By Continuity. Since $(0, 1)$ isn't compact then this is impossible.

b. Possible. Use $f (x) = | \sin (\frac{1}{x}) |$ .

c. Possible. A similar function $g (x) = \frac{| \sin (\frac{1}{x}) | (1 - x)}{2} + \frac{1}{2}$ gets us the range we wanted.

☐

4.4.13

Continuous Extension Theorem

a. Show that a uniformly continuous function preserves Cauchy Sequences; that is if $f : A \to R$ is uniformly continuous and $(x_{n}) \subseteq A$ is a Cauchy sequence, then show that $f (x_{n})$ is a Cauchy sequence.
b. Let $g$ be a continuous function on the open interval $(a, b)$ . Prove that $g$ is uniformly continuous on $(a, b)$ iff it is possible to define values $g (a)$ and $g (b)$ at the endpoints so that the extended function $g$ is continuous on $[a, b]$ .
(Hint: for (b)'s forward direction, first produce candidates for $g (a), g (b)$ and then show the extended $g$ is continuous).

Proof

a. Let $(x_{n}) \subseteq A$ be Cauchy. Then it converges to some point $x$ . Further let $f : A \to R$ be uniformly continuous.

To show that $f (x_{n})$ is Cauchy it is enough to show it converges. Now because $f$ is uniformly continuous then by the Characterization of Continuity then since $(x_{n}) \subseteq A$ and converges to $x$ then $f (x_{n}) \to f (x)$ and thus $f (x_{n})$ must be Cauchy.

( $\Rightarrow$ ): Suppose that $g$ is uniformly continuous. Choose $g (a)$ by constructing $(a_{n}) = \frac{(2^{n} - 1) a + b}{2^{n}} \subseteq (a, b)$ . Notice that $a_{n} \to a$ so then via (a) then $g (a_{n})$ converges. Denote the value it converges to as $g (a)$ . Similarly construct $(b_{n}) = \frac{a + (2^{n} - 1) b_{n}}{2^{n}} \subseteq (a, b)$ . Again $b_{n} \to b$ so then via (a) then $g (b_{n})$ converges. Define this convergence point $g (b)$ .

Now we will show that this redefinition of $g$ is continuous on $[a, b]$ . Clearly it is continuous on $(a, b)$ . For the endpoints notice that we've shown that the since now $a, b$ are in our domain, then all we have to show is that any convergent sequence $(x_{n}) \subseteq [a, b]$ where $x_{n} \to a$ implies that $g (x_{n}) \to g (a)$ as we've constructed (the argument for $b$ is a near carbon copy).

Let $(x_{n}) \to a$ be such a sequence. Notice that for contradiction if $f (x_{n}) \to c \neq g (a)$ then by Existence of Functional Limits then our $x_{n}, a_{n}$ sequences are enough to show that $g$ is not continuous at $a$ , which is a contradiction as $g$ is supposedly continuous. As such then the only valid value that $x_{n}$ can converge to is $g (a)$ , completing the proof.

( $\Leftarrow$ ): Since $g$ is continuous and $[a, b]$ is a compact set, then our range must be a compact set $f ([a, b])$ . As a result, then because Uniformly Continuous iff Compact and Continuous then $g$ must be uniformly continuous.

☐