43 Continuity Practice

#todo/school : Read §4.3; 📅 2024-11-06 ✅ 2024-11-06
#todo/school : Do problems 4.3.1, 3(a or b), 6, 7, 8, 9, 11 📅 2024-11-10 ✅ 2024-11-11

4.3.1

Question

Let $g (x) = \sqrt[3]{x}$
a. Prove that $g$ is continuous at $c = 0$
b. Prove that $g$ is continuous at a point $c \neq 0$ . (The identity $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ will be helpful).

Scratch:

a. See 42 Functional Limits Practice#4.2.8 (d) for showing the limit exists at $0$ .

b. If we have some $ε > 0$ we'd want to show that a given $δ > 0$ produces, when $| x - c | < δ$ that:

\begin{aligned} | g (x) - g (c) | & = | \sqrt[3]{x} - \sqrt[3]{c} | \\ = | x^{\frac{1}{3}} - c^{\frac{1}{3}} | \\ = \frac{| x - c |}{| x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} |} & (a = x^{\frac{1}{3}}, b = c^{\frac{1}{3}}) \end{aligned}

Clearly we'll have the $| x - c | < δ$ come in on the numerator. For the denominator we can increase it:

| x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} | \leq | x^{\frac{2}{3}} | + | x^{\frac{1}{3}} c^{\frac{1}{3}} | + | c^{\frac{2}{3}} | \Rightarrow | c^{\frac{2}{3}} | \geq | x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} |

As such:

\begin{aligned} < \frac{| x - c |}{| c^{\frac{2}{3}} |} & (c \neq 0) \\ < \frac{δ}{| c^{\frac{2}{3}} |} \\ = ε \end{aligned}

This ends well if $δ = | c^{\frac{2}{3}} | ε$ .

Proof

a. See 42 Functional Limits Practice#4.2.8 (d) for the proof that the limit is $g (0) = \sqrt[3]{0} = 0$ .

b. Let $ε > 0$ . Choose $δ = | c^{\frac{2}{3}} | ε > 0$ . First notice that:

| x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} | \leq | x^{\frac{2}{3}} | + | x^{\frac{1}{3}} c^{\frac{1}{3}} | + | c^{\frac{2}{3}} | \Rightarrow | c^{\frac{2}{3}} | \geq | x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} |

Then we will use this in the following:

\begin{aligned} | g (x) - g (c) | & = | \sqrt[3]{x} - \sqrt[3]{c} | \\ = | x^{\frac{1}{3}} - c^{\frac{1}{3}} | \\ = \frac{| x - c |}{| x^{\frac{2}{3}} + x^{\frac{1}{3}} c^{\frac{1}{3}} + c^{\frac{2}{3}} |} & (a = x^{\frac{1}{3}}, b = c^{\frac{1}{3}}) \\ < \frac{| x - c |}{| c^{\frac{2}{3}} |} & (c \neq 0) \\ < \frac{δ}{| c^{\frac{2}{3}} |} \\ = | c^{\frac{2}{3}} | ε \cdot \frac{1}{| c^{\frac{2}{3}} |} \\ = ε \end{aligned}

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4.3.3

Question

a. Supply a proof for Composition of Continuous Functions using the definition of continuity for functions.
b. Give another proof of this theorem using the Characterization of Continuity (use (3))

Scratch:

a. Let $ε > 0$ . We want to show that:

| (g \circ f) (x) - (g \circ f) (c) | < ε

Suppose we chose such a $δ$ such that we suppose $0 < | x - c | < δ$ . Working through it:

\begin{aligned} | (g \circ f) (x) - (g \circ f) (c) | & = | g (f (x)) - g (f (c)) | \end{aligned}

Now because $g$ is continuous at $f (c)$ then $lim_{x \to f (c)} g (x) = g (f (c))$ , so then for any $ε > 0$ then $\exists δ_{1} > 0$ where for any $0 < | x - f (c) | < δ_{1}$ then $| g (x) - g (f (c)) | < ε$ . Similarly since $f$ is continuous at $c$ then $lim_{x \to c} f (x) = f (c)$ , so then for any $ε > 0$ then $\exists δ_{2} > 0$ where if $0 < | x - c | < δ_{2}$ then $| f (x) - f (c) | < ε$ .

Notice that we want the $x$ part in the first given to be that $f (x)$ . Notice the cascading nature of the limit definitions that will allow this. Namely, since $f$ is continuous and our $ε > 0$ then $\exists δ_{2} > 0$ where if $0 < | x - c | < δ_{2}$ (which is our supposition) then we get:

| f (x) - f (c) | < ε

Now because $ε > 0$ then we get that $\exists δ_{1} > 0$ such that since our supposition $| x - f (c) | < δ_{2} = δ_{1}$ then we get that:

| g (f (x)) - g (f (c)) | < ε

as desired. Thus to have both be true then choose $δ = δ_{2}$ to be able to use both suppositions.

b. The proof is actually pretty simple here.

Proof

a. Let $ε > 0$ . Since $g$ is continuous at $f (c)$ then $\exists δ_{1} > 0$ such that $0 < | x - f (c) | < δ_{1} \to | g (x) - g (f (c)) | < ε$ . Now using $ε = δ_{1} > 0$ in the definition of continuity for $f$ at $c$ gives that $\exists δ_{2} > 0$ such that $0 < | x - c | < δ_{2} \to | f (x) - f (c) | < δ_{1}$ .

Now choose $δ = δ_{2}$ . Suppose that $0 < | x - c | < δ = δ_{2}$ . Then $0 < | x - c | < δ_{2}$ , so then we immediately get that $| f (x) - f (c) | < δ_{1}$ . Using the $g$ continuity definition (see above, where now $x$ is replaced with $f (x)$ ), then that implies that $| g (f (x)) - g (f (c)) | < ε$ as we so wanted.

b. Since $f$ is continuous at $c \in A$ then $\forall (x_{n}) \subseteq A$ where $x_{n} \to c$ then $f (x_{n}) \to f (c)$ . Similarly, since $g$ is continuous at $f (c) \in f (A) \subseteq B$ then $\forall (y_{n}) \subseteq f (A)$ where $y_{n} \to f (c)$ then $g (y_{n}) \to g (f (c))$ .

Now we need to show that $g \circ f$ is continuous at $c$ , so let $(x_{n}) \subseteq A$ be arbitrary where $x_{n} \to c$ . Then since $f$ is continuous then $f (x_{n}) \to f (c)$ . Now notice that the sequence $(y_{n}) = f (x_{n}) \subseteq f (A)$ so then clearly $y_{n} \to f (c)$ , which is just that $f (x_{n}) \to f (c)$ . This implies that $g (y_{n}) = g (f (x_{n})) \to g (f (c))$ . This is exactly what we wanted!

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4.3.6

Question

Provide an example of each or explain why the request is impossible.
a. Two functions $f$ and $g$ , neither of which is continuous at $0$ but such that $f (x) g (x)$ and $f (x) + g (x)$ are continuous at $0$ .
b. A function $f (x)$ continuous at $0$ and $g (x)$ not continuous at $0$ such that $f (x) + g (x)$ is continuous at $0$ .
c. A function $f (x)$ continuous at $0$ and $g (x)$ not continuous at $0$ such that $f (x) g (x)$ is continuous at $0$ .
d. A function $f (x)$ not continuous at $0$ such that $f (x) + \frac{1}{f (x)}$ is continuous at $0$ .
e. A function $f (x)$ not continuous at $0$ such that $[f (x)]^{3}$ is continuous at $0$ .

Proof

a. Possible. Use:

f (x) = {\begin{cases} 1 & x \geq 0 \\ - 1 & x < 0 \end{cases}, g (x) = - f (x) = {\begin{cases} - 1 & x \geq 0 \\ 1 & x < 0 \end{cases}

both of which are discontinuous at $0$ . However:

f (x) + g (x) = f (x) - f (x) = 0

is continuous at $0$ and furthermore:

f (x) g (x) = {\begin{cases} 1 & x \geq 0 \\ 1 & x < 0 \end{cases} = 1

is also continuous at $0$ .

b. Impossible. If $f (x) + g (x)$ is continuous at $0$ and $f (x)$ as well, then $lim_{x \to 0} f (x) + g (x) - f (x) = lim_{x \to 0} g (x)$ must exist (see Algebraic Continuity Theorem), which contradicts $g$ not being continuous at $0$ .

c. Possible. Use $f (x) = x$ (continuous at 0) and $g (x) = \frac{1}{x}$ (not continuous at 0). Then $lim_{x \to 0} f (x) g (x) = lim_{x \to 0} \frac{x}{x} = lim_{x \to 0} 1 = 1$ exists, implying $f (x) g (x)$ is continuous at $0$ .

d. Possible. If:

f (x) = {\begin{cases} 2 & x \geq 0 \\ \frac{1}{2} & x < 0 \end{cases} \Rightarrow \frac{1}{f (x)} = {\begin{cases} \frac{1}{2} & x \geq 0 \\ 2 & x < 0 \end{cases}

then:

f (x) + \frac{1}{f (x)} = {\begin{cases} 2 + \frac{1}{2} & x \geq 0 \\ \frac{1}{2} + 2 & x < 0 \end{cases} = \frac{5}{2}

which is continuous at $0$ .

e. Impossible. If $lim_{x \to 0} [f (x)]^{3}$ exists then:

lim_{x \to 0} f (x) = lim_{n \to 0} \sqrt[3]{f (x)^{3}}

exists via 43 Continuity Practice#4.3.1 (a). As such this contradicts $f (x)$ not being continuous at $0$ .

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4.3.7

Question

a. Referring to the proper theorems, give a formal argument that the Dirichlet's Function is nowhere-continuous on $R$
b. Review the definition of Thomae's Function. Demonstrate that it fails to be continuous at every rational point.
c. Use the Characterization of Continuity (3) to show that Thomae's Function is continuous at every irrational point in $R$ . (Given $ε > 0$ , consider the set of points ${x \in R : t (x) \geq ε}$ ).

Proof

a. Let $c \in R$ be arbitrary. We want to show that $g$ given above is not continuous at $c$ , so namely that we need to choose some $ε > 0$ such that $\forall δ > 0$ then $\exists x \in A$ where $| f (x) - f (c) | \geq ε \to | x - c | \geq δ$ . We can construct a sequence $(q_{n}) \subseteq Q$ where $q_{n} \to c$ and also $(x_{n}) \subseteq I$ where $x_{n} \to c$ (because The Density of Q in R implies that this works for both $Q$ and $I$ ). Now notice that:

g (q_{n}) = 1 \to 1, g (x_{n}) = 0 \to 0

so by the Existence of Functional Limits then $lim_{x \to c} g (x)$ doesn't exist. Since $c$ was arbitrary, then this happens for each point in the domain.

b. Let $c \in Q$ . We'll show that $t$ is not continuous here by constructing two sequences that converge to $c$ but when composed on $t$ approaches different values. Let $a_{n} \subseteq Q - {0}$ such that each $a_{n}$ is in the lowest terms (see the definition of Thomae's Function for this construction) and $a_{n} \to c$ . Clearly $t (a_{n}) = \frac{1}{k}$ for some integers $k$ , where since $c = \frac{m}{n}$ for some $m, n \in N$ then $t (a_{n}) \to \frac{1}{n}$ .

However, construct $b_{n} \subseteq I$ converging to $c$ (because $I$ is dense in $R$ ). Then because each $t (b_{n}) = 0$ then $t (b_{n}) \to 0$ as $b_{n} \to c$ .

Now because $t (a_{n}) \to \frac{1}{n} \neq 0 \leftarrow t (b_{n})$ then by the Existence of Functional Limits then $lim_{x \to c} t (x)$ doesn't exist for any $c \in Q$ .

c. As the hint suggests, there are finitely many points in the set proposed, so we can use that to show that the rest of the points (infinitely many of them) must be less than $ε$ .

Let $c \in I$ . Let $ε > 0$ . Now notice that ${x \in V_{1} (c) | t (x) \geq ε}$ is finite because the requirement $t (x) \geq ε$ is the same as $x = \frac{m}{n} \land \frac{1}{n} \geq ε$ , and we can count all of the rational $x$ 's here (since it's the subset of $Q$ which is countable). As a result the set is countable and especially finite $\equiv {x_{1}, x_{2}, \dots, x_{n}}$ . Thus set $δ = min (| x_{i} - c | : i \in {1, \dots, n})$ . Then every $x \in V_{δ} (c)$ has $t (x) \leq ε$ .

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4.3.8

Question

Decide if the following claims are true or false, providing either a short proof or counterexample to justify each conclusion. Assume throughout that $g$ is defined and continuous on all $R$ :
a. If $g (x) \geq 0$ for all $x < 1$ then $g (1) \geq 0$ as well.
b. If $g (r) = 0$ for all $r \in Q$ then $g (x) = 0$ for all $x \in R$ .
c. If $g (x_{0}) > 0$ for a single point $x_{0} \in R$ then $g (x)$ is in fact strictly positive for uncountable many points.

Proof

a. True. If we suppose $g (x) \geq 0$ for all $x < 1$ and $g (1) < 0$ then by the definition of continuity then $\exists g (1)$ and by the Characterization of Continuity we have that any sequence $(x_{n}) \subseteq R$ where $x_{n} \to 1$ implies that $g (x_{n}) \to g (1)$ . Consider constructing the sequence $x_{n} = 1 - \frac{1}{n} < 1$ for all $n$ . Then since $x_{n} \to 1$ then that implies that $g (x_{n}) \geq 0$ then approaches $g (1)$ . Namely $g (x_{n}) \to g (1)$ . By the Limits and Order (Order Limit Theorem) then since $g (x_{n}) \geq 0$ then $g (1) \geq 0$ .

b. True. Let $x \in R$ be arbitrary. If $x \in Q$ then $g (x) = 0$ by construction and we are done, so suppose $x \notin Q$ , so $x \in I$ . Because The Density of Q in R (except using irrationals here) then there is some sequence of rationals that approach $x$ from above and below. Let $(x_{n}) \subseteq Q^{< x}$ and $(y_{n}) \subseteq Q^{> x}$ where $x_{n}, y_{n} \to x$ . Since $g$ is continuous then to avoid being discontinuous then we must have it that $g (x_{n}) \to g (x)$ and $g (y_{n}) \to g (x)$ . But notice that $g (x_{n}) = 0 = g (y_{n}) \to 0$ , so then that implies that $g (x) = 0$ .

c. True. If this is the case then since $g$ is continuous at $x_{0}$ then using $ε = g (x_{0}) > 0$ gives that $\exists δ > 0$ where for any $x \in R$ where $x \in V_{δ} (x_{0}) \to g (x) \in V_{g (x_{0})} (g (x_{0}))$ . Notice that $V_{δ} (x_{0})$ is an uncountable set, so only consider $x$ in that neighborhood. Since $g (x)$ is in this neighborhood notice that:

⟺ g (x) \in (g (x_{0}) - g (x_{0}), 2 g (x_{0})) \equiv (0, 2 g (x_{0})) \Rightarrow g (x) > 0

Thus $g (x)$ is positive for uncountably many points.

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4.3.9

Question

Suppose $h : R \to R$ is continuous on $R$ . Let $K = {x : h (x) = 0}$ . Show that $K$ is a closed set.

Proof

Let $x$ be a limit point of $K$ . Thus then $\exists (x_{n}) \subseteq K - {x}$ such that $x_{n} \to x$ . We want to show that $x \in K$ , or equivalently that $h (x) = 0$ .

Now because $h$ is continuous on $R$ then by the Characterization of Continuity then $h (x_{n}) \to h (x)$ . Now notice that since each $x_{n} \in K - {x}$ then by definition then $h (x_{n}) = 0$ for all $n$ . Thus then $h (x_{n}) = 0 \to h (x) = 0$ as desired.

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4.3.11

Contraction Mapping Theorem

Let $f$ be a function defined on all of $R$ and assume there is a constant $c$ such that $0 < c < 1$ and

| f (x) - f (y) | \leq c | x - y |

for all $x, y \in R$ .
a. Show that $f$ is continuous on $R$ .
b. Pick some point $y_{1} \in R$ and construct the sequence

{y_{1}, f (y_{1}), f (f (y_{1})), \dots}

In general if $y_{n + 1} = f (y_{n})$ , show that the resulting sequence $(y_{n})$ is a Cauchy Sequence. Hence, we may let $y = lim_{n \to \infty} y_{n}$ .
c. Prove that $y$ is a fixed point of $f$ (ie: $f (y) = y$ ) and that it is unique in this regard.
d. Finally, prove that if $x$ is any arbitrary point in $R$ then the sequence ${x, f (x), f (f (x)), \dots}$ converges to $y$ defined in (b).

Proof

a. We'll use the sequence Characterization of Continuity here. Let $(x_{n}) \subseteq R$ where $x_{n} \to x$ be arbitrary. We want to show that $f (x_{n}) \to f (x)$ . First notice that since $f$ is defined on all of $R$ then $f (x)$ is well defined here.

Let $ε > 0$ . Since $x_{n}, x \in R$ then notice that $\exists 0 < c < 1$ where:

| f (x_{n}) - f (x) | \leq c | x_{n} - x |

Now choose $δ = \frac{ε}{c} > 0$ . Then since $x_{n} \to x$ then $\exists N \in N$ where for all $n \geq N$ then $| x_{n} - x | < δ$ . Now with this supposition then:

| f (x_{n}) - f (x) | \leq c | x_{n} - x | < c \cdot \frac{ε}{c} = ε

as desired.

b. We want $| y_{n} - y_{m} | < ε$ . Since $0 < c < 1$ we have:

| y_{n + 1} - y_{m + 1} | \leq c | y_{n} - y_{m} | \leq \dots \leq c^{m} | y_{n - m + 1} - y_{1} |

where we are assuming that $n > m$ here. Thus if we can bound $| y_{k} - y_{1} | \leq M$ for some constant $M$ then we will be done by choosing $m$ large enough such that $c^{m} M < ε$ .

In general, notice that $| y_{a + 1} - y_{a} | \leq c | y_{a} - y_{a - 1} |$ so then:

\begin{aligned} | y_{k} - y_{1} | & \leq | y_{1} - y_{2} | + | y_{2} - y_{3} | + \dots + | y_{k - 1} - y_{k} | \\ \leq | y_{1} - y_{2} | + c | y_{1} - y_{2} | + \dots + c^{m} | y_{1} - y_{2} | \\ = | y_{1} - y_{2} | \sum_{i = 0}^{m} c^{i} \\ < | y_{1} - y_{2} | \sum_{i = 0}^{\infty} c^{i} \\ = \frac{| y_{1} - y_{2} |}{1 - c} = M \end{aligned}

c. Since $f$ is continuous over all $R$ (see (a)) then:

y = lim_{n \to \infty} y_{n + 1} = lim_{n \to \infty} f (y_{n}) = f (lim_{n \to \infty} y_{n}) = f (y)

as desired. To show that it is unique, consider two sequences $a_{n}, b_{n}$ where $a_{1}, b_{1} \in R$ and $a_{n + 1} = f (a_{n})$ and $b_{n + 1} = f (b_{n})$ . By the Limit Laws (Algebraic Limit Theorem) then:

lim_{n \to \infty} (b_{n} - a_{n}) \leq lim_{n \to \infty} | b_{n} - a_{n} | \leq lim_{n \to \infty} c^{n - 1} | b_{1} - a_{1} | = 0

since $0 < c < 1$ . Therefore $lim_{n \to \infty} b_{n} = lim_{n \to \infty} a_{n}$ so $a = b$ . This implies that no matter what our starting choice is, that it will reach the same value.

d. See (c).

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