Continuity

Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$. We say $\lim_{x \to a} f(x) = L$ if for Every $\varepsilon > 0$There exists $\delta > 0$ such that

$0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \varepsilon$

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. In epsilon-delta form: For every $\varepsilon > 0$There exists $\delta > 0$ such that

$|x - a| \lt \delta \implies |f(x) - f(a)| \lt \varepsilon$

Remark. A function is continuous on a set $E$ if it is continuous at every point of $E$. A function is globally continuous (or “continuous”) if it is continuous on its entire domain.

Definition. $f$ is discontinuous at $a$ if it is not continuous at $a$. Discontinuities are Classified as:

• Removable: $\lim_{x \to a} f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined).
• Jump: $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist but are unequal.
• Essential (or infinite/oscillatory): At least one one-sided limit does not exist.

Proposition 4.3. Polynomials are continuous on $\mathbb{R}$. Rational functions $p(x)/q(x)$ are Continuous wherever $q(x) \neq 0$. The functions $\sin x$, $\cos x$, $e^x$, $\ln x$ are continuous On their domains.

Theorem 4.1 (Algebra of Continuous Functions). If $f$ and $g$ are continuous at $a$Then $f+g$ $f-g$, $fg$And (where defined) $f/g$ are continuous at $a$.

Theorem 4.2. Compositions of continuous functions are continuous: if $f$ is continuous at $a$ and $g$ is continuous at $f(a)$Then $g \circ f$ is continuous at $a$.

The epsilon-delta definitions can be reformulated in terms of sequences, which is often more Convenient for …/1-number-and-algebra/3_proof-and-logics.

Proposition 4.2a (Sequential Criterion for Limits). $\lim_{x \to c} f(x) = L$ if and only if For every sequence $(x_n)$ with $x_n \to c$ and $x_n \neq c$ for all $n$We have $f(x_n) \to L$.

Proof. ($\Rightarrow$) Let $\varepsilon > 0$. Choose $\delta > 0$ from the $\varepsilon$-$\delta$ definition. Since $x_n \to c$There exists $N$ with $|x_n - c| \lt \delta$ for $n \geq N$. Then $|f(x_n) - L| \lt \varepsilon$ for $n \geq N$.

($\Leftarrow$) Suppose the $\varepsilon$-$\delta$ condition fails. Then there exists $\varepsilon > 0$ such That for every $n \in \mathbb{N}$There exists $x_n$ with $0 \lt |x_n - c| \lt 1/n$ but $|f(x_n) - L| \geq \varepsilon$. Then $x_n \to c$ but $f(x_n) \not\to L$Contradicting the hypothesis. $\blacksquare$

Corollary 4.2b. $f$ is continuous at $c$ if and only if for every sequence $(x_n)$ with $x_n \to c$ We have $f(x_n) \to f(c)$.

This is especially useful for proving that a function is not continuous: find one sequence Converging to $c$ whose image does not converge to $f(c)$.

Theorem 4.3 (IVT). If $f : [a,b] \to \mathbb{R}$ is continuous and $f(a) \lt y \lt f(b)$ (or $f(b) \lt y \lt f(a)$), then there exists $c \in (a,b)$ such that $f(c) = y$.

Proof. Assume $f(a) \lt y \lt f(b)$. Let $S = \{x \in [a,b] : f(x) \lt y\}$. Since $a \in S$ $S$ is non-empty and bounded above by $b$. Let $c = \sup(S)$. We show $f(c) = y$.

If $f(c) \lt y$Then by continuity at $c$There exists $\delta > 0$ such that $f(x) \lt y$ for $x \in (c - \delta, c + \delta)$. But then $c + \delta/2 \in S$Contradicting that $c = \sup(S)$.

If $f(c) > y$Then by continuity, there exists $\delta > 0$ such that $f(x) > y$ for $x \in (c - \delta, c + \delta)$. But then $c - \delta/2$ is an upper bound for $S$Contradicting That $c = \sup(S)$.

Therefore $f(c) = y$. $\blacksquare$

Alternative …/1-number-and-algebra/3_proof-and-logic (bisection). Set $a_0 = a$, $b_0 = b$. Given $[a_n, b_n]$ with $f(a_n) \lt y \lt f(b_n)$ Let $m_n = (a_n + b_n)/2$. If $f(m_n) \geq y$Set $a_{n+1} = a_n$, $b_{n+1} = m_n$. If $f(m_n) \lt y$ Set $a_{n+1} = m_n$, $b_{n+1} = b_n$. Either way, $f(a_n) \lt y \leq f(b_n)$ and $b_n - a_n = (b-a)/2^n \to 0$. By the nested interval property, $a_n \to c$ and $b_n \to c$. By continuity, $f(c) = \lim f(a_n) \leq y$ And $f(c) = \lim f(b_n) \geq y$So $f(c) = y$. $\blacksquare$

Theorem 4.4 (EVT). If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and Minimum on $[a,b]$: there exist $c_1, c_2 \in [a,b]$ such that $f(c_1) \leq f(x) \leq f(c_2)$ for all $x \in [a,b]$.

Proof. We first show $f$ is bounded. Suppose not; then for each $n \in \mathbb{N}$There exists $x_n \in [a,b]$ with $|f(x_n)| > n$. By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$. By continuity, $f(x_{n_k}) \to f(c)$So $(f(x_{n_k}))$ is bounded. But $|f(x_{n_k})| > n_k \to \infty$A contradiction.

Now we show $f$ attains its supremum. Let $M = \sup\{f(x) : x \in [a,b]\}$. For each $n$Choose $x_n \in [a,b]$ with $f(x_n) > M - 1/n$. By Bolzano-Weierstrass, $(x_n)$ has a subsequence $x_{n_k} \to c \in [a,b]$. By continuity, $f(c) = \lim f(x_{n_k})$. Since $M - 1/n_k \lt f(x_{n_k}) \leq M$ for all $k$The squeeze theorem gives $f(c) = M$. The argument for the infimum is similar (consider $-f$). $\blacksquare$

Definition. $f$ is uniformly continuous on $D$ if for every $\varepsilon > 0$There exists $\delta > 0$ such that for all $x, y \in D$:

$|x - y| \lt \delta \implies |f(x) - f(y)| \lt \varepsilon$

The key distinction: for ordinary continuity, $\delta$ may depend on both $\varepsilon$ and the point $a$; for uniform continuity, $\delta$ depends only on $\varepsilon$.

Theorem 4.5 (Heine-Cantor). If $f : [a,b] \to \mathbb{R}$ is continuous on the closed, bounded Interval $[a,b]$Then $f$ is uniformly continuous on $[a,b]$.

Proof. Suppose $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\varepsilon > 0$ such that for every $n \in \mathbb{N}$There exist $x_n, y_n \in [a,b]$ with $|x_n - y_n| \lt 1/n$ but $|f(x_n) - f(y_n)| \geq \varepsilon$.

By the Bolzano-Weierstrass theorem, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$. Since $|x_{n_k} - y_{n_k}| \lt 1/n_k \to 0$We have $y_{n_k} \to c$ as well.

By continuity of $f$ at $c$: there exists $\delta > 0$ such that $|x - c| \lt \delta$ implies $|f(x) - f(c)| \lt \varepsilon/2$. For $k$ sufficiently large, $|x_{n_k} - c| \lt \delta$ and $|y_{n_k} - c| \lt \delta$So:

$|f(x_{n_k}) - f(y_{n_k})| \leq |f(x_{n_k}) - f(c)| + |f(y_{n_k}) - f(c)| \lt \varepsilon/2 + \varepsilon/2 = \varepsilon$

Contradicting $|f(x_{n_k}) - f(y_{n_k})| \geq \varepsilon$. $\blacksquare$

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

Solution. For $x, y \geq 0$: $|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq |x - y|^{1/2}$.

Given $\varepsilon > 0$Choose $\delta = \varepsilon^2$. Then $|x - y| \lt \delta$ implies $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x-y|} \lt \sqrt{\delta} = \varepsilon$. Since $\delta$ depends Only on $\varepsilon$The continuity is uniform. $\blacksquare$

:::caution Common Pitfall Continuity on $(a, b)$ does not imply uniform continuity. The function $f(x) = 1/x$ on $(0, 1)$ is Continuous but not uniformly continuous. The Heine-Cantor theorem requires a closed and bounded Interval. Also, a function can be uniformly continuous on an unbounded domain (e.g., $f(x) = \sqrt{x}$ On $[0, \infty)$) --- boundedness of the domain is sufficient but not necessary.

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