Continuity & Global Theorems on Intervals

1. The Three-Part Definition of Continuity at a Point

If any of these three conditions fails, $f$ is said to be discontinuous at $c$.

2. The $\varepsilon$-$\delta$ Characterization of Continuity

Notice that unlike the limit definition where $x \ne c$ was required ($0 < |x - c|$), here if $x = c$, $|f(c) - f(c)| = 0 < \varepsilon$ holds trivially.

3. Continuity on an Interval

A function $f$ is continuous on an open interval $(a, b)$ if it is continuous at every $c \in (a, b)$. It is continuous on a closed interval $[a, b]$ if it is continuous on $(a, b)$, right-continuous at $a$ ($\lim_{x \to a^+} f(x) = f(a)$), and left-continuous at $b$ ($\lim_{x \to b^-} f(x) = f(b)$).

1. Removable Discontinuities (Holes)

A point $c$ is a removable discontinuity of $f$ if $\lim_{x \to c} f(x) = L$ exists as a finite real number, but either $f(c)$ is undefined or $f(c) \ne L$.

The discontinuity can be "removed" by redefining $f(c) = L$. Typical example: $f(x) = \frac{x^2 - 1}{x - 1}$ at $c = 1$.

2. Jump (First Kind) Discontinuities

A point $c$ is a jump discontinuity if both one-sided limits exist as finite real numbers, but they are unequal:

The quantity $J = L_2 - L_1$ is called the jump of $f$ at $c$. Classic example: the signum function $\operatorname{sgn}(x)$ at $c = 0$, where $L_1 = -1, L_2 = +1, J = 2$.

3. Essential (Second Kind) Discontinuities

A discontinuity at $c$ is essential if at least one of the one-sided limits fails to exist (either diverging to $\pm\infty$ or oscillating indefinitely):

• Infinite Discontinuity: $f(x) = 1/(x - 2)$ at $c = 2$.
• Oscillatory Discontinuity: $f(x) = \sin(1/x)$ at $c = 0$. As $x \to 0$, $1/x \to \infty$, and the sine function oscillates between $-1$ and $+1$ infinitely often, so neither one-sided limit exists.

1. Algebraic Combinations

Theorem: If $f$ and $g$ are continuous at $c$, then for any scalar $k \in \mathbb{R}$:

2. Continuity of Composite Functions

Proof: Let $\varepsilon > 0$. Since $f$ is continuous at $u_0 = g(c)$, $\exists \eta > 0$ such that $|u - u_0| < \eta \implies |f(u) - f(u_0)| < \varepsilon$. Since $g$ is continuous at $c$, for this $\eta > 0$, $\exists \delta > 0$ such that $|x - c| < \delta \implies |g(x) - g(c)| < \eta$. Substituting $u = g(x)$ establishes $|f(g(x)) - f(g(c))| < \varepsilon \quad \blacksquare$

3. Continuity of Elementary Functions

All polynomials, rational functions, power functions, trigonometric functions, inverse trigonometric functions, exponential functions, logarithmic functions, and hyperbolic functions are continuous on their entire natural domains.

1. The Intermediate Value Theorem (Bolzano-Cauchy)

Geometrically, the graph of a continuous function on $[a, b]$ has no breaks or jumps, so it must cross any horizontal line $y = u$ lying between the endpoint values $y = f(a)$ and $y = f(b)$.

2. Bolzano's Theorem (Existence of Roots)

Corollary: If $f \in C[a, b]$ and $f(a) \cdot f(b) < 0$ (opposite signs at endpoints), then there exists at least one root $c \in (a, b)$ such that $f(c) = 0$.

3. The Bisection Algorithm

Bolzano's theorem provides a constructive numerical method to compute roots to arbitrary precision:

1. Set interval $[a_0, b_0]$ with $f(a_0) f(b_0) < 0$.
2. Evaluate midpoint $m_k = \frac{a_k + b_k}{2}$. If $f(m_k) = 0$, $m_k$ is the exact root.
3. If $f(a_k) f(m_k) < 0$, set $[a_{k+1}, b_{k+1}] = [a_k, m_k]$; otherwise set $[a_{k+1}, b_{k+1}] = [m_k, b_k]$.
4. After $n$ iterations, the error is bounded by $|c - m_n| \le \frac{b_0 - a_0}{2^{n+1}}$.

1. The Extreme Value Theorem (Weierstrass)

2. Necessity of the Hypotheses

Both hypotheses—continuity and a closed bounded interval—are strictly necessary:

• Failure on Open Interval: $f(x) = x$ on $(0, 1)$ is continuous, but attains neither a minimum (infimum is $0 \notin (0, 1)$) nor a maximum (supremum is $1 \notin (0, 1)$).
• Failure on Unbounded Domain: $f(x) = x^2$ on $[0, \infty)$ is continuous, but attains no maximum as $x \to \infty$.
• Failure if Discontinuous: $f(x) = \begin{cases} 1/x, & x \in (0, 1] \\ 0, & x = 0 \end{cases}$ on $[0, 1]$ is defined on a closed bounded interval, but discontinuous at $x = 0$, having no maximum.