Continuity of Functions and Theorems About Their Properties

What Does “Continuous Function” Mean?

Intuitively, a continuous function is one that can be drawn without lifting your pen from the paper. The formal definition translates this intuition into strict language.

A function $f$ is called continuous at point $a$ if:

$f$ is defined at $a$
The limit $\lim_{x \to a} f(x)$ exists
$\lim_{x \to a} f(x) = f(a)$

All three conditions are important. Violation of any of them yields a discontinuity. The function $f(x) = \sin(x)/x$ for $x \neq 0$ and $f(0) = 1$ is continuous everywhere. But if we set $f(0) = 0$, then at zero there is a first-kind discontinuity (removable).

Classification of Discontinuities

Removable discontinuity: the limit exists, but is not equal to the function value. It is enough to redefine the function at this point.

First-kind discontinuity: both one-sided limits are finite, but not equal to each other (jump).

Second-kind discontinuity: at least one of the one-sided limits is infinite or does not exist.

Properties of Continuous Functions

The sum, difference, product, and quotient (where the denominator is nonzero) of continuous functions are continuous. The composition of continuous functions is continuous.

This means: all elementary functions (polynomials, rational, trigonometric, exponential, logarithmic) are continuous on their domains of definition.

Weierstrass Theorem

If $f$ is continuous on the closed interval $[a, b]$, then it attains its maximum and minimum on this interval.

Proof: the image of $[a, b]$ is a bounded set (a continuous function maps bounded closed sets into bounded closed sets). Therefore, it has a supremum $M$. We need to show that $M$ is attained—that follows from closedness of the image.

Why is closedness of the interval important? The function $f(x) = x$ on the open interval $(0,1)$ does not attain $\sup = 1$ and $\inf = 0$. Closedness is not just a technical condition.

Bolzano Theorem (Intermediate Value Theorem)

If $f$ is continuous on $[a, b]$ and $f(a) \cdot f(b) < 0$ (the function has different signs at the endpoints), then there exists $c \in (a, b)$ such that $f(c) = 0$.

More generally: if $f(a) = A$ and $f(b) = B$, then $f$ takes all values between $A$ and $B$.

Proof by bisection. Consider the midpoint $c = (a + b)/2$. If $f(c) = 0$—done. Otherwise, the sign of $f(c)$ coincides with one of $f(a), f(b)$—choose the half where the signs are different. Iterate. Obtain a convergent sequence of intervals—their limit is the root.

Consequence: the equation $x^5 - 3x + 1 = 0$ has a root on $(0, 1)$, because $f(0) = 1 > 0$, $f(1) = -1 < 0$. This is an existential statement—we know that the root exists without finding it explicitly.

Uniform Continuity

A function $f$ is uniformly continuous on set $E$ if for any $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x, y \in E$, with $|x - y| < \delta$, we have $|f(x) - f(y)| < \varepsilon$.

Difference from ordinary continuity: $\delta$ does not depend on $x$—the same $\delta$ works for the entire set.

Cantor’s Theorem: A continuous function on a closed interval is uniformly continuous.

This is important for integration theory: uniform continuity guarantees Riemann integrability.

Theorem on Preservation of Sign

If $f$ is continuous at point $a$ and $f(a) > 0$, then there exists a neighborhood of point $a$ where $f > 0$.

This is “stability” of the sign property: small perturbations of the argument do not change the sign of the function. The theorem is used in proofs of mean value theorems.

Continuity in Economics and Engineering

Continuity of a function is not just a mathematical property; it is a guarantee of predictable behavior. In economics, the demand function is considered continuous: a small change in price leads to a small change in demand—without sharp jumps. In engineering, the transfer function of a regulator must be continuous so that the system does not give instantaneous failures.

Bolzano’s intermediate value theorem explains why an economist can claim: “Somewhere in the interval between these prices, demand equals supply”—market equilibrium exists if the demand and supply functions are continuous and have suitable boundary values. The theorem does not say where exactly—but guarantees existence.

Reflection question: Can a continuous function map the open interval $(0, 1)$ into the closed interval $[0, 1]$? Give an example or prove impossibility.

Uniform Continuity and Cantor’s Theorem

A continuous function on a closed bounded interval is necessarily uniformly continuous (Cantor’s theorem). The difference: with ordinary continuity, $\delta$ may depend on $x$; with uniform continuity, $\delta$ depends only on $\varepsilon$, not on the point. Example of contrast: $f(x) = 1/x$ is continuous on $(0, 1]$, but not uniformly—when $x \to 0$, arbitrarily small $\delta$ yield large $\Delta f$. On $[0.1, 1]$—uniformly continuous (Cantor). Consequence: a uniformly continuous function can be extended to a continuous one on the closure. In numerical analysis, this means that error in the argument is controllably limiting the error in the function value—basis of numerical stability.