Question

Let \(f(x+y)=f(x)+f(y)\) for all \(x\) and \(y\) and suppose that \(f\) is continuous at \(x=0\). (a) Prove that \(f\) is continuous everywhere. (b) Prove that there is a constant \(m\) such that \(f(t)=m t\) for all \(t\)

Short answer

(a) \(f\) is continuous everywhere; (b) \(f(t) = mt\) for some constant \(m\).

Step-by-step solution

Verify the given functional equation

We begin with the given functional equation \(f(x + y) = f(x) + f(y)\). This is the definition of a function being additive.

Use the continuity at 0

We know that the function \(f\) is continuous at \(x = 0\). By the definition of continuity, we have \(\lim_{x \to 0} f(x) = f(0)\).

Show continuity at an arbitrary point

To show continuity at any arbitrary point \(c\), consider \(\lim_{h \to 0} f(c + h) = \lim_{h \to 0} [f(c) + f(h)] = f(c) + \lim_{h \to 0} f(h)\). Since \(f\) is continuous at 0, we have \(\lim_{h \to 0} f(h) = f(0)\). Thus, \(\lim_{h \to 0} f(c + h) = f(c) + f(0) = f(c)\). Hence, \(f\) is continuous at \(c\).

Prove the existence of a constant \(m\)

Consider \(f(y) = f\left(\frac{1}{n} \cdot n y\right) = n \cdot f\left(\frac{y}{n}\right)\). Let \(y = 1\), then \(f(1) = n \cdot f\left(\frac{1}{n}\right)\). As \(f(x+y)=f(x)+f(y)\), it implies \(f(1) = f\left(n \cdot \frac{1}{n}\right) = n \cdot f\left(\frac{1}{n}\right)\). Thus \(f(x)=f(1) \cdot \frac{x}{1}\), so \(f(x) = f(1)x\). Hence \(m = f(1)\) and \(f(x) = mx\).

Conclusion of the function form

We have shown that for all \(x\), there exists a constant \(m = f(1)\) such that \(f(x) = mx\). Therefore, \(f(t) = mt\) for all \(t\), where \(m\) is constant for the function.

Key concepts

Continuity

Continuity is a fundamental concept when studying functions, particularly in calculus and real analysis. A function, say \(f(x)\), is continuous at a point \(x = a\) if small changes in \(x\) around \(a\) result in small changes in \(f(x)\). In mathematical terms, \(f(x)\) is continuous at \(x = a\) if \(\lim_{x \to a} f(x) = f(a)\). This means that as \(x\) gets closer to \(a\), \(f(x)\) approaches \(f(a)\).
One of the powerful implications of continuity is that it ensures no abrupt jumps or gaps in the function's behavior at the point of continuity. In the given exercise, it's crucial to show that a function defined by a specific rule is continuous, not just at one point but everywhere. We start with knowing \(f\) is continuous at 0, and use this to demonstrate that it's continuous for all \(x\). This concept is essential because if \(f\) were not continuous everywhere, applying the formula \(f(x + y) = f(x) + f(y)\) systematically might lead to inconsistencies or undefined values.

Additive Function

An additive function is a type of function that satisfies the equation \(f(x + y) = f(x) + f(y)\). This property is known as additivity. It simplifies many operations by allowing a sum of inputs to translate into a sum of outputs. The concept is not limited to linear functions, but in many cases, especially under conditions like continuity and well-defined domains, additive functions often exhibit linear behaviors.
The exercise begins by verifying the property of additivity, which then becomes a stepping stone to establish further properties of the function, such as its linearity and continuity. Understanding additivity allows us to further decompose functions into simpler parts, making it easier to analyze their properties and behaviors under different mathematical operations.

Constant Function

A constant function is one where the output value of the function does not change regardless of the input. In other words, for a constant function \(f(x) = c\), the same result \(c\) is produced for any input \(x\). This is a specific and simple case of a function where variance in \(x\) results in no change in \(f(x)\).
In the context of the exercise, we derive that there is a constant \(m\) such that \(f(t) = mt\) for any \(t\). Although not a constant function per se, the constant \(m\) defines the slope of the line represented by the function. This linearity stems from the combination of the additive property and continuity, culminating in the form \(f(x) = mx\).
Understanding the concept of constant functions can provide insight into how more complex functions behave, especially when such functions revert to a simpler linear form.

Proof Technique

Proof techniques are essential in mathematics, providing systematic ways to build arguments that are logically sound. In proving mathematical statements, especially in analysis, we frequently rely on properties like continuity, limits, and algebraic identities.
In the original exercise, we used a proof technique that leveraged the continuity at a specific point \(x = 0\) and generalized it to any arbitrary point. This involved substituting into the functional equation and invoking limits through continuity. By decomposing \(c+h\) into the sum \(f(c) + f(h)\), and applying limit definitions, we could demonstrate continuity at any point by rigorously showing \(f(c+h)\) approaches \(f(c)\) as \(h\) approaches zero.
Another part of the technique involved showing the function's form. By methodically breaking down the equation \(f(1) = n \cdot f\left(\frac{1}{n}\right)\), we found that the linearity follows by substituting arguments into the additive function. Thus, proof techniques help transition from specific known values to general truths about functions, ensuring a deep and consistent understanding of their behaviors.