Is this theorem true?

Question

If $f(x)+f(y)=f(x+y)$, then:

$f(x)=a x$

where $a$ is a constant.

Is the above statement true? Is there a way of proving it?

The application of this theorem is in the last part of page 52 (second page of the chapter)

Answer 1

If we are considering functions $\mathbb{R} \to \mathbb{R}$, this is known as Cauchy's functional equation.

To summarize the Wikipedia article, if you also require $f$ to satisy any of the below conditions, then $f(x) = ax$ is the only family of solutions:

• $f$ is continuous at at least one point
• $f$ is monotonic on any interval
• $f$ is bounded on any interval
• $f$ is Lebesgue measurable

As this is being used for an application, I would guess that it is being assumed $f$ is continuous at at least one point, implying that $f(x) = ax$.

The wikipedia article also mentions that any other solutins are highly pathological; if you were to graph them, the graph would look like you are just shading the paper. More formally, if you take a disk of any size anywhere on the paper, that disk will contain a point $(x, f(x))$ of the function.

Let me know if anything needs clarification.

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EDIT:

1) I haven't looked at all the proofs, but this seems to be a great resource.

2) There are many inintuitive things in mathematics as I'm sure there are in physics :). One example of a function which is discontinous everywhere is

$$f(x) = \begin{cases}0, \text{ if x is irrational} \\1, \text{ if x is rational}\end{cases}$$

If you were to draw this function it would look like two continous horizontal lines; but of course they're riddled with infinitesimal holes.

To prove this function is discontinous at every point, you have to prove that $\lim_{x \to a} f(x) \not = f(a)$ for any $a$.

Bounded just means that the function is cut off height-wise; for example $\sin x$ is bounded because it doesn't go higher/lower than $\pm1$, which $x^2$ and $x^3$ are not bounded.

Lebesgue measurable is a bit more complicated; the Lebesgue measure, $m$, of a subset of $\mathbb{R}$ is its length. For example the measure of $[3, 5]$ is $2$; the measure of a single point is $0$. But stranger things happen too; it turns out that the measure of $\mathbb{Q}$, the set of all the rational numbers on the number line, is also $0$.

Now the "$f$ is Lebesgue measurable" condition is very very weak; to find a function which is not Lebesgue measurable, it requires you to find a set which is not Lebesgue measuable, which means a set with no length. For a long time mathematicians thought such a set is impossible, until a guy named Vitali came up with his Vitali set.

Answer 2

Hard to say anything unless you state the sets the function is defined on.