# Is every monotonic additive function $f \colon \mathbb{R} \to \mathbb{R}$ continuous?

Let a function $f \colon \mathbb{R} \to \mathbb{R}$ have the following two properties:

(1) For all $x_1, x_2 \in \mathbb{R}$ such that $x_1 < x_2$, we have $$f \left( x_1 \right) \leq f \left( x_2 \right).$$

(2) For all $x_1, x_2 \in \mathbb{R}$, we have $$f \left( x_1 + x_2 \right) = f \left( x_1 \right) + f \left( x_2 \right).$$

Is such a function $f$ continuous at every point $c$ of $\mathbb{R}$?

My Effort:

We can show that for every rational number $q$, we have $$f(q) = q f(1).$$

As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable.

How to proceed from here?

If we could show that $f$ is continuous at every rational point $c \in \mathbb{Q}$, then $f$ would also be continuous at every point of $\mathbb{R}$. Am I right?

What next?

Context:

Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition.

• Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions. Sep 5, 2018 at 11:25

If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives

$$f(x) \ge f(r_n) = r_n f(1) \to x f(1),$$

so we conclude $$f(x) \ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) \le xf(1)$. Hence, $$f(x) = xf(1) \quad \forall x\in\mathbb R.$$

(edit) I note the relevant portions of the book, and, it seems that $(a)$ you misread the book, which says you only need to prove continuity at one point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $\mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.

As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_n\to y$ then $x_n := y_n - y + x \to x$ and the additive property $(2)$ implies that $$f(y_n) = f(x_n) + f(y) - f(x) \to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.

• Sir, I get your answer Upto $f(x)=xf(1)$ for every $x\in\mathbb{R}$. But from this, how Continuity of $f$ on $\mathbb{R}$ is proved? Where you had used additive property? Please elaborate. Dec 1, 2020 at 3:46
• @AkashPatalwanshi, this was years ago, but $f(x) = cx$ looks like a continuous function to me? Dec 1, 2020 at 4:26
• yes sir. But where "Theorem 5.6.4" and "Exercises 5.2.12" is used? Dec 1, 2020 at 4:40
• @AkashPatalwanshi That is exactly what I said 2 years ago, I think Dec 1, 2020 at 4:46
• @AkashPatalwanshi I am not sure I understand you. Point $(a)$ was specific to the OP (Saaqib). He didn't understand the book, which was outlining the above proof. I still don't know if Saaqib understood me; he didn't accept my answer. You don't need to read $(a)$. Dec 1, 2020 at 4:55

Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n \rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$ Hence as $f$ is continuous at $0$, $\lim_{n \rightarrow \infty } f(x_n-c) = f(0)$. Hence $\lim_{n \rightarrow \infty } f(x_n-c) - f(-c)= \lim_{n \rightarrow \infty } f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.