# Cauchy's functional equation without the assumption of continuity

I have been reading the book Functional Equations and How to Solve Them by Christopher G. Small. In chapter 2 the first section about Cauchy's equation $$f(x+y) = f(x)+f(y)$$ has a theorem that states the following

Let $$f : \Bbb R → \Bbb R$$ satisfy Cauchy’s equation. Suppose in addition that there exists some interval $$[c,d]$$ of real numbers, > where $$c, such that $$f$$ is bounded from below on $$[c,d]$$. In other words, there exists a real number $$A$$ such that $$f(x) ≥ A$$ for all $$c ≤ x ≤ d$$. Then there exists a real number $$a$$ such that $$f(x) = ax$$ for all real numbers $$x$$.

Here continuity is not assumed. In the exercises, problem 2 is supposed to lead you through the proof of this theorem. However the problem states :

This problem asks the reader to fill in the details of the proof of Theorem 2.4, above. Let $$f :\Bbb R → \Bbb R$$ be a continuous function satisfying Cauchy’s equation. Suppose in addition that there exists some interval $$[c, d]$$ of real numbers, where $$c, such that $$f$$ is bounded from below on $$[c, d]$$. a) Show that $$f(nx) = nf(x)$$ for all real $$x$$. b) Define $$p = d − c$$. Show that $$f$$ is bounded from below on the interval $$[0, p]$$. (However, it need not be bounded below by the same constant as on the interval $$[c, d]$$.) c).......

My confusion here is the fact that the problem says $$f$$ is supposed to be continuous unlike the actual theorem also going by that logic part b of this problem is almost trivial since I just have to say that a continuous function on a closed and bounded interval must be bounded below. But then it feels like the other assumption given in this problem that the function is bounded on $$[c,d]$$ is completely redundant so there must be something I'm missing but I can't seem to find a way to prove part b if continuity was not given.

Sso to boil it down to what's important, my question is how to prove b if $$f$$ was not stated to be continuous but the assumption about boundedness was given.

Assume that $$f : \mathbb R \to \mathbb R$$ is additive, and bounded below by the constant $$A$$ on the non-degenerate interval $$[ c , d ]$$. Then, for $$p = d - c$$ and any $$x \in \mathbb R$$ we have: \begin {align*} & & & 0 \le x \le p \\ \implies & & & c \le x + c \le d \\ \implies & & & f ( x + c ) \ge A \\ \implies & & & f ( x ) + f ( c ) \ge A \\ \implies & & & f ( x ) \ge A - f ( c ) \end {align*} Therefore $$f$$ is bounded below by $$A - f ( c )$$ on $$[ 0 , p ]$$.
You can see that no property of $$f$$ other than additivity and boundedness on $$[ c , d ]$$ is used above.