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Out of curiosity, I was wondering whether if a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ has the property that in any 2 intervals $[a,b]$ and $[c,d]$ on which it is defined, $(a-b = c -d) \implies (f(a)-f(b) = f(c)-f(d))$, then the graph of this function is a straight line.

So far, I have tried reasoning the contrapositive of this- that not being a straight line means there exist intervals of equal length where the function changes a different amount over them. While this seemed more intuitive, I didn't see a rigorous way to prove this.

So is this claim true, and if so, what is the proof?

Princess Mia
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  • If you think that $f$ has to be a straight line, see if you can figure out what the gradient and $y$-intercept are, and see if you can show that all points satisfy the usual formula (you will need to use continuity for some of these points). – jses Jan 28 '24 at 09:47
  • What is the domain of $f$? All real numbers? Or it can be defined on some intervals as you write? Then what does graph being a straight line mean? Is a union of segments a “straight line”? – Aig Jan 28 '24 at 13:15
  • Hint: Can you show that your assumptions imply that, both, $f$ and $-f,,$ are midpoint convex? The rest then follows from the properties of midpoint convex functions. – Kurt G. Jan 28 '24 at 16:38

1 Answers1

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Letting $a=x+y$ and $b=y$, $c=x$, and $d=0$, you have $f(x+y)-f(y)=f(x)-f(0)$, which you can re-write:

$$f(x+y)= f(x)+f(y)-f(0),$$ $$f(x+y)-f(0)= f(x)-f(0)+f(y)-f(0),$$ and so defining $g\colon \mathbb R\to \mathbb R$ by $g(t)= f(t)-f(0)$, we see that $g$ is continuous and satisfies $$g(x+y)= g(x)+g(y)\tag{1}$$ for all $x,y\in\mathbb R$.

In particular, we also have $g(nx)= g(x)+\dots+g(x) = ng(x)$, for each $n\in \mathbb N$, and also $g(-x)+g(x)=g(0)=0$, so $g(zx)=zg(x)$ for all $z\in \mathbb Z$.

Therefore for any rational $\lambda= \frac{p}{q}\in \mathbb Q$, we have $$q g(\lambda x)= g(px) = pg(x),$$ giving $$ g(\lambda x)=\lambda g(x), \tag{2}$$ and since (2) holds for every rational $\lambda$, and rationals are dense in $\mathbb R$, (2) holds for all $\lambda\in \mathbb R$, by continuity of $g$.

Let $m=g(1)$, and $b=f(0)$. From (2), we have $g(x)=x g(1)=mx$, thus by definition of $g$ we have $$f(x)= g(x)+f(0)=mx + b.$$

M W
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