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We know that
$f(x + 19) \leq f(x) + 19$
and
$f(x + 94) \geq f(x) + 94$
$\forall x \in \mathbb{R} $
And I have to prove that $f(x + 1) = f(x) + 1$.
I know that this question has been asked already, but I couldn't understand how to use the hints that were given there. I have got the inequalities to this point:
$\frac{f(x+19)-f(x)}{19} \leq 1$

$\frac{f(x+94)-f(x)}{94} \geq 1$
But I have not idea how to proceed. Any help would be greatly appreciated.

harold jefferson
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2 Answers2

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The conclusion holds if $$f(x+a)\le f(x)+a,\qquad f(x+b)\ge f(x)+b,$$ and $a,b$ are relatively prime natural numbers. Indeed there are $n,n'$ and $m,m'$ such that $$am=bn+1, \qquad bm'=an'+1$$ Let $k=am.$ Then $$ f(x+k)\le f(x)+k,\qquad f(x+k-1)\ge f(x)+k-1.$$ Therefore $ f(x+k)\le f(x+k-1)+1$ for any $x,$ and equivalently $$f(x+1)\le f(x)+1\qquad x\in \mathbb{R}$$

Similarly interchanging the roles of $a$ and $b$ we get $$f(x+1)\ge f(x)+1\qquad x\in \mathbb{R}$$

Ryszard Szwarc
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You have those inequalities given for all real $x$. In particular for $y + 19$ for any $y ∈ ℝ$. But since we change names, also for $x + 19$. So: $$ \begin{align} f(x + 19) &\leq f(x) + 19 \\ ⇒ f(x) & \geq f(x+19) - 19 \geq f(x+19 + 19) - 19 - 19 \\ & \geq \dots \geq f(x + 19a) - 19a \end{align} $$ for any $a ∈ ℕ$. When plugging in $x + 1$ in the second condition we have $f(x+95) \geq f(x + 1) + 94$. Can you use those two equations together?

flukx
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