# $f(x)\leq f(x+\frac{1}{n} )$ for all $x\in \mathbb{R}$ and $n\geq 1$ . Prove that $f$ is non-decreasing .

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(x)\leq f(x+\frac{1}{n} )$ for all $x\in \mathbb{R}$ and $n\geq 1$ . Prove that $f$ is non-decreasing .

I realy don't have any ideas. My first try was that by archemddian and well ordering property there exists a smallest $n$ such that $n(y-x)\geq 1$ where it was chosen such that $y>x$ . Now if $y=x+\frac{1}{n}$ Then $f(y)=f(x+\frac{1}{n}) >f(x)$ . Now if $y>x+\frac{1}{n}$ i have no idea how to proceed .

So provide a solution . Thank you .

• Is $n$ integer? – kingW3 May 10 '17 at 9:05
• Hint: $\sum_{n} \frac{1}{n}$ diverges to infinity ... Now, consider using the inequality successively for integers betweens 1 and an integer N: $f(x) \leq f(x + \sum_{n} \frac{1}{n})$ – mvggz May 10 '17 at 9:05

Observe that for any $m,n \in \mathbb{N}$, we have $$f(x) \le f(x+\frac1n)\le f(x+\frac2n) \le \cdots f(x+\frac{m}{n})$$ So, $f(x) \le f(x+r)$ for all positive rationals $r$. Take $y>x$. Get a sequence of postive rationals $r_n$ such that $r_n \to (y-x)$. We have $$f(x)\le f(x+r_n) \ \ \forall n \in \mathbb{N}$$ Taking limits and using the fact that $f$ is continuous we have $$f(x) \le \lim_{n\to\infty} f(x+r_n) =f(y)$$