Proving that a continuous $\frac{1}{\sqrt{n}}-$periodic function is constant

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function with the property $f(x)=f(x+\frac{1}{\sqrt{n}}),\forall x \in \mathbb{R},\forall n \in \mathbb{N}$.Prove that $f$ is constant.

Here is my attempt:

Let $x \in \mathbb{R}$.

Firstly we notice that $\forall n \in \mathbb{N}$ we have $f(-\frac{1}{\sqrt{n}})=f(0)=f(\frac{1}{\sqrt{n}})$

From this we can deduce that $$f(\frac{m}{\sqrt{n}})=f(0),\forall n \in \mathbb{N},\forall m \in \mathbb{Z}$$

Now $x=\frac{x \sqrt{n}}{\sqrt{n}}$ and $x-\frac{1}{\sqrt{n}}=\frac{x \sqrt{n}-1}{\sqrt{n}} \leqslant \frac{[x \sqrt{n}]}{\sqrt{n}} \leqslant x$

The sequence $x_n=\frac{m_n}{\sqrt{n}} \rightarrow x$ where $m_n=[x\sqrt{n}]$

and $f(x_n)=f(0), \forall n \in \mathbb{N}$

From continuity by taking limits we have that $f(x)=f(0)$

Thus $f(x)=f(0),\forall x \in \mathbb{R}$ proving that $f$ is constant.

Is my argument correct?

If not can someone provide me a hint?

Thank you in advance!

The problem statement is overkill. Periods $1$ and $\sqrt 2$ alone suffice — Hagen von Eitzen, Jul 23 '17 at 19:15
The result you proved is actually a special case of the following theorem "If $f$ is a continuous, $T_1,T_2$-periodic function with $T_2 >0$ and $\frac{T_1}{T_2}\notin \Bbb{Q}$, then $f$ is constant". This itself is a special case of the following theorem "Any subgroup of $(\mathbb{R},+)$ is either cyclic, or dense. — Maxime Ramzi, Jul 23 '17 at 19:16
@HagenvonEitzen i found this problem in a final exam in Introduction to Analysis,a course of the 3rd semester in our university. — Marios Gretsas, Jul 23 '17 at 19:24
@Max i've never heard of this theorem.Thank you for mentioning it. — Marios Gretsas, Jul 23 '17 at 19:25
To put the comments of @HagenvonEitzen and @_Max in simple terms: if a continuous function is constant on a dense subset, it is constant everywhere. — Gyro Gearloose, Jul 23 '17 at 19:46
I know that but i want to prove that for any periods $T_1,T_2$ where $T_1/T_2$ is irrational. — Marios Gretsas, Jul 23 '17 at 19:51
"I know that but i want to prove that for any periods T1,T2 where T1/T2 is irrational." that is a different question ;-) — Gyro Gearloose, Jul 23 '17 at 19:55
i pressume that this is a far more difficult question then.. — Marios Gretsas, Jul 23 '17 at 19:58
No, not far more, but more general, deeper, and not included in the original question. And I've been away from math far too long to guaranty I can answer this, in moderate time. — Gyro Gearloose, Jul 23 '17 at 20:01
Here for instance :https://math.stackexchange.com/questions/153058/how-can-we-find-and-categorize-the-subgroups-of-mathbbr . It's a classical theorem of real analysis/group theory — Maxime Ramzi, Jul 23 '17 at 20:04

DanielWainfleet · Accepted Answer · 2017-07-24T05:21:51.170

For every $r>0$ there exist integers $A,B$ with $0<A+B\frac {1}{\sqrt 2}<r.$

Let $x<y.$ Given $\epsilon >0,$ take $r\in (0,y-x)$ small enough that $z\in (y-r,y]\implies |f(z)-f(y)|<\epsilon.$ Take integers $A,B$ with $0<A+B\frac {1}{\sqrt 2}<r.$

Let $c=A+B\frac {1}{\sqrt 2}.$ There exists $n\in \mathbb N$ with $nc\leq y-x<(n+1)c.$ We now have $f(x)=f(x+nc)$ and $x+nc\in (y-r,y],$ so $|f(x)-f(y)|=|f(x+nc)-f(y)|<\epsilon.$

As $|f(x)-f(y)|<\epsilon$ for any $\epsilon >0 $, we have $f(x)=f(y).$

In general if $f:\mathbb R\to \mathbb R$ is continuous and periodic with non-zero periods $P_1,P_2$ where $P_1/P_2$ is irrational, then by the above method, $f$ is constant.

Proving that a continuous $\frac{1}{\sqrt{n}}-$periodic function is constant

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