# Math Help - Continuous & Periodic

1. ## Continuous & Periodic

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and $f(x)=f(x+1)=f(x+\sqrt{2}),$ for all $x \in \mathbb{R}.$ Prove that $f$ is constant.

Source: Berkely Preliminary Exam

2. $f(x)=f(x+1)$

Also, $f(x)=f(x+\sqrt{2})$

If T is periodic with period $1$ and also periodic with period $\sqrt{2}$ then $\sqrt{2}$ must be an integral multiple of $1$ which is impossible.

This means that $f(x)$ is periodic but its period is undefined.

The situation is similar to
$f(x)=c$
where $c$ is constant.Here $f(x)$ is periodic but its period is not defined.

Thus $f(x)$ may be a constant function.

But my solution lacks rigour.

3. Originally Posted by NonCommAlg
Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and $f(x)=f(x+1)=f(x+\sqrt{2}),$ for all $x \in \mathbb{R}.$ Prove that $f$ is constant. Is there a real analogue of Liouville's Theorem? Because $\mathbb{R} \subseteq \mathbb{C}$.

Source: Berkely Preliminary Exam
We can say that $f$ is bounded and uniformly continuous. Maybe we can say that $f$ is a doubly periodic function (e.g. $f(z) = f(z+u) = f(z+v)$). And somehow use Liouville's theorem to show that it is constant? The imaginary parts would be $0$.

Or going from the first idea...if $f$ is bounded, then $|f(x)| \leq M$ for some $M>0$ and for all $x \in \mathbb{R}$. So suppose for contradiction that $f$ was not constant. Then it is monotonic increasing or decreasing. But this contradicts the fact that $f$ is bounded. Hence $f$ is constant.

4. I think I have it.

$f(x)=f(x+1)$

$x+1=x+nT$,where $T$ is the fundamental period or the smallest period and $n$ is any positive integer.

$T=\frac{1}{n}$

Similarly, $f(x)=f(x+\sqrt 2)$ results in

$x+\sqrt 2=x+mT$,where $m$ is a positive integer and $T$ is the fundamental period.

$T=\frac{\sqrt 2}{m}$

$\frac{1}{n}=\frac{\sqrt 2}{m}$

$m=\sqrt 2 n$

which is impossible since m and n are integers.Therefore $T$ is undefined.But since the function repeats itself it can only be constant

5. Originally Posted by pankaj
I think I have it.

$f(x)=f(x+1)$

$x+1=x+nT$,where $T$ is the fundamental period or the smallest period and $n$ is any positive integer.

$T=\frac{1}{n}$
that's not always true. for example consider the function $f(x)=\begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}.$ here we have $f(x+r)=f(x), \ \forall x \in \mathbb{R}, \ \forall r \in \mathbb{Q}.$ of course, $f$ is not continuous!

my point is that i don't see how "continuity" of $f$ was used in your solution!

6. Consider the set $S=\{a+b\sqrt{2}\: : \: a,b \in \mathbb{Z}\}$. This set is dense in $\mathbb{R}$, and $f$ must be equal to $f(0)$ everywhere in $S$. Since $f$ is continuous and $S$ is dense, $f(x)=f(0)$ for all real $x$ and $f$ is constant.

To see that $S$ is dense in $\mathbb{R}$, notice that it is a subgroup of $\mathbb{R}$, hence it suffices to prove that it is dense on some interval; we choose the interval $[0,1]$. Consider $J=\{s-\lfloor s\rfloor\: : \: s\in S\}$; this is a subset of $S \cap [0,1]$. Moreover if $s_1-\lfloor s_1\rfloor = s_2-\lfloor s_2\rfloor$ then $s_1=s_2$ because $\sqrt{2}$ is irrational. Hence there are infinitely many points of $S$ in $[0,1]$, hence two of them are arbitrarily close; but then since $S$ is closed under addition, $S$ is dense in $\mathbb{R}$, Q.E.D.

7. Originally Posted by Bruno J.

Moreover if $s_1-\lfloor s_1\rfloor = s_2-\lfloor s_2\rfloor$ then $s_1=s_2$
this is not true! counter-examples: $s_1=a_1+b\sqrt{2}, \ s_2=a_2+b\sqrt{2}, \ \ a_1 \neq a_2.$

Hence there are infinitely many points of $S$ in $[0,1]$, hence two of them are arbitrarily close
what do you mean by "them"? whatever it means, obviously $S \cap [0,1]=S_1$ having infinitely many points doesn't imply that $S_1$ is dense in $[0,1].$

8. 'S' is dense in R

What does this statement mean

9. Originally Posted by pankaj
'S' is dense in R

What does this statement mean
Dense set - Wikipedia, the free encyclopedia

10. Originally Posted by NonCommAlg
this is not true! counter-examples: $s_1=a_1+b\sqrt{2}, \ s_2=a_2+b\sqrt{2}, \ \ a_1 \neq a_2.$
Oops. What I was thinking is that if $s_1=b_1\sqrt{2}, s_2=b_2\sqrt{2}$ and $s_1-s_2 \in \mathbb{Z}$ then $s_1=s_2$; since there are infinitely many choices of $b$, there are infinitely many points in $S_1$.

what do you mean by "them"? whatever it means, obviously $S \cap [0,1]=S_1$ having infinitely many points doesn't imply that $S_1$ is dense in $[0,1].$
The fact that $S_1$ contains infinitely many points does indeed imply that $S$ is dense in $\mathbb{R}$ : it implies that $S_1$ (and hence $S$) has an accumulation point, and any additive subgroup of $\mathbb{R}$ having an accumulation point is dense in $\mathbb{R}$.

11. Originally Posted by Bruno J.
Oops. What I was thinking is that if $s_1=b_1\sqrt{2}, s_2=b_2\sqrt{2}$ and $s_1-s_2 \in \mathbb{Z}$ then $s_1=s_2$; since there are infinitely many choices of $b$, there are infinitely many points in $S_1$.

The fact that $S_1$ contains infinitely many points does indeed imply that $S$ is dense in $\mathbb{R}$ : it implies that $S_1$ (and hence $S$) has an accumulation point, and any additive subgroup of $\mathbb{R}$ having an accumulation point is dense in $\mathbb{R}$.
Why is $S_1$ compact in $\mathbb{R}$?, or why is it closed? That's what you're using right?, that $S_1$ is sequentially compact

12. Originally Posted by Jose27
Why is $S_1$ compact in $\mathbb{R}$?, or why is it closed? That's what you're using right?, that $S_1$ is sequentially compact
Where do you see that I said $S_1$ is compact or closed? It is neither. I said that $S_1$ is dense in $[0,1]$.

To see this rigorously consider the map

$S\rightarrow [0,1]$
$s\mapsto s-\lfloor s \rfloor$

This map, restricted to the subset of $S$ consisting of elements of the form $b\sqrt{2},\ b \in \mathbb{Z}$, is an injection; moreover when $s \in S$, we have $s-\lfloor s \rfloor \in S$; hence there are infinitely many points of $S$ in the bounded set $[0,1]$. Immediately from the Bolzano-Weierstrass theorem we know that $S$ has an accumulation point. It is then an easy matter to show that any additive subgroup of $\mathbb{R}$ having an accumulation point is dense in $\mathbb{R}$. To see this, suppose you have an open interval $I$ of length $L$ in $\mathbb{R}$. Then by the above there is an element of $S$ with 0<|s|<L; taking all its multiples (which are all elements of $S$) we see that one of them must fall in $I$.

13. Originally Posted by Bruno J.
Where do you see that I said $S_1$ is compact or closed? It is neither. I said that $S_1$ is dense in $[0,1]$.

To see this rigorously consider the map

$S\rightarrow [0,1]$
$s\mapsto \lfloor s \rfloor$

This map, restricted to the subset of $S$ consisting of elements of the form $b\sqrt{2},\ b \in \mathbb{Z}$, is an injection; moreover when $s \in S$, we have $\lfloor s \rfloor \in S$; hence there are infinitely many points of $S$ in the bounded set $[0,1]$. Immediately from the Bolzano-Weierstrass theorem we know that $S$ has an accumulation point. It is then an easy matter to show that any additive subgroup of $\mathbb{R}$ having an accumulation point is dense in $\mathbb{R}$. To see this, suppose you have an open interval $I$ of length $L$ in $\mathbb{R}$. Then by the above there is an element of $S$ with 0<|s|<L; taking all its multiples (which are all elements of $S$) we see that one of them must fall in $I$.
I still don't get it, for example in $I=[0,1]$ there are only two integers $0,1$ and no elements of the form $b\sqrt{2}$ with $b\neq 0$, the element $\sqrt{2} -1 \in [0,1]$ and is the only one that falls in $I$ from the set $A= \{ \sqrt{2} +a : a \in \mathbb{Z} \}$ and the set $B= \{ b\sqrt{2} -1 : b \in \mathbb{Z} \}$ and $S=A+B$, but how do you conclude that there are infinitely many elements from $S$ in $[0,1]$

14. You should read carefully what I wrote.

$\sqrt{2} = 1.41421...$
$2\sqrt{2} = 2.82843...$
$3\sqrt{2} = 4.24264...$
$4\sqrt{2} = 5.65685...$

and the sequence

$\{0.41421...,\: 0.82843...,\: 0.24264...,\: 0.65685...,\: ...\}$

is a sequence of distinct elements of $S$, all in $[0,1]$.

15. I think it is time that solution be revealed

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