Let \(f:\mathbb R\to\mathbb R\) be continuous and suppose \[ f(x)=f(x+1)=f(x+\sqrt2) \] for every \(x\in\mathbb R\). Prove that \(f\) is constant.
Solution
Let \[ H\coloneqq\{m+n\sqrt2:m,n\in\ZZ\}. \]
1. Every element of \(H\) is a period of \(f\):\[ f(x+h)=f(x) \]for all \(x\in\RR\) and \(h\in H\).
The hypotheses say that \(1\) and \(\sqrt2\) are periods. If \(p\) is a period, then so is \(-p\), because replacing \(x\) by \(x-p\) in \[ f(x+p)=f(x) \] gives \(f(x)=f(x-p)\). Integer multiples and sums of periods are again periods by repeated substitution. Hence every \[ m+n\sqrt2 \] with \(m,n\in\ZZ\) is a period.
2. For every positive integer \(N\), there exists\[ h_N\in H \]such that\[ 0<h_N<\frac1N. \]
Consider the \(N+1\) fractional parts \[ \{j\sqrt2\}, \qquad j=0,1,\ldots,N, \] in \([0,1)\). Partition \([0,1)\) into \(N\) half-open intervals of length \(1/N\). By the pigeonhole principle, two distinct fractional parts, say those for \(0\leq i<j\leq N\), lie in the same interval. Therefore their difference has absolute value less than \(1/N\).
Writing \[ j\sqrt2=\lfloor j\sqrt2\rfloor+\{j\sqrt2\} \] and similarly for \(i\), this difference equals \[ (j-i)\sqrt2-m \] for some \(m\in\ZZ\). It is nonzero because \(\sqrt2\) is irrational. Thus \(H\) contains a nonzero element of absolute value less than \(1/N\). Replacing it by its negative if necessary gives the required positive \(h_N\).
3. The subgroup \(H\) is dense in \(\RR\).
Fix \(t\in\RR\). For each \(N\), choose \(h_N\) as in step 2 and let \(q_N\in\ZZ\) be an integer nearest to \(t/h_N\). Then \[ \abs{q_Nh_N-t} \leq \frac{h_N}{2} < \frac{1}{2N}. \] Since \(q_Nh_N\in H\), the sequence \((q_Nh_N)\) consists of elements of \(H\) and converges to \(t\). Thus every real number lies in the closure of \(H\).
4. For arbitrary \(x,y\in\RR\),\[ f(y)=f(x). \]
Apply step 3 to \[ t=y-x. \] There is a sequence \(p_N\in H\) with \[ p_N\longrightarrow y-x. \] Hence \[ x+p_N\longrightarrow y. \] By step 1, \[ f(x+p_N)=f(x) \] for every \(N\). Continuity of \(f\) gives \[ f(y) = \lim_{N\to\infty}f(x+p_N) = f(x). \]
5. Therefore\[ \boxed{f\text{ is constant on }\RR}. \]
Step 4 shows that any two values of \(f\) are equal.
6. Q.E.D.
Step 5 is the required conclusion.