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Find all continuous functions $f:\mathbb R\to \mathbb R$ such that : $$f(\sqrt 2 \, x)=2f(x)$$ $$f(x+1)=f(x)+2x+1$$ for all $x\in \mathbb R$.

Clearly $x^2$ is a valid solution but how can i find other solutions or prove that there is no other solution? And please give me ressources for studying functional equations and some advanced techniques cause the techniques I use are still for a beginner level

lulu
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mohamed raed
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Jan 08 '24 at 12:48

2 Answers2

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To summarize the discussion in the comments:

We note that $x\mapsto x^2$ is a possible solution for $f(x)$.

Working on the theory that this is the only possible solution, we define $$G(x)=f(x)-x^2$$ and note that $$G(\sqrt 2\,x)=2G(x)\quad \&\quad G(x+1)=G(x)$$

We'd like to show that $G(x)$ was identically $0$.

Now, $G(x)$ is continuous and periodic. That means that it attains a global maximum, $M$. Let $G(x_0)=M$. We remark that $M≥0$ since $G(0)=0$. But then $$G(\sqrt 2\,x_0)=2M\implies 2M≤M\implies M=0$$ A similar argument shows that the global minimum is also $0$ and we are done.

lulu
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  • I have already read your last comment, but could you tell me if in your opinion it is possible to prove that $G(x)=0$ for any $x\in\Bbb Q(\sqrt2)$ and, in the case that it is possible, how can we do it? Moreover, what do you think about my answer? Is there an easier way to prove that $G(x)=0$ for any $x\in\Bbb Q,?;$ – Angelo Jan 05 '24 at 11:38
  • @Angelo Yes...my guess (which might well be wrong) is that all of $\mathbb Q(\sqrt 2)$ is in one equivalence class. Maybe it is a whole equivalence class. Anyway, that's what I suggest proving (or disproving). As you noticed, my argument that there were infinitely many equivalence classes was purely set theoretic...I certianly didn't show how to produce them. – lulu Jan 05 '24 at 11:42
  • @Angelo But first, it's necessary to prove that my vague sketch makes sense. Maybe it doesn't. It's not that easy to "meld" two equivalence classes...suppose $x\sim y$ because they differ by an integer and $y\sim z$ because their ratio is a power of $\sqrt 2$. Is it true that $x\sim z$? Maybe not. If that fails, then my counterexample doesn't work. – lulu Jan 05 '24 at 11:45
  • Lulu, please could you try to prove that $G(x)=0$ for any $x\in\Bbb Q(\sqrt2),?;$ I have tried but I have not managed to get it. – Angelo Jan 05 '24 at 11:47
  • I'm focussed on something else at the moment, but I'll come back to this. – lulu Jan 05 '24 at 11:47
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Let $\,G\,$ be the function defined as $\,G(x)=f(x)-x^2$ for all $\,x\in\Bbb R\,.\,$ Without using continuity, we can prove that $\,G(r)=0\,$ for any $\,r\in\Bbb Q\,.$

Let $\,\varphi\,$ be the Euler’s totient function.

For any $\,r\in\Bbb Q\,$ there exist $\,a,d\in\Bbb Z\,,\,$ $b,c\in\Bbb N_0=\Bbb N\cup\{0\}\,$ such that

$r\!=\!\dfrac a{2^b(2c\!+\!1)}\;\;,\qquad$ $2^{\varphi(2c+1)}\!=\!1\!+\!(2c\!+\!1)d\;.$

Since $\,G(2^nx)\!=\!4^nG(x)\,$ and $\,G(x\!+\!m)\!=\!G(x)\,$ for any $\,n,m\in\Bbb Z\,$ and for any $\,x\in\Bbb R\,,\,$ it follows that$$4^{\phi(2c+1)+b}G(r)=G\!\left(2^{\phi(2c+1)+b}r\right)=G\left(2^br[1\!+\!(2c\!+\!1)d]\right)=G\left(2^br\!+\!ad\right)=G\left(2^br\right)=4^bG(r)$$Consequently , $\;G(r)=0\,.$

It means that $\,f(x)=x^2\,$ for any $\,x\in\Bbb Q\,$ even if the function $\,f:\Bbb R\to\Bbb R\,$ is not continuous.

Angelo
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