Function Satisfying $f(x)=f(2x+1)$

If $f: \mathbb{R} \to \mathbb{R}$ is a continual function and also pleases $f(x)=f(2x+1)$, after that its not to tough to show that $f$ is a constant.

My inquiry is intend $f$ is continual and also it pleases $f(x)=f(2x+1)$, after that can the domain name of $f$ be limited to make sure that $f$ does not continue to be a constant. If of course, after that offer an instance of such a function.

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2019-05-09 11:32:21
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Answers: 2

Let $f$ have value $1$ on $[0,\infty)$ and also value $0$ on $(-\infty,-1]$. This function is not constant (although it is in your area constant), and also pleases $f(x)=f(2x+1)$ whenever $x$ remains in its domain name.

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2019-05-10 08:40:40
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As in the previous proof of $f$ being constant on $\mathbb{R}$, specify $g(x) = f(x-1)$, to make sure that $g(x) = g(2x)$ ; the domain names of $f$ and also $g$ are simply changed variations of each various other.

Absolutely, if the domain name of $g$ is tiny sufficient, claim $[2,3]$, after that $g$ can be any kind of continual function, due to the fact that the domain name has no $x$ and also $2x$ at the very same time. An even more intriguing inquiry is : just how huge can we make the domain name to make sure that $g$ will still not be constant? The response to this is recommended by JDH's answer : if we remove just the solitary factor 0, making the domain name $\mathbb{R} \setminus \\{0\\}$, it is separated right into 2 parts which can individually have constant values.

Just how large can a domain name get on which $g$ is not also in your area constant? Remove a randomly tiny period around $0$. Take any kind of non - constant continual function $h$ which is routine with device duration, and also allow $g(x) = h(\log_2 x)$. After that $g(x) = g(2x)$ for all $x$, and also is continual almost everywhere.

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2019-05-10 08:32:48
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