Timeline for Does function $f(x)=f(2x)$, $f(x)$ - non const, exist? ($f(x)$ - continuous function on real numbers) [closed]
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 28, 2020 at 14:58 | comment | added | Dmitriy Shekhmatov | Thank you! Function really has problem when $x = 0$ | |
| Nov 28, 2020 at 14:55 | vote | accept | Dmitriy Shekhmatov | ||
| Nov 28, 2020 at 12:39 | comment | added | Pietro Majer | After all $(\mathbb{R}_+,\cdot)$ is isomorphic as a topological group to $(\mathbb{R},+)$. So, up to notation, the question is whether there exist nonconstant, continuous, $2$-periodic functions: yes on $\mathbb{R}$, no if one also wants a limit at $-\infty$, that is at $0$ in the multiplicative formulation. | |
| Nov 28, 2020 at 7:33 | history | closed |
Nate Eldredge LSpice Piotr Hajlasz abx GH from MO |
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| Nov 28, 2020 at 6:48 | answer | added | Wlod AA | timeline score: 1 | |
| Nov 28, 2020 at 6:25 | answer | added | Thomas Browning | timeline score: 4 | |
| Nov 28, 2020 at 4:42 | review | Close votes | |||
| Nov 28, 2020 at 7:35 | |||||
| Nov 28, 2020 at 4:41 | comment | added | LSpice | What is the value of your function at $x = 0$? | |
| Nov 28, 2020 at 4:17 | comment | added | Thomas Browning | Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ is continuous and satisfies $f(x)=f(2x)$. Let $\varepsilon>0$ and pick $\delta>0$ such that $\lvert f(x)-f(0)\rvert<\varepsilon$ for all $\lvert x\rvert<\delta$. Then for each $x\in\mathbb{R}$, $\lvert f(x)-f(0)\rvert=\lvert f(x/2^k)-f(0)\rvert<\varepsilon$ (where $k$ is chosen large enough so that $\lvert x/2^k\rvert<\delta$). Since this holds for all $x\in\mathbb{R}$ and all $\varepsilon>0$, this forces $f(x)=f(0)$ for all $x\in\mathbb{R}$. | |
| Nov 28, 2020 at 4:03 | review | First posts | |||
| Nov 28, 2020 at 5:38 | |||||
| Nov 28, 2020 at 4:01 | history | asked | Dmitriy Shekhmatov | CC BY-SA 4.0 |