Is it possible to construct a compactly supported smooth function $\phi\geq 0$ such that $\operatorname{Supp} \phi\subseteq[1/2,2]$ and $\phi(t)+\phi(t/2)=1$ for all $t\in [1,2]$?
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asked Mar 6 at 11:10
- 1$\begingroup$ I read the requirement for $\phi$ that $\phi:\mathbb R\to [0,\infty)$ should be smooth with support in $[1/2,2]$. This excludes the constant function $\phi=1/2$. $\endgroup$Jochen Wengenroth– Jochen Wengenroth2025-03-06 12:09:35 +00:00Commented Mar 6 at 12:09
- $\begingroup$ @Claude. I require the support to be in [1/2,2]. This excludes your trivial example. $\endgroup$A beginner mathmatician– A beginner mathmatician2025-03-06 12:18:24 +00:00Commented Mar 6 at 12:18
- 9$\begingroup$ Consider a positive smooth bump function $f:\mathbb{R}\to\mathbb{R}$ with support $[1/2,0.9]$ and with $\int_{0}^1f=1$. Let $F:\mathbb{R}\to\mathbb{R}$ $F(x)=\int_0^xf(x)$ and define $G:\mathbb{R}\to\mathbb{R}$ by $G(x)=F(x)$ for $x\leq1$ and $G(x)=1-F(x/2)$ for all $x\geq1$. It seems $G$ is what you want (if it makes sense I can make this an answer) $\endgroup$Saúl RM– Saúl RM2025-03-06 12:19:46 +00:00Commented Mar 6 at 12:19
- 1$\begingroup$ Every $C^\infty$ function with support in $[1/2,1]$ has all derivatives vanishing at $1/2$, $\varphi^{(j)}(1/2)=0$, and if is satisfies the equation, $\varphi(1)=0$ and $\varphi^{(j)}(1)=0$ for $j\ge1$. Conversely, every function $\varphi$ on $[1/2,1]$, with $\varphi(1)=1$ and $\varphi^{(j)}(1/2)=\varphi^{(j+1)}(1)=0$ for all $j\ge0$ extends uniquely to a $C^\infty$ solution of the functional equation defining $\phi(x):=1-\phi(x/2)$ for $x\in[1,2]$. $\endgroup$Pietro Majer– Pietro Majer2025-03-06 20:59:37 +00:00Commented Mar 6 at 20:59
- 1$\begingroup$ If it suffices that $f$ is smooth on $(\frac12,2)$, then use $$f(x)=\sin^2 \left(\pi\,\log_4 2\,x\right).$$ $\endgroup$Karl Fabian– Karl Fabian2025-03-07 08:54:15 +00:00Commented Mar 7 at 8:54
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2 Answers
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5 My comment turned answer.
Let $f:\mathbb{R}\to\mathbb{R}$ be a positive, $C^\infty$ bump function with support $[1/2,0.9]$ and $\int_{[0,1]}f(x)dx=1$, let $F:\mathbb{R}\to\mathbb{R};F(x)=\int_0^xf(x)dx$ and let $G:\mathbb{R}\to\mathbb{R}$ be given by $G(x)=F(x)$ for $x\leq1$ and $G(x)=1-F(x/2)$ for $x\geq1$. Then $G$ answers the question.
- 1$\begingroup$ Why the $0.9$ not $1$? $\endgroup$Adayah– Adayah2025-03-07 13:14:26 +00:00Commented Mar 7 at 13:14
- $\begingroup$ @Adayah This ensures smoothness at 1. If the support of $f$ were [1/2,1], then $G$ may have a kink at 1. $\endgroup$quarague– quarague2025-03-07 13:30:51 +00:00Commented Mar 7 at 13:30
- 1$\begingroup$ @quarague It won't have a kink. $\endgroup$Adayah– Adayah2025-03-07 13:37:44 +00:00Commented Mar 7 at 13:37
- 1$\begingroup$ @Adayah that way the answer is a bit shorter (as one doesn't need to check that there is no problem at $1$) $\endgroup$Saúl RM– Saúl RM2025-03-07 14:02:37 +00:00Commented Mar 7 at 14:02
- 2$\begingroup$ Indeed one can use a $C^\infty$ function $f$ with support in $[1/2,1]$, and $\int fdx=1$; it is trivial that joining $F(x)$ and $1-F(x/2)$ one gets a $C^\infty$ function, since all derivatives vanish at 1/2. This represents all solutions (see comment above) $\endgroup$Pietro Majer– Pietro Majer2025-03-07 19:57:22 +00:00Commented Mar 7 at 19:57
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First of all, if we are given a compactly supported smooth function $\psi \geq 0$ with $\operatorname{supp}(\psi) \subseteq [-1, 1]$ and $\forall s \in [-1, 0] : \psi(s) + \psi(s + 1) = 1$, then we may easily construct a compactly supported smooth function $\varphi \geq 0$ with $\operatorname{supp}(\varphi) \subseteq \left[ \frac{1}{2}, 2 \right]$ and $\forall t \in \left[ \frac{1}{2}, 1 \right] : \varphi(t) + \varphi(2t) = 1$: Simply take $\varphi(t) = \psi(\operatorname{log}_2(t))$.
Thus the question reduces to finding a compactly supported smooth function $\psi \geq 0$ with $\operatorname{supp}(\psi) \subseteq [-1, 1]$ and $\forall s \in [-1, 0] : \psi(s) + \psi(s + 1) = 1$.
Now, if we are given a smooth function $0 \leq \eta \leq 1$ with $\eta |_{(-\infty, -1]} = 0$ and $\eta |_{[1, \infty)} = 1$ and $\forall r \in \mathbb{R} : \eta(r) + \eta(-r) = 1$, then we may easily construct a compactly supported smooth function $\psi \geq 0$ with $\operatorname{supp}(\psi) \subseteq [-1, 1]$ and $\forall s \in [-1, 0] : \psi(s) + \psi(s + 1) = 1$: Simply take $$ \psi(s) = \begin{cases} 0, & s \leq -1 \\ \eta(2s + 1), & -1 \leq s \leq 0 \\ \eta(1 - 2s), & 0 \leq s \leq 1 \\0, & 1 \leq s \end{cases} $$
Thus the question reduces to finding a smooth function $0 \leq \eta \leq 1$ with $\eta |_{(-\infty, -1]} = 0$ and $\eta |_{[1, \infty)} = 1$ and $\forall r \in \mathbb{R} : \eta(r) + \eta(-r) = 1$.
