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$$(1) F:[0,1] \rightarrow\, [0,1]\,$$

Where a $F$ is a $(C)$ continuous convex function $$\forall t\in [0,1];\forall(x,y)\in[0,1];\,F(tx+(1-t)y)\leq tF(x)+(1-t)\times F(y)$$. where $F$ satisfies $(E)$ . $$(E)\,F(1)\geq 1\,\,, F(0)\leq 0\,\,,\&\, \exists \,\,\text{some third interior fixed point} \,m\in\text{dom}(F);\,F(m)\geq m\,\,\text{where}; 0 \neq m \neq 1$$.

(2) $$F:[0,1] \leftrightarrow\, [0,1]\,$$. Where $F$ is a bi-jective self map ,of the unit interval that satisfies. $(A), (B) \land (SC)$ below.

$$(A) \text{F is 1-auto-diffeo-morphophism of unit interval}$$, Once continuous derivatives of $F$ and its inverse function. $$(A)F(0)=0,\, F(1)=1\land F(m)=m\,\text{where} \,m\in (0,1)$$.

$$(SC)\text{and star convex at 0}: $$. $$(SC)\forall(x\in(0,1)):F(tx)\leq t\times F(x)$$.

Can function (1) $F(x)$ meet with a line segment in three distinct points without $F(x)$ being $F(x)=x$?

Answer no. What about star convex function $(2)$ My answer: $$\forall x\in [0,1];x\geq m):F(x)=x$$

Caveat: If ,yes, Suppose we replace $(SC)$ with midpoint -star convexity at $0$

$$\forall(x)\in [0,1]:F(\frac{1}{2}\times x)\leq \frac{1}{2}\times F(x)$$. where $$m\in \text{Q}\cap [0,1]$$?.

I presume, but do know that the answer is also no here as it is for a strictly monotonic increasing that is continousthat is 2- midpoint convex function with three fixed points.

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    $\begingroup$ In geometric terms, convexity of a function means that, for all $x<y$ the graph $\lbrace (t,f(t)): x\le t\le y\rbrace$ is below the line segment joining $(x,f(x))$ and $(y,f(y))$.Assume that there is $z$ with $f(z)<z$ and apply the above to one of the intervals $[0,z]$ or $[z,1]$ which contains $1/2$. $\endgroup$ – Jochen May 4 '17 at 11:21
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The only possibility is $F(x) = x$. Here is an argument without assuming differentiability. For greater generality, replace $x^* \in (0,1)$ as the third point where $F(x^*) = x^*$ instead of $x^*=1/2$.

By convexity, for any $x$ in $(0,1)$ with $x\ne 1/2$, the incremental ratio $$R(x) = \frac{F(x) - F(1/2)}{x-(1/2)}$$ is increasing, so $R(0) \le R(x) \le R(1)$.

When $F(x) = x$ for $x=0, 1/2, 1$, we have $$R(0) = R(1) = 1$$ which squeezes $R(x)=1$ and implies $F(x) = x$.

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    $\begingroup$ @William Balthes The argument above does not assume differentiability and holds for $F(0.6) = 0.6$ replacing $0.5$ with $0.6$. $\endgroup$ – mlc May 4 '17 at 12:02
  • $\begingroup$ when you edited the post and removed the redundant assumptions; was bi-jectivity redundant already given the convexity,and strict monotone increasingness and the other conditions, (that either differentiability or convexity alone implies continuity and as a result of the conditions surjectivity and thus F bijective as its already injective) or was this because i used $\Leftrightarrow$ $\endgroup$ – William Balthes May 4 '17 at 12:40
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    $\begingroup$ @William Balthes: Strict increasing implies injective. If you add $F(0)=)$ with $F(1)=1$, you have subjectivity as well. $\endgroup$ – mlc May 4 '17 at 12:46
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    $\begingroup$ @William Balthes I fail to get your comment, sorry --- you have assumed increasing in the title of your OP. What are you referring to? $\endgroup$ – mlc Jun 3 '17 at 15:10
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    $\begingroup$ This string of comments is way too long. Please post a new question, perhaps with a reference to the present one. $\endgroup$ – mlc Jun 6 '17 at 9:19

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