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Solve for $f:\mathbb{R}\to\mathbb{R} \ \ \ $ s.t.

$$f(n)=n \ \ \forall n\in\mathbb{N}$$ $$f^{(n)}(x)\geq0 \ \ \forall n\in\mathbb{N} \ , \ x\in\mathbb{R} $$

Could you please prove that there exists an unique solution: $f(x)=x$ ?

High GPA
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1 Answers1

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Suppose we have $f(n)=n$ for all integer $n$, and $f''(x)\ge0$ for all real $x$. That means $f$ is convex, i.e. $$f(\lambda u+(1-\lambda)v)\le\lambda f(u)+(1-\lambda)f(v)$$ for all real $u,v$ and all real $\lambda\in[0,1]$ (Jensen's inequality). Now take an $x\in(n-1,n).$ Since $$x=(n-x)(n-1)+(x-n+1)n,$$ Jensen's inequality with $\lambda=n-x,$ $u=n-1$ and $v=n$ gives $f(x)\le x.$ Using the same inequality for $\lambda=1/(n+1-x),$ $u=x$ and $v=n+1$ gives $$f(n)=n\le\lambda f(x)+(1-\lambda)(n+1),$$ so $f(x)\ge x$.

  • But $f(n)=n$ was only given for $n \in \mathbb N.$ – zhw. Jun 11 '17 at 20:31
  • @ProfessorVector Thank you for making me understand this. Your first inequity is intuitive and easy to see. But how did you come up with the second (brilliant) inequity which is more abstract and hard to reach? – High GPA Jun 11 '17 at 22:40
  • @zhw. I think Vector's proof is complete and $n\in N$ is considered? – High GPA Jun 11 '17 at 22:45
  • Actually this doesn't answer the OP's question. It is given only that $f(n)=n, n=1,2,3,\dots ,$ not for all integers $n.$ – zhw. Jun 14 '17 at 21:52
  • @zhw The OP explicitly said it does, but since you know better... In that case, the above $f(x)=x$ is still valid for $x\le1$. Let's additionally assume $f'''(x)\ge0$ for all $x$. For some $x<1,$ we have $0=f''(1)=f''(x)+\int^1_xf'''(t),dt\ge f''(x).$ This means $f''(x)\le0$ and thus $f''(x)=0$ for $x<1$, $f'(x)$ must be constant, $=f'(1)$, and thus, we must have $f(x)=x$ for $x<1,$ too. –  Jun 15 '17 at 03:27