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I have asked this question on MathStackExchange. My question: is there any non-constant real analytic function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ such that, $$\nabla f(x_0)=0 \Rightarrow \nabla^2 f(x_0)=0$$ and $$f(x+m)=f(x),\quad\forall m\in\mathbb{Z}^n,x\in\mathbb{R}^n.$$

I could find a smooth function with the given property. For example, let $$ f(x)=(2-x^4)\exp(1/(x^4-1))\quad x\in[-1,1],$$ then copy and translate it to fill the whole $\mathbb{R}$. I guess that existence of such functions is impossible, but I don't know how to use the analyticity to prove it

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    $\begingroup$ $\cos^4(\frac\pi 2\sin^4 x)$ $\endgroup$
    – fedja
    Commented Mar 16, 2023 at 3:26
  • $\begingroup$ @fedja Thanks for the example! $\endgroup$
    – Jianxing
    Commented Mar 16, 2023 at 8:47

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