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I am looking for a sequence of smooth functions $f_n: \mathbb R^2 \rightarrow \mathbb R$such that the

  • 1.) the nodal set of $f_n$, i.e. $N_n:=f_n^{-1}(0)$ converges to the graph of the absolute value function $\{x_2 =\vert x_1 \vert \}.$

  • 2.) In addition, I require that $\vert \nabla f_n \vert_{N_n}\vert=1$ and

  • 3.) $\nabla^2 f_n \nabla f_n \vert_{N_n} = 0.$

I was almost able to construct such a sequence by choosing

$$f_n (x) = \Big(x_2 -\sqrt{x_1^2+ 1/n}\Big) \frac{\sqrt{x_1^2+1/n}}{\sqrt{2x_1^2 + 1/n}}.$$

This one satisfies 1 and 2 but not 3. Any thoughts?

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    $\begingroup$ $\nabla f_n$ is a vector. What does it mean for it to equal 1? $\endgroup$
    – Willie Wong
    Commented May 11, 2021 at 19:29
  • $\begingroup$ @WillieWong sorry, its norm $\endgroup$
    – Dagobert Duck
    Commented May 11, 2021 at 19:47
  • $\begingroup$ @WillieWong I accidentally swapped numerator and denominator. But the derivation of a function satisfying 1 and 2 is easy, you just take an ansatz $f_n(x)=(x_2-\sqrt{x_1^2+1/n})g_n(x)$ and normalize. $\endgroup$
    – Dagobert Duck
    Commented May 11, 2021 at 20:09
  • $\begingroup$ ... and to satisfy 3, you also just need to normalize. To satisfy 2 the normalization gives you what $g_n(x)|_{N_n}$ must be. To satisfy 3 also, you need to choose the normal derivative appropriately. Try the ansatz $$ f_n(x) = (x_2 - \sqrt{x^2_1 + 1/n}) g_n(x_1) + (x_2 - \sqrt{x^2_1 + 1/n})^2 h_n(x_1) $$ (You may also need to cut-off away from $N_n$ to makes sure that the quadratic doesn't generate another branch of $f_n^{-1}(\{0\})$. $\endgroup$
    – Willie Wong
    Commented May 11, 2021 at 20:22
  • $\begingroup$ @WillieWong thank you, that worked out nicely. $\endgroup$
    – Dagobert Duck
    Commented May 11, 2021 at 22:32

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