Let $u:\mathbb{R}^2\rightarrow\mathbb{R}$ be a continuous function

Let $\phi$ be a $C^{2,2}$ function such that at $x_0$ such that

  1. $u(x_0)=\phi(x_0)$

  2. $\phi(x) \geq u(x)$ in a neighbourhood of $x_0$

question: does there exist a function $\psi(x) \geq u(x)$ such that all the derivatives of

i.e. $\partial_{n_1,n_2}\phi(x_0)=\partial_{n_1,n_2}\psi(x_0)$ for all $n_1\leq 2$ and $n_2\leq 2$ and $\psi(x)\geq u(x)$ for all $\in\mathbb{R^2}$?

My thoughts: There is an $\epsilon>0$ such that $\phi(x)\geq u(x)$ for $x\in B(x_0,\epsilon/2)$, where $B$ is a closed ball. Let $\psi(x)=\phi(x)$ for $x\in B(x_0,\epsilon/2)$, then

choose a function has the required regularity which grows much much fast than $u$, we can do this because u is continuous, so we can control how fast it grows

How do I do this rigorously? Which theorem can I invoke to say there exists a smooth function which dominates a continuous one?

Here is also a nice observation: we can add $|x-x_0|^4$ onto $x_0$ to make $\phi(x)+|x-x_0|^4>u(x)$ to be strict on $B(x_0,\epsilon/2)$, this prevents the problem on the boundary and the derivatives do not change at $x_0$.

I am interested in $\mathbb{R}^n$, but I get feeling this is a straightforward extention.

Motivation: this is something related to viscosity solutions - a way of defining solutions for PDEs, for which there are no classical solutions. The trick is to compare it against 'test functions' which agree with the function locally. Indeed, the conditions I give is the definition for a supersolution. $\phi$ is a test function which dominated $u$ locally. However, it seems we can replace such test functions by test functions which dominates $u$ globally. I am trying to show this is the case. I cannot find any literature on this

  • $\begingroup$ Should your first $\psi$ be a $\phi$? $\endgroup$ – TonyK Jan 14 '14 at 19:36
  • $\begingroup$ @TonyK yes thank you. $\endgroup$ – Lost1 Jan 14 '14 at 19:37
  • $\begingroup$ What is $\alpha$? And your question ("does there exist...") is an unfinished sentence. $\endgroup$ – TonyK Jan 14 '14 at 19:38
  • $\begingroup$ @TonyK not meant to be there $\endgroup$ – Lost1 Jan 14 '14 at 19:39

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