# smooth extensions with unique critical point

Let $f:B^{2}\rightarrow\mathbb{R}$ be a continuous function on the unit disk $B^{2}$ which is smooth in $B^{2}\backslash\{0\}$ and has no critical points there. May we find a smooth function $g:B^{2}\rightarrow\mathbb{R}$ such that $g(x)=f(x)$ for $\left\Vert x\right\Vert \geq1/2$ and $g$ has $\leq1$ critical points (possibly degenerated) in $B^{2}$?

References are welcome. I need this technical proposition for the estimation of the critical points set of a smooth function with given boundary conditions. Same question may be asked in the $n$-dimensional case.

• What exactly do you mean by 'possibly degenerate'? – Thomas Apr 22 '14 at 5:54
• And: have you heard about degree theory? – Thomas Apr 22 '14 at 6:04
• @Thomas: It means that the unique critical point of g(x) may be degenerate, i.e. the Hessian may equal 0 there. – t22 Apr 22 '14 at 6:14
• Degree theory says that the index of the critical set equals some number k, it doesn't claim that we have |k| geometrically different critical points. It would be the case if they were all non-degenerate. – t22 Apr 22 '14 at 6:22
• If $f(x) = |x|$, then it is continuous and smooth away from $0$. But you cannot get a smooth extension. If your $f$ does satisfies some elliptic PDE, then there might be some smooth extensions. – user99914 Apr 22 '14 at 6:35