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Let $B^+$ be the open half ball $\{(x,y)\in \mathbb{R}^2 |x^2+y^2<1,y>0\}$. Assume that a function $f\in C^2(B^+)$ is $f=0$ on the $x$-axis such that $f,\partial_{x} f\partial_{y} f,\partial_{x^2} f, \partial_{xy} f,\partial_{x^2} f$ are uniformly continuous (thus they all continuously extends to the boundary).

Set $g(x)\in C(B)$ as $f(x,y)$ if $y \ge0$ and $-f(x,-y)$ is $y<0$ for $(x,y) \in B=\{(x,y)\in \mathbb{R}^2 |x^2+y^2<1\}$. How can we prove that $f\in C^2(B)$?

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What have you done so far? –  Pragabhava Oct 17 '12 at 21:21
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1 Answer

Well, it follows directly from the definition that $g$ and all its partial derivatives are uniformly continuous on the closed ball $\{(x,y) \in \mathbb{R}^2 | x^2+y^2 <= 1, y <= 0\}$. It also follows immediatly that $g \in C^2(\{(x,y) \in \mathbb{R}^2 | x^2+y^2 < 1, y \neq 0\})$. So what remains to shows is that the partial derivatives up to the second order exists at $y=0$.

For the first-order partial derivative $\partial_y g$ you have $g_y(x,0) = \lim_{\epsilon \rightarrow 0^+} \frac{g(x,\epsilon)-g(x,-\epsilon)}{2\epsilon}$. Since $-g(x,-\epsilon)=g(x,\epsilon)=f(x,\epsilon)$ this is the same as $\lim_{\epsilon \rightarrow 0^+} \frac{f(x,\epsilon)}{\epsilon} = f_y(x,0)$. A similar approach should work for the other partial derivatives.

To convince yourself that $f_y(x,0)$ (i.e. the continuous extension of $f_y$) coincides with the one-sided derivative of $f$ at $(x,0)$, write the former out as a double limit $$ f_y(x,0) = \lim_{y \rightarrow 0^+} \lim_{\epsilon \rightarrow 0^+} \frac{f(x,y+\epsilon)-f(x,y)}{\epsilon}. $$ Note that since $f_y$ is bounded, the inner limit converges uniformly with respect to $y$. You may thus swap the limits, whith shows that $$ f_y(x,0) = \lim_{\epsilon \rightarrow 0^+} \frac{f(x,\epsilon)}{\epsilon}. $$

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Thank you for the answer. $f_{y}(x,0)$ is the limit of $f_{y}(x,y)$ as $y$ approaches $0$ (because the derivative does not make sense there), and we want to show that it can be regarded as the $y$-derivative of $f$ there. If I understand correctly, we have $\lim_{\epsilon\to0^+}\frac{g(x,\epsilon)-0}{\epsilon}=\lim_{\epsilon\to0^+} \frac{0-g(x,-\epsilon)}{\epsilon}$, but does this coincide with $f_{y}(x,0)$? –  Pooya Oct 8 '12 at 23:36
    
In other words, $\lim_{y\to 0}f_{y}(x,y)=\lim_{y\to \epsilon^+}\frac{f(x,\epsilon)-0}{\epsilon}$? –  Pooya Oct 8 '12 at 23:39
    
@Pooys This follows from the uniform continuity of $f_y$. I've extended my answer to include this argument. –  fgp Oct 9 '12 at 1:54
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