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Let $M\subset \mathbb R^2$ be a halfplane and $X$ be a Banach space.

We say that a function $f:M\rightarrow X$ is a of class $C^\infty$ on $M$ if it is of class $C^\infty$ on $int M$, for each multiindex $\alpha$ and for each $x$ from boundary of $M$ there exists $\lim_{y\rightarrow x \atop y\in int M} D^\alpha f(y)$, and function $D^\alpha f$ defined on $int M$ in usual sense and for $x$ from boundary by $D^\alpha f(x) =\lim_{y\rightarrow x \atop y\in int M} D^\alpha f(y)$ is continuous.

Let $M_1=(-\infty, 0] \times \mathbb R$, $M_2=[0,\infty) \times \mathbb R$.

Let's assume that functions $f:M_1\rightarrow X$, $g: M_2 \rightarrow X$ are of class $C^\infty$.

Is it the function $h:\mathbb R^2 \rightarrow X $ defined by $h(x,y)=f(x,y)$ for $(x,y) \in M_1$ and $h(x,y)=g(x,y)$ for $(x,y)\in M_2$ of clas C^\infty$?

Edit:

I forgot to add that for each $(x,y)$ from the boundary of M we assume $f(x,y)=g(x,y)$ and for each boundary point $(x,y)$ and each multindex $\alpha$ we assume that $D^\alpha f(x,y)=D^\alpha g(x,y)$.

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What about taking $X=\mathbb{R}^2$ and $h(x,y)=(|x|,y)$? –  Mark Grant Oct 13 '12 at 8:09
    
Sorry. I forgot to write that in each boundary point $(x,y)$ $f(x,y)=g(x,y)$ and for each $\alpha$ and $(x,y)$ from boundary $D^\alpha f(x,y)=D^\alpha g(x,y)$. –  user 123 Oct 13 '12 at 10:01
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With your edit, I think the function $h$ is smooth by definition - you are forcing all derivatives of all orders to be continuous. –  Mark Grant Oct 13 '12 at 10:31
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