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1.(Riesz-Dunford Functional Calculus)

Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ simultaneously in every neighborhood of zero which is impossible. So the answer to the first problem is: No, in general there won't be an extension continuously differentiable in a whole neighborhood.

2.(Manifolds with Boundary)

Besides this still not resolves the second problem as one can choose the following extension:

1.(Riesz-Dunford Functional Calculus)

Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ simultaneously in every neighborhood of zero which is impossible. So the answer to the first problem is: No, in general there won't be an extension continuously differentiable in a whole neighborhood.

2.(Manifolds with Boundary)

Besides this example still doesn't resolve(!) the second problem as one can choose the following extension:

Consider the function: $f:\{z\in\mathbb{C}:z\text{ lies either on the axes or on the diagonals}\}\to\mathbb{C}:z\mapsto|z|^2$

Then for any point in its domain there exists an extension that is differentiable there: $$z=0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=|z|^2$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z>0,f_E(z):=|z|^2\text{ for }\Re z\leq 0$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z<0,f_E(z):=|z|^2\text{ for }\Re z\geq 0$$

What is left to prove is that there is no extension that is simultaneously differentiable for any point:

1.(Riesz-Dunford Functional Calculus)

Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ simultaneously in every neighborhood of zero which is impossible. So the answer to the first problem is: No, in general there won't be an extension continuously differentiable in a whole neighborhood.

2.(Manifolds with Boundary)

Besides this still not resolves the second problem as one can choose the following extension:

Consider the function: $f:\{z\in\mathbb{C}:z\text{ lies either on the axes or on the diagonals}\}\to\mathbb{C}:z\mapsto|z|^2$

Then for any point in its domain there exists an extension that is differentiable there: $$z=0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=|z|^2$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z>0,f_E(z):=|z|^2\text{ for }\Re z\leq 0$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z<0,f_E(z):=|z|^2\text{ for }\Re z\geq 0$$

What is left to prove is that there is no extension that is simultaneously differentiable for any point...

Consider the function: $f:\{z\in\mathbb{C}:z\text{ lies either on the axes or on the diagonals}\}\to\mathbb{C}:z\mapsto|z|^2$

Then for any point in its domain there exists an extension that is differentiable there: $$z=0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=|z|^2$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z>0,f_E(z):=|z|^2\text{ for }\Re z\leq 0$$ $$z> 0:\quad f_E:\mathbb{C}\to\mathbb{C}:f_E(z):=z^2\text{ for }\Re z<0,f_E(z):=|z|^2\text{ for }\Re z\geq 0$$

What is left to prove is that there is no extension that is simultaneously differentiable for any point: