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During a course of functional analysis, we gave the following definition:

$$\mathcal{C}^{\infty}_{c}(\Omega)=\mathcal{C}^{\infty}(\Omega)\cap\mathcal{C}_{c}(\Omega)\qquad\Omega\subseteq\mathbb{R}^{d}\hspace{1.5mm}\text{open}$$

That is, the set of smooth function with compact support. We pointed out that such functions need not to be analytical. My question is, how do we know such functions are not also analytical? Given $f\in\mathcal{C}^{\infty}_{c}(\Omega)$, what is $f$ missing in order to be analytical?

Davide Trono
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Analytic means that there is a power series that converges to that function (locally at every point).

The only power series that is identically zero in a region is the zero series.

That is, nonzero analytic functions cannot have compact support (ignoring arbitrary domain restrictions).

obscurans
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    For reference: This is called the 'identity principle' - it holds both for holomorphic functions and for real analytic functions as here. – Jan Bohr Jan 03 '21 at 20:37
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    Not quite correct: analytic means analytic at all points, which means that for all $x\in\Omega$ there is an open set $U\ni x$ such that $f(y)$ coincides with the value of a power series centered at $x$ for all $y\in U$. Compare with $\frac1{1+x^2}$ on $\Bbb R$, which is analytic at all points but no single power series represents it everywhere. This is a relevant technical detail in the economy of a proof. –  Jan 03 '21 at 20:40
  • Thanks, corrected – obscurans Jan 03 '21 at 20:41