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I have given a functional $l$ on $C_c^\infty(\mathbb{R}^n)$. Now let's assume that for any $p \in \mathbb{R}^n$ we have a neighborhood $V_p$ and a $2\pi$-periodic $C^\infty$-function $u_p$ on $\mathbb{R}^n$, such that

$ \forall \varphi \in C_c^\infty(V_p) $ (compact support in $V_p$)$ \colon \, l(\varphi) = \langle u_p , \varphi \rangle := 1/(2\pi)^n \int u_p \varphi$

So locally the functional is given by $u_p$. If I have overlapping neighborhoods $V_p$ and $V_q$ one can easily conclude that $l = \langle u_p , \cdot \rangle = \langle u_q , \cdot \rangle$ on $C_c^\infty(V_p \cap V_q)$. But since $u_p,u_q$ are not compactly supported on $V_p \cap V_q$ I can not conclude directly $u_p = u_q$ on $V_p \cap V_q$.

Am I right so far? How can I show that $u_p = u_q$ on the overlapping area?

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For every point $x\in V_p\cap V_q$, you can use bump functions $\varphi_n$ (with integral $1$) supported in $(1/n)$-neighborhood of $x$ to conclude that $$u_p(x) = \lim_{n\to\infty} \langle u_p, \varphi_n\rangle = \lim_{n\to\infty} \langle v_p, \varphi_n\rangle = v_p(x)$$ Put another way, $\varphi_n$ converge to the Dirac delta at $x$ in the sense of distributions, and thus, passing to the limit, we get equality of pointwise values at $x$.

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  • $\begingroup$ Ah thanks! Distributions are relatively new to me, so two questions come to my mind: What's with the factor $1/(2\pi)^n$ in front of the integral, don't we need to multiply through? And which theorem assures $u_p(x) = \lim \langle u_p , \varphi_n \rangle$? $\endgroup$ – T'x Jun 15 '16 at 8:18
  • $\begingroup$ The factor is just for convenience of something. The convergence is explained in math.stackexchange.com/a/859776 $\endgroup$ – user147263 Jun 15 '16 at 8:21
  • $\begingroup$ Yes, but doesn't the factor give $1/(2\pi)^n u_p(x)$ in the limit? Thanks for the link :) $\endgroup$ – T'x Jun 15 '16 at 8:27
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In fact there is an easier way to look at the problem: Let $U$ be such an overlapping area.

Consider the pairing $\, C^\infty(U) \times C_c^\infty(U) \to \mathbb{C}\\ \, (u,\phi) \mapsto \langle u,\phi \rangle$

All we need to show is that this is non degenerate: (by contraposition) If $u \neq 0$, then $u$ does not vanish on an open set $V$. Take any bump function $\phi$ with compact support in $V$, then $\langle u,\phi \rangle \neq 0 $.

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