Extension of Pseudodifferential Operator to the real line When defining Pseudodifferential-Operators on manifolds, there is the 
subtlety that one theoretically has to consider all possible charts instead of just a chosen atlas. I am trying to understand how to overcome this difficulty in praxis and thus have come up with a toy problem that I hope will give me some useful insights on the matter.
Suppose we are given a linear and continuous operator $P: C_c^\infty(\mathbb{R}) \rightarrow C^\infty(\mathbb{R})$ with the following properties:


*

*The restriction of $P$ to the interval $(0,1)$ is a (say properly supported) Pseudodifferential-Operator. I.e. there is a symbol $\sigma$, defined on $(0,1)\times\mathbb{R}$ (lying in some symbol class), such that for any $u\in C_c^\infty(0,1)$ we have
$$ Pu(x) = \int_\mathbb{R} e^{ix\xi}\sigma(x,\xi)\hat u(\xi) d\xi \quad \text{for all } \quad 0<x<1$$ 

*Outside of the interval $(0,1)$, $P$ annihilates all functions. I.e.
if $u\in C_c^\infty(\mathbb{R})$ has support disjoint from $(0,1)$, then $Pu=0$. 


Is it true that $P$ is a Pseudodifferential-Operator on all of $\mathbb{R}$?
Of course the symbol of $P$ should vanish outside of $(0,1)$, so I hoped to show that $\sigma$ vanishes in an appropriate manner as $x\rightarrow 0$ and $x\rightarrow 1$, so we could smoothely extend it by $0$. Of course one can compute the value of the symbol at $(x,\xi_0)$ by applying $P$ to $u(y)=e^{ix\xi_0}$, but I don't see how one can use property 2 to get information about the boundary behaviour.
I am thankful for any comment.
 A: Thanks for your comment felipeh, your idea allowed me to make some progress. I first intended to prove the following:
Property 2' There is some $\epsilon > 0$ such that the Schwartz-kernel $k$ of $P$ is supported in $(\epsilon,1-\epsilon)^2$.
While it is easy to show that $\text{supp}k \cap (0,1)^2$ is contained in such a box  and that there is no contribution from 
outside of $[0,1]^2$, I had difficulties to rule out the case that there is some support contained on the boundary of the box $[0,1]^2$. However, for now I'm happy to replace the Annihilation condition with this property, so let's assume it is true.
Lemma Fix $x\in (0,1)$. Let $\zeta_{x}^n \overset{*}{\rightharpoonup} \delta_x$ be an approximation of $1$ centered at $x$ and define the tempered function $\Sigma_x \in \mathcal{S}'(\mathbb{R})$ by $\Sigma_x(\xi) = e^{ix\xi} \sigma(x,\xi)$. Then for any $u\in\mathcal{S}(\mathbb{R})$ with support in $(0,1)$ we have $$\langle k \vert \zeta^n_x \otimes u \rangle \rightarrow \langle \hat \Sigma_x \vert u \rangle $$ 
Proof of Lemma: $\langle k \vert \zeta^n_x \otimes u \rangle = \langle Pu \vert \zeta^n_x \rangle \rightarrow Pu(x) = \int_{\mathbb{R}} e^{ix\xi}\sigma(x,\xi) \hat u(\xi) d\xi = \langle \hat \Sigma_x \vert u \rangle $
Property 2' and the Lemma include the result:
Note that if $0 < x < \epsilon$ or $1-\epsilon < x < 1$, the support of $\zeta^n_x \otimes v $ can be assumed to eventually lie outside of $(\epsilon,1-\epsilon)^2$. Hence $\hat \Sigma_x = 0 $ and consequently $\sigma(x,\xi) = 0 $ for all $\xi\in \mathbb{R}$. This obviously means that we can  extend the symbol by $0$.

Edit (31st August) I've been pondering about the question again and found two things. 
First, it is necessary to have property 2' as opposed to property 2. As a counterexample consider the operator $P$ defined by $Pu(x) = u(1) \phi(x)$, where $\phi$ is some fixed test function with support in $(0,1)$. This satisfies properties 1 and 2, but has a Schwartz-kernel with support $\text{supp}\phi \times \{1\}$. But kernels of pseudodifferential-operators are smooth away from the diagonal in $\mathbb{R}^2$, thus $P$ is not a pseudodifferential-operator on all of $\mathbb{R}$.
Second, there is another proof for the statement, whose adaption to higher dimensions is more straightforward. Assuming property 2', one takes a cutoff function $\chi\in\mathcal{D}(0,1)$, which is $\equiv 1$ on a neighbourhood of $(\epsilon,1-\epsilon)$. One can then argue that $P = \chi P\chi$, the latter of which having $\chi(x)\sigma(x,\xi)\chi(y)$ as an amplitude on $(0,1)$. This amplitude on the other hand can obviously be extended to the whole real line, which gives an amplitude for $P$.
