Proposition: Smooth off-set data need not extend

There is a smooth function on $(0,1)$ that has no distributional extension to $(-1,1)$.

Proof. Take $u(x)=e^{1/x^2}$. Suppose a distribution $T$ on $(-1,1)$ restricts to $u$ on $(0,1)$.

On a fixed compact neighborhood of zero, continuity of $T$ gives constants $C$ and $m$ such that

for tests supported in that neighborhood.

Choose a nonnegative nonzero $\phi \in {C_c}^{\infty }((-1/4,1/4))$, and for small $\epsilon >0$ put

Its support lies in $(0,1)$, so

For sufficiently small $\epsilon$, points in this support satisfy $0<x<2\epsilon$. Hence

On the other hand,

The exponential lower bound contradicts the polynomial upper bound as $\epsilon \to 0$.