Proposition: Bounded continuous coefficients extend distributionally

Every bounded continuous function on $(-1,1)\setminus \{0\}$ has an $L^{\infty }$ representative on $(-1,1)$ and therefore an order-zero distributional extension. A continuous extension need not exist.

Proof. Assign any value at zero. The resulting measurable function is bounded, and

is a distribution extending the punctured one. Its definition is independent of the value assigned at one point. The bounded smooth function $\sin (1/x)$ has no limit at zero, so it has no continuous extension although its distributional extension exists.