Why the extension of distribution T in $(D(\Omega))^{'}$ to dual of sobolev space $W^{k,p}(\Omega)$ not unique?

Question

I am trying to understand section about dual of sobolev spaces from Adam's book "Sobolev Spaces". As I understand it he states that every element in the dual of $W^{k,p}(\Omega)$, $\Omega \subset \mathbb{R}^{n}$ open, can be represented by a vector $v=(v_1,v_2,....,v_N)$ where each $v_i$ is in $L^{q}$ and q is the conjugate exponent of p.N is the number of multi derivate indices that depend on k and n. This also represents an element in the space of distributions $(D(\Omega))^{'}$. But however he says that this representation need not be unique. Can you give an example where there are more than one such representation for elements in the dual $(W^{k,p}(\Omega))^{'}$?

Edited:

Is it possible that there are two different elements in $(W^{k,p}(\Omega))^{'}$, $T_1$ and $T_2$ such that they agree on $C_0^{\infty}(\Omega)$. ie is it possible that $T_1{\phi}=T_2{\phi}$ for all $\phi$ in $C_0^{\infty}(\Omega)$ but $T_1{u} \neq T_2{u}$ for some $u$ in $W^{k,p}(\Omega)$.

The relevant section can be found in chapter 3 of the book Adams, R.A. and Fournier, J.J.F, Sobolev Spaces, 2nd ed., Academic Press, 2003.

Answer 1

The non-uniqueness refers to the vector $v$. Note that for $f\in W^{1,p}(\Omega)$ and $\phi \in C_0^1(\Omega)$ we have $$\int_\Omega \nabla f \cdot \phi d x = - \int_\Omega f \mathop{div} \phi d x.$$ In other words, if you find a function $\phi \in C_0^1(\Omega)$ with $\mathop{div} \phi = 0$, (and there are many of those) you can add it to the respective components of $v$ without changing the operator. The same of course also can be done in the higher derivatives.

Answer 2

Copying mlk's answer, let $\phi(x) = e^{-1/(1-x^2)}1_{|x| < 1} \in C^\infty_c[-1,1]$ then $$\int_{-1}^1 f'(x)\phi(x)dx=\int_{-1}^1 f(x)\phi'(x)dx$$

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Also $$\int_0^a f'(x)x dx = a f(a)-\int_0^1 f(x)dx$$ So that $a f(a) =\int_0^a f'(x)x dx+\int_0^1 f(x)dx$ and $$\int_0^2 f'(x)x dx = 2 f(2)-\int_0^2 f(x)dx=2 f(1)+ 2\int_1^2 f'(x)dx-\int_0^2 f(x)dx\\= 2\int_0^1 f'(x)x dx+2\int_0^1 f(x)dx+2\int_1^2 f'(x)dx-\int_0^2 f(x)dx$$