# The dual space of the Sobolev space $W^{k,p}(\Omega)$.

Let $$\Omega$$ be a nice domain in $$\Bbb R^n$$. It is known that any element $$T\in\left( W^{k,p}(\Omega)\right)^*$$ admits a (possibly non-unique) representation of the form $$Tu = \sum_{|a|\le k} \int_\Omega f_\alpha D^\alpha u\ dx, \tag{0}$$ where $$f_\alpha \in L^{p'}$$ and $$\frac 1p + \frac 1{p'}=1$$. The functional $$T$$ can be identified with $$T =\sum_{|a|\le k} (-1)^{|\alpha|}D^\alpha f_\alpha \tag{1}\label{eq1}$$ as a distribution in $$\mathcal D(\Omega)$$.

In the book Weakly Differentiable Functions by Ziemer, there is a claim that confused me. The book said (modulo some paraphrasing) the following:

... However, not every distribution $$T$$ of the form $$(1)$$ is necessarily in $$\left( W^{k,p}(\Omega)\right)^*$$. In case one deals with $$W^{k,p}_0(\Omega)$$ instead of $$W^{k,p}(\Omega)$$, distribution of the form $$(1)$$ completely describes the dual space...

I am not sure if I fully understand what the passage means. I know that such a $$T$$ can be uniquely extend to an element in $$W^{-k,p'} = \left( W_0^{k,p}\right)^*$$ by the standard density argument, whereas $$T$$ in $$(1)$$ have more than one extension to an element of $$\left( W^{k,p}\right)^*$$. Perhaps this is what the passage means?

It seems weird to me to say that $$T$$ in $$(1)$$ is not necessary in $$\left( W^{k,p}\right)^*$$ since $$(0)$$ is obviously one way to define $$T$$ on $$W^{k,p}$$, I would rather mention the non-uniqueness explicitly. Is there any other deeper interpretation of the passage that I may have missed?

I think that the problem is that functionals in $$(1)$$ are not necessarily in $$(W^{k,p})^*$$ rather than non-uniqueness.
Take for example the simple setting $$\Omega=(0,1)$$, $$W^{k,p}=H^1$$.
Then $$f=x^{-\frac{1}{3}} \in L^2$$ so $$T=-\partial_x x^{-\frac{1}{3}}=\frac{1}{3}x^{-\frac{4}{3}}$$ satisfies $$(1)$$.
But it is not in $$(H^1)^*$$, as for example $$Tu$$ is ill-defined if $$u(0) \neq 0$$.