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 $u(0) \neq 0$.


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