Let $X$ be a real Hilbert space, and let $K$ be a closed convex cone.

The dual cone is defined by $K^*=\{x^*\in X \mid (\forall k\in K)\, \langle x^*,k\rangle \geq 0 \}$.

I am looking for some interesting examples of $K$ and $K^*$, especially when $X$ is infinite-dimensional.

For example,

  • $X=\ell_2$ and $K=\ell_2^+ = K^*$
  • $X=L_2[0,1]$ and $K=L_2^+[0,1]=K^*$

Note that these examples satisfy $K-K=X$. I wonder if there are some concrete examples where

$$K-K\neq X\quad \text{but}\quad \overline{K-K}=X$$

and $K^*$ is actually known?

Any examples/comments/references would be greatly appreciated.


1 Answer 1


Let us take the Sobolev space $X = H_0^1(\Omega)$ and $K = X^+ = \{ v \in H_0^1(\Omega) \mid v \ge 0 \text{ a.e.}\}$. Then, we have $X = K - K$. Note that $X$ is a Hilbert space.

However, the dual cone $K^*$ coincides with the non-negative functionals in $H^{-1}(\Omega) = X^*$. This are precisely the functionals in $X^*$ which can be represented by measures which are finite on compact sets, i.e., for each $\mu \in K^*$, there is a measure $\hat\mu$ such that $$ \langle \mu, v\rangle = \int_\Omega v \, \mathrm{d}\hat\mu\qquad \forall \mu \in H_0^1(\Omega) \cap C_c(\Omega).$$

Now, one can check that we have $X^* = \overline{K^* - K^*} \ne K^* - K^*$, see also Decomposition of functionals on sobolev spaces and Decomposition of measures acting on sobolev spaces.

  • $\begingroup$ Thanks! Can you add some details on whether X is a Hilbert space here? A reference would be great. $\endgroup$
    – max_zorn
    Commented Jun 14, 2021 at 7:27
  • $\begingroup$ Yes $X$ is a Hilbert space, I have added this information and I also added that $X$ is called a Sobolev space. As reference, you can take any book on Sobolev spaces. $\endgroup$
    – gerw
    Commented Jun 14, 2021 at 9:11
  • $\begingroup$ Any favourite reference easily accessible to outsiders, gerw? Also, you say that $X^*=H^{-1}(\Omega)$. But $X^*=X$ for a Hilbert space - what am I missing? $\endgroup$
    – max_zorn
    Commented Jun 14, 2021 at 16:09
  • $\begingroup$ Maybe "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Brezis; in German there is also "Lineare Funktionalanalysis" by Alt. In $X^* = H^{-1}(\Omega)$ (and similar statements), the "=" is actually a "is isometrically isomorphic"; and all separable Hilbert spaces are isometrically isomorphic. For function spaces, one typically does not use the isomorphism which makes $X = X^*$, since this involves the inner product and this depends on the space. For example, we still identify $L^2(\Omega)^* = L^2(\Omega)$, but not $H^1(\Omega)$ and its dual. $\endgroup$
    – gerw
    Commented Jun 15, 2021 at 5:51

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