1

Let $X$ be a Banach space, by Hahn-Banach we know that for all $x\in X$ there exists $x^*\in X^*$ such that $||x^*||=||x||$ and $x^*(x)=||x||^2$. There might be different candidates for $x^*$, so the application $\cdot^*$ is not uniquely defined, and might not be linear. Is it always possible to pick $\cdot^*$ such that it is Lipschitz? (So for all $x,y\in X$, $||x^*-y^*||\leq K ||x-y||$).

I know it is when $X$ is a Hilbert. Are there some more exotic examples?

Tobsn
  • 564
  • 1
  • 5
  • 13
Papagon
  • 315
  • Try doing this when $X$ is $\mathbb R^2$ with the $l^1$ norm. What happens when $x$ goes around a corner of the unit ball? – GEdgar Jul 11 '22 at 17:08

1 Answers1

0

The answer is no: there are cases where the Lipschitz condition isn't satisfied.

The mapping $x\mapsto x^*$ is called a dualuty mapping. It's generally multivalued. It's definition is as follows: $$ J\colon X\multimap X^*,\quad J(x)=\{x^*\in X^*:x^*(x)=\|x\|^2=\|x^*\|^2\}.$$

Lp spaces

Let's check the situation for the $L^p$ spaces. Let $X=L^p(\Omega)$, $1<p<\infty$. We know that $X^*\equiv L^q(\Omega)$, where $p^{-1}+q^{-1}=1$.

Let $x\in L^p(\Omega)$. Then the element $x^*\in X^*$ that $\|x^*\|=\|x\|$ and $x^*(x)=\|x\|^2$ is unique and is given by the formula: $$x^* = \|x\|_{p}^{2-p}\cdot x\cdot |x|^{p-2}.\tag{*}$$ Let's check it:

  • $\displaystyle \|x^*\|_q^q = \|x\|_{p}^{(2-p)q}\cdot \int_\Omega |x|^{(p-1)q} = \|x\|_{p}^{(2-p)q}\cdot \int_\Omega |x|^{p} = \|x\|_{p}^{q-p}\cdot \|x\|_p^p = \|x\|_p^q$. Therefore $\|x^*\|_q=\|x\|_p$.
  • $\displaystyle x^*(x)=\|x\|_{p}^{2-p}\cdot \int_\Omega x\cdot |x|^{p-2}\cdot x = \|x\|_{p}^{2-p}\cdot \int_\Omega |x|^{p} = \|x\|_{p}^{2-p}\cdot \|x\|_{p}^p = \|x\|_{p}^2$.

lp space

Let's now consider the simpler case: $\Bbb R^2$ with norm $\ell^p$. Then, using the formula (*) we get that for $x,y\geq 0$ we have

$$(x,y)^*=\frac 1{(x^p+y^p)^{(p-2)/p}}\cdot (x^{p-1},y^{p-1}) = (f(x,y),g(x,y)). $$

Now take $p=1.5$. Then the function $f$ isn't Lipschitz near $(0,1)$ (one can calculate the derivative).

Multivalued case

If the space isn't strictly convex, the duality mapping can be multivalued.

Example Let $X=\Bbb R^2$ with $\|\cdot\|_1$ norm. Then $J((1,0)) = \{1\}\times[-1,1]$.

Example Let $X=\Bbb R^2$ with $\|\cdot\|_\infty$ norm. Then $J((1,1)) = \mathrm{conv}\{(0,1),(1,0)\}$.

These are the situations where there are no continuous (hence Lipschitz) selectors $f(x)\in J(x)$.

Duplicate

Alright. I've just found that the question about the Lipschitz condition of $J$ was already present on MSE. Here's a link. Maybe I should remove my answer and just give a flag duplicate. I don't remove my answer since maybe it'll be helpful (I added some additional information).

Mateo
  • 4,956