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It is well known that there exists a unique minimum norm vector over a closed convex set.

Suppose we have a Banach space X (if it needs to be more concrete we can think of $L_2$, the space of square integrable functions). Suppose $U \subset X$ and $U$ is open, bounded, $U$ contains the 0 function, and $U$ convex. Let $\partial U$ denote the boundary of $U$. My question is this:

Consider $L= \inf_{u \in \partial U}\| u\|$. Is there a $u^*\in \partial U$ such that $\|u^*\|=L$? In other words, is there a function in the boundary of $U$ that attains the boundary's infimum? I know it won't be unique, but does it exist?

If it's any easier we could also consider the same problem but $L = \inf_{u \in U^c} \|u\|$ where $U^c$ stands for the complement of $U$. Thanks for your thoughts!

Jason_optControlnerd
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  • A related but helpful question might be this: Suppose X is a banach space. Let U be an open, convex, bounded, nonempty set. Let $u \in U$. Does there always exist a point on the boundary of $U$ that minimizes the distance between $u$ and the boundary of $U$? – Jason_optControlnerd Oct 03 '13 at 15:56

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It is well known that there exists a unique minimum norm vector over a closed convex set.

In what kind of space? In a general Banach space both uniqueness and existence may fail. See, for example, A vector without minimum norm in a Banach space. In a reflexive space, existence holds. In a strictly convex space, uniqueness holds.

is there a function in the boundary of $U$ that attains the boundary's infimum?

The answer is negative, even in a Hilbert space.

The example in the post linked above is a hyperplane in $C[0,1]$. It separates the space $C[0,1]$ into two convex open subsets, one of which contains $0$. Intersecting this subset with a large open unit ball, you get a bounded open set $U\subset C[0,1]$ such that $0\in U$ and $\partial U$ has no vector of smallest norm.

And here is a counterexample in a Hilbert space; I'll take $\ell^2$ for convenience. Let $$U=\left\{(x_n)\in \ell^2: \sum_{n=1}^\infty \frac{n}{n+1}|x_n|^2<1\right\}$$ The set $U$ is open, bounded, and convex. It contains the closed unit ball; therefore any $x\in \partial U$ has $\|x\|>1$. On the other hand, $\sqrt{1+1/n}\,e_n\in \partial U$ for every $n$, where $\{e_n\}$ is the standard basis. Thus, $\inf_{\partial U}\|x\| =1$ and the infimum is not attained.

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