Let's say that a subset $C$ of a real Hilbert space $H$ is thin if for every $\epsilon>0$ there exists a hyperplane $M$ such that $C$ is contained in the $\epsilon$-neighborhood of $M$. Equivalently, $C$ is thin if for every $\epsilon>0$ there exists a unit-norm linear functional $f$ on $H$ such that $\operatorname{diam} f(C)<\epsilon$.

Suppose $C\subset H$ is closed, convex, and has empty interior. Does it follow that $C$ is thin?


  • Such $C$ need not be contained in a hyperplane: the Hilbert cube is an example.
  • Every compact subset of $H$ is thin, but $C$ need not be compact: e.g., it could be a hyperplane itself.
  • It's necessary to assume that $C$ is closed; otherwise, $\ell^1 \subset \ell^2$ (or any other dense subspace) would be a counterexample.
  • The same question could be posed in other normed spaces.

1 Answer 1


I think here is a counterexample: the non-negative cone in $L_2[0,1]$ $$ C=\{x\in L_2\colon x\ge 0\text{ a.e.}\}. $$ Clearly, it is a closed convex set with an empty interior.

On the other hand, $L_2=C-C$ since any $x\in L_2$ can be split as $x=x_+-x_-$ where $$ x_+=\max\{x,0\},\quad x_-=\max\{-x,0\} $$ are the positive resp. negative part of the function $x$. If $|f(C)|\le\epsilon$ then $$ |f(L_2)|\le |f(C)|+|f(C)|\le 2\epsilon<1 $$ for small enough $\epsilon$, which contradicts the unity norm assumption on $f$.

  • 1
    $\begingroup$ Oh... and the same example works in the sequence space $\ell^2$. And you don't need small $\epsilon$, since $f(L_2)$ has to be all of $\mathbb R$ for any nonconstant $f$... a nonconstant functional cannot be uniformly bounded on the space. $\endgroup$
    – user357151
    Commented Apr 13, 2017 at 22:37
  • $\begingroup$ @Gerry My first reaction was that in $\ell^2$ the interior of the non-negative cone is not empty, but you are right, it works since $x_n\to 0$. $\endgroup$
    – A.Γ.
    Commented Apr 13, 2017 at 22:51

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