Is the following statement true?

"Let $H$ be a Hilbert space and $C\subset H$ a convex set. If $C$ has empty interior then there exist $a$ and a proper subspace $V\subset H$ such that $C\subset(a+V)$."

I guess this is false due to the following counterexample. Let $$C=\{(x_n)_{n\in\mathbb N}:|x_n|\leq 1/2^n,\ \forall n\in\mathbb N\}\subset \ell_2(\mathbb N).$$ Clearly, $C$ is convex. Moreover, it is easy to see that $C^\perp=\{0\}$, and consequently, there are no proper subspace $V$ and $a$ such that $C\subset(a+V)$.

Finally, given some $(y_n)_{n\in\mathbb N}\in C$ and $r>0$, fix $n_0$ such that $1/2^{n_0}<r$. Then $$|y_{n_0+2}-1/2^{n_0}|\geq1/2^{n_0}-1/2^{n_0+2}>1/2^{n_0+2}.$$ Define $$z_n=\left\{\begin{array}{r} y_n,\ if\ n\neq n_0+2 \\ y_{n_0+2}-1/2^{n_0},\ if\ n= n_0+2\end{array}\right..$$ Then $(z_n)_{n\in\mathbb{N}}\in B((y_n)_{n\in\mathbb N},r)$ and $(z_n)_{n\in\mathbb{N}}\notin C$. This way we proved that no ball is contained in $C$, so $C$ has empty interior.

Is everything correct? Am I missing something here?

  • $\begingroup$ There are easier counterexamples. What about the line $x=1$ in $\mathbb{R}^2$? $\endgroup$ – ChocolateAndCheese Mar 28 '17 at 23:55
  • $\begingroup$ @ChocolateAndCheese That's true! I'm so sorry, there was a mistake in my question. Actually I meant "in the translation of a proper subspace of $H$". $\endgroup$ – André Porto Mar 29 '17 at 0:16
  • $\begingroup$ Your proof that int (C) is empty is fine. $\endgroup$ – DanielWainfleet Mar 29 '17 at 6:58

Your counterexample is correct, assuming that "subspace" (as often) means a closed subspace of $H$. Otherwise, it would not work because the set $C$ is contained in $\ell^1(\mathbb{N})$ which is a dense subspace of $\ell^2(\mathbb{N})$. So I proceed by assuming the subspaces are closed.

The justification is slightly lacking in the following: the property $C^\perp = \{0\}$ only implies that $C$ is not contained in any proper linear subspace; it does not preclude $C$ from being contained in a proper affine subspace. For example, the set $A=e_1+e_1^\perp$, which is an affine hyperplane, satisfies $A^\perp = \{0\}$ since its linear span is all of $\ell^2$.

However, the above is easy to repair: since $C$ contains $0$, any affine subspace containing it would be a linear subspace.

A simpler example

Let $C$ be the set of all sequences $x$ such that $x_n=0$ except for finitely many $n$. Then $C$ is convex and has empty interior, since adding an arbitrarily small multiple of the vector $(1/2^n)_{n\in\mathbb{N}}$ to an element of $C$ takes one out of $C$. It's also dense in $\ell^2$, so can't be contained in a proper closed subset of any kind.

(Your example has the additional property of being closed, which however wasn't required.)

  • $\begingroup$ the statement "the property $C^\perp={0}$ only implies that $C$ is not contained in any proper linear subspace" is not true. The counterexample is the set $C$ defined in the counterexample gave by you. Clearly, $C^\perp=\{0\}$ and it is a proper linear subspace of $H$. $\endgroup$ – André Porto Apr 17 '17 at 19:38
  • $\begingroup$ @AndréPorto See the first paragraph: "I proceed by assuming the subspaces are closed." $\endgroup$ – user357151 Apr 17 '17 at 19:43
  • $\begingroup$ Yeah, ok then. My bad. $\endgroup$ – André Porto Apr 17 '17 at 20:37

As @Gerry pointed out, $C^\perp=\{0\}$ does not imply that $C$ is not contained in a proper affine subspace of $H$, so the example I gave in my question may not hold.

Fortunately, I found another example that really works this time. It is a much simpler one and is a counterexample to the statement even when $C$ is assumed to be closed.

Let $C$ be the set of vectors in $\ell_2(\mathbb N)$ with all its coordinates greater than or equal to $0$.

Clearly, $C$ is a closed convex. Let us see that $C$ has empty interior. Given $a=(a_n)_{n\in\mathbb N}\in C$ and $r>0$, since $a_n\to 0$, fix $m$ such that $a_m<r/2$ and define $b=(b_n)_{n\in\mathbb N}$ to be the sequence obtained by changing $a_m$ by $a_m-r/2$. Clearly, $b\in B(a,r)$ and, since $b_m=a_m-r/2<0$, $b\notin C$. Then $B(a,r)$ is not contained in $C$.

Finally, let us see that $C$ is not contained in a proper affine subspace. Pick $a\in \ell_2(\mathbb N)$ and a subspace $V$ such that $C\subset(a+V)$. Since $0\in C$, we get that $a\in V$, so actually $C\subset V$. We will conclude below that $V=\ell_2(\mathbb N)$.

Given $a=(a_n)_{n\in\mathbb N}\in \ell_2(\mathbb N)$, define $a^+=(a^+_n)_{n\in\mathbb N}$ by $$ a^+_n=\left\{\begin{array}{r}a_n,\ \mbox{if}\ a_n\geq0 \\ 0,\ \mbox{if}\ a_n<0\end{array}\right.. $$ and $a^-=(a^-_n)_{n\in\mathbb N}$ by $$ a^-_n=\left\{\begin{array}{r}-a_n,\ \mbox{if}\ a_n<0 \\ 0,\ \mbox{if}\ a_n\geq0\end{array}\right.. $$ Then, $a^+,a^-\in C\subset V$ and consequently $a=a^+-a^-\in V$.


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