# A variation of the Riesz Lemma

Given a normed space $$X$$, a closed proper subspace $$Y$$ and $$\alpha\in (0,1)$$, the Riesz Lemma states that there is $$x\in X$$ such that $$\|x\|=1$$ and $$d(x,Y)>\alpha$$. Observe that also $$d(-x,Y)=d(x,Y)>\alpha$$. That is, the distance from the unit sphere of $$\langle x \rangle$$ (where $$\langle A\rangle$$ denotes the linear span of $$A$$) to $$Y$$ is at least $$\alpha$$.

I'm looking for a proof or counterexample for the following variation of the Riesz Lemma:

Given a normed space $$X$$, a closed proper subspace $$Y$$ and $$\alpha\in (0,1)$$, there is $$x\in X$$ such that the distance from $$\langle x\rangle$$ to the unit sphere of $$Y$$ is at least $$\alpha$$.

I am mostly interested in positive answers, so partial positive answers would be nice if the result is false. Particularly, I'm only concerned with Banach spaces, so this can be assumed without problem.

EDIT: Pietro presented a clever construction in finite dimension and with $$\mathrm{codim}(Y)=1$$. Is there also a counterexample when $$\dim(X)=\infty$$ and $$\mathrm{codim}(Y)=\infty$$?

• Notice that $\|x\|,$ coded as \|x\|, looks different from $||x||,$ and $\|x\|\|y\|,$ coded as \|x\|\|y\|, looks very different from $||x|| ||y||,$ coded as ||x|| ||y||. The former notation is standard usage. \begin{align} & ||x|| ||y|| \\ {} \\ & \|x\|\|y\| \end{align} Commented May 2 at 21:19
• @MichaelHardy, re, \lVert\rVert can work even better than \|\|. Compare, for example, \|-x\| $\|-x\|$ to \lVert-x\rVert $\lVert-x\rVert$. Although a bit awkward, this can be swept up in a macro like \newcommand\norm[1]{\lVert#1\rVert}, and even repeated such macro definition can be semi-automated using \DeclarePairedDelimiter from mathtools. Commented May 3 at 0:04

Let's denote $$B$$ the unit ball of $$X$$. Geometrically, we are asking if there is a line $$\langle x\rangle$$ such that the ball $$\alpha B$$ can be translated along $$\langle x\rangle$$ and pass through the hole of $$Y\setminus B$$, that is we seek a non-zero $$x$$ such that $$\big( \alpha B+ \langle x\rangle\big) \cap \big(Y\setminus B\big)=\emptyset.$$
I think the simplest counterexample is: $$X:=\mathbb R^3$$ with the box norm $$\|(a,b,c)\|_\infty:=\max(|a|,|b|,|c|\}$$ and $$Y:=(1,1,1)^\perp=\{(a,b,c):a+b+c=0\}$$. The section $$Y\cap B$$ is a regular hexagon with vertices in mid-points of some edges, but every projection of $$\alpha B$$ onto $$Y$$ is a polygon with vertices in the projection of some vertices of $$\alpha B$$, thus never included in $$Y\cap B$$ for $$\alpha$$ too close to $$1$$. (Or just consider the angles between $$Y$$ and the faces of the cube).
• Thanks for your answer, @PietroMajer. I'm just confused with a thing. The set $Y=\{(a,b,c) : a+b+c=1\}$ isn't a linear subspace, right? It isn't closed by scalar multiplication. Commented May 2 at 20:37
• That is a clever construction, thank you @PietroMajer. As you said, the worst possible case is the closed hyperplanes. But, in the specific case where I am interested in applying that result, I have $Y$ with infinite codimension. Actually, I have $\dim(X)$ infinite and $\mathrm{codim}(Y)$=$\dim(X)$. Do you know a possible counterexample in that setting, where $\mathrm{codim}(Y)=\infty$? Should I post another question or just edit it with this add-on? Commented May 2 at 22:20
• Maybe a product of countably many copies of the above counterexample? So now $X$ is $\ell_\infty$ and $Y$ the subspace defined by $x_{3n}+x_{3n+1}+x_{3n+2}=0$ for all $n\ge0$. Commented May 2 at 23:12
• I get it. This can even be made with an arbitrarily large number of copies. In that case, we have the space $l_\infty(M)$, where $M$ is a set with any cardinality. Thanks, @Pietro Commented May 3 at 2:50