Riesz's Lemma says the following: Let $X$ be a normed vector space and $Y$ a proper closed subspace of $X$. Pick $\alpha \in (0,1)$. Then $\exists x\in X$ such that $|x|=1$ and $d(x,y) \geq \alpha$ for all $y \in Y$.
I am investigating why we cannot take $\alpha=1$ in general. Wikipedia says "the space $l^\infty$ of all bounded sequences shows that the lemma does not hold for $\alpha = 1$". I believe there's a simple example but I don't see what to take as $Y$… So what exactly is the example (i.e. $Y$)?
Edit: It is unfortunate that Wikipedia states the lemma with the version $>\alpha$. With this version, I do not need the specific space of $l^\infty$ to show that the lemma does not hold for $\alpha = 1$. Let $X$ be any normed vector space, and take any $x$ with $|x|=1$. Then $\inf_{y\in Y} d(x,y) \leq d(x,0) = 1$.
As Bombyx pointed out, $l^\infty$ is not reflexive so there should be an example for the version I have stated Riesz's Lemma. Since Wikipedia raised $l^\infty$, I was thinking that maybe there is a (simple) constructive proof for this example. Hence, I was asking if
anyone has any idea on what $Y$ should be.
Best Answer
I think the answer is no, because $l^{\infty}$ is not reflexive. A related discussion can be found at here. The credit should be given to the answerer in the other post.