Answer to Question #213039 in Functional Analysis for smi

Question

If dim Y< ∞ in Riesz's lemma, show that one can even choose

 θ= 1.

Answer 1

Let $X$ be a normed space, $Y$ its closed proper subspace of a finite dimension. As $Y$ is proper, there is at least one $v\in X\setminus Y$, let us denote $F:=Span(Y,v)$ the subspace of $X$ generated by $Y$ and $v$. This is a finite dimensional normed space, as its dimension is $\dim Y+1$ and the norm is inherited from $X$. Therefore we consider now the following problem : $F$ is a finite dimensional normed space and $Y$ is its proper subspace. Therefore, the unit sphere in $F$, i.e. $S:=\{x\in F\;|\; ||x||=1 \}$, is compact. We define a function $d(\:.\:,Y):S\to \mathbb{R}^+$, it associates to a point on the sphere its distance to $Y$. This function is continuous and the sphere is compact, so the image $d(\:. \: ,Y) (S)$ is a compact interval of $\mathbb{R}^+$. In addition it contains the open interval $(0;1)$ by the original Riesz's lemma, and therefore we conclude that the image contains $1$. By rewriting this we find the result : $(\exists x\in S\subset F\subseteq X)\; (||x||=1)(\forall y\in Y)(||x-y||\geq1)$

Which is exactly the Riesz's lemma for $\theta=1$.