Question #213039

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

 θ= 1.


1
Expert's answer
2021-07-05T15:22:09-0400

Let XX be a normed space, YY its closed proper subspace of a finite dimension. As YY is proper, there is at least one vXYv\in X\setminus Y, let us denote F:=Span(Y,v)F:=Span(Y,v) the subspace of XX generated by YY and vv. This is a finite dimensional normed space, as its dimension is dimY+1\dim Y+1 and the norm is inherited from XX. Therefore we consider now the following problem : FF is a finite dimensional normed space and YY is its proper subspace. Therefore, the unit sphere in FF, i.e. S:={xF    x=1}S:=\{x\in F\;|\; ||x||=1 \}, is compact. We define a function d(.,Y):SR+d(\:.\:,Y):S\to \mathbb{R}^+, it associates to a point on the sphere its distance to YY. This function is continuous and the sphere is compact, so the image d(.,Y)(S)d(\:. \: ,Y) (S) is a compact interval of R+\mathbb{R}^+. In addition it contains the open interval (0;1)(0;1) by the original Riesz's lemma, and therefore we conclude that the image contains 11. By rewriting this we find the result : (xSFX)  (x=1)(yY)(xy1)(\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 θ=1\theta=1.


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