1
$\begingroup$

I need to prove the following statement:

"$X$ is a normed vectorspace and $Y$ is a finite dimensional subspace. For all $x \in X$ the distance is defined as $d(x,Y)=\inf\{ \|x-y\|:\ y \in Y\}$"

So I need to prove that there is an specific y in Y such that $d(x,Y) = \|x-y\|$.

I used the Corollary that every finite dimensional subspace of a normed vectorspace is also closed, and then I tried to apply the Riesz Lemma. But I am not sure.

Can anyone give me a hint for this exercise?

$\endgroup$
  • $\begingroup$ In a finite-dimensional normed vector space (or subspace) every closed bounded subset is compact. Any $n$-dimensional real normed vector space has a linear homeomorphism to $\mathbb R^n.$ $\endgroup$ – DanielWainfleet Oct 9 '16 at 16:51
1
$\begingroup$

Use the definition of infimum to find a sequence $\left(y_n\right)_{n\geqslant 1}$ of elements of $Y$ such that $\lVert x-y_n\rVert\leqslant d(x,Y)+1/n$ for each $n\geqslant 1$. The sequence $\left(y_n\right)_{n\geqslant 1}$ is bounded (why?) and lies in a normed vector space of finite dimension: you can extract a converging subsequence.

$\endgroup$
  • $\begingroup$ What exactly is that showing me? Sorry I don't get the point of doing that,.. $\endgroup$ – Yuhe Oct 9 '16 at 9:40
  • $\begingroup$ The wanted $y$ will be the limit of the converging subsequence. $\endgroup$ – Davide Giraudo Oct 9 '16 at 9:41
  • $\begingroup$ Ah I see, thanks :) the sequence is bounded because it is in a finite space I guess? or is there a better reason? $\endgroup$ – Yuhe Oct 9 '16 at 9:52
  • $\begingroup$ No, juse use $||y_n||\leqslant ||y_n-x|| +d(x,Y) \leqslant 1/n+||x|| \leqslant 1+||x ||$. $\endgroup$ – Davide Giraudo Oct 9 '16 at 9:56
  • $\begingroup$ How do you get to this inequality? $\endgroup$ – Yuhe Oct 9 '16 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.