# Math Help - Riesz's lemma

1. ## Riesz's lemma

With Riesz's lemma, I know that if i let X be a normed space. and Y and Z be subspaces of a normed space X then if Y is closed proper subset of Z then for every real number theta on interval (0,1) there exist a ||z||=1 s.t.
||z-y||>= theta

My question:
When Y is finite dimensional then theta can be (0,1]

i am having trouble proofing this.
I believe that i want to show that since Y is bounded and closed it is compact. Therefore
their exists value v in Z-Y s.t.
a=inf||v-y||
since Y is compact there exists y_o in Y such that
a=||v-y_o||

and if this is true i could easily finish the proof from there.

Basically my two questions are:
How do i Show that Y is bounded. Is it automatically implied because it is finite dimensional and closed?
If Y is compact can i assume that there exists a y_0 in Y s.t. a=||v-y_o||?

2. Originally Posted by macrone
With Riesz's lemma, I know that if i let X be a normed space. and Y and Z be subspaces of a normed space X then if Y is closed proper subset of Z then for every real number theta on interval (0,1) there exist a ||z||=1 s.t.
||z-y||>= theta

My question:
When Y is finite dimensional then theta can be (0,1]

i am having trouble proofing this.
I believe that i want to show that since Y is bounded and closed it is compact. Therefore
their exists value v in Z-Y s.t.
a=inf||v-y||
since Y is compact there exists y_o in Y such that
a=||v-y_o||

and if this is true i could easily finish the proof from there.

Basically my two questions are:
How do i Show that Y is bounded. Is it automatically implied because it is finite dimensional and closed?
If Y is compact can i assume that there exists a y_0 in Y s.t. a=||v-y_o||?
A subspace of a vector space can never be compact because it is not bounded. What is true is that if the subspace is finite-dimensional then its closed unit ball is compact. More generally, any closed ball in a finite-dimensional space is compact.

The proof of Riesz's lemma shows that you can take $\theta = 1$ provided that the distance $\inf_{y\in Y}\|v-y\|$ is attained. It easily follows from the triangle inequality that the smallest values of $\|v-y\|$ must occur inside the closed ball $\|y\|\leqslant2$. The function $y\mapsto\|v-y\|$ is continuous, and will therefore attain its infimum on any compact set.

Put those facts together and you have your proof.