Consider the normed space $\left(\mathbb{R}^{3},\|\cdot\|\right)$ where $\|\cdot\|=|\cdot|_{\infty}$ is the usual maximum norm. Consider the 2 -dimensional vector subspace $\left \{ (x,y,z):x+y+z=0 \right \}$. Prove that there exists $\epsilon>0$ such that for every linear projector $P: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ onto $W$ one has $\|P\| \leq 1+\epsilon$. By a "linear projector onto $W$ we mean a map $P$ that is (i) linear (ii) $P\left(\mathbb{R}^{3}\right)=W$ and (iii) for every $u \in W$ : $P(u)=u$. Here, the operator norm $\|P\|$ of $P$ refers to the maximum norm $\|\|$ on the domain and range of $P$. I want to estimate $\epsilon$. And then I want to compare my result with the case when $\mathbb{R}^{3}$ is equipped with the Euclidean norm. in the Euclidean space, the operator has norm 1 because this space is an inner product space, and in the maximum norm case I want to first write down an orthonormal basis of that plane and then write down the complement of that plane but I don't know how to write down the general form of the complement, and I also want to know how to estimate the operator norm after writing down the general form of the complement
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I don't think, this is true. Each projector is determined by its kernel. Take one vector $v\notin W$ and project it with $P$, where the kernel of $P$ has an angle to W that becomes ever smaller. Then the norm of $P(v)$ tends to inifity. $\endgroup$– user473423Commented Apr 18, 2022 at 10:18
-
$\begingroup$ It's like the shadow of a person becomes longer when the sun sets. $\endgroup$– user473423Commented Apr 18, 2022 at 10:18
Add a comment
|