Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I got stuck at the following problem.

Let $X,Y$ be normed spaces. A bounded linear operator $\tau\in\mathcal{B}(X,Y)$ is called strictly coisometric if $$ \tau(\operatorname{Ball}_X(0,1))=\operatorname{Ball}_Y(0,1)) $$ which is equivalent to that $\Vert\tau\Vert=1$ and for all $y\in Y$ there exist $x\in X$ so that $\tau(x)=y$ and $\Vert x\Vert= \Vert y\Vert$.

Is it true that the functor $\mathcal{B}(Z,-)$ preseve coisometries, i.e. is true that given $\tau$ strictly coisometric operator $$ \mathcal{B}(Z,\tau):\mathcal{B}(Z,Y)\to\mathcal{B}(Z,X):\varphi\mapsto\tau\circ\varphi $$ is strictly coisometric too.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Taking another look, I see that you only wanted some hints, so ignore the second part of this answer until you have solved this problem yourself.

Your definition of strict coisometry implies surjectivity but not injectivity. Take advantage of this. There could be a subspace of $X$ isometric to $Y$ and another subspace of $X$ not isometric to $Y$, also mapped to $Y$ by $\tau$ but where the norm is not preserved.

Further details: Fix $0<\alpha<1$ and $\beta>1$ such that $\alpha\beta<1$.

Let $X=\mathbb C \oplus \mathbb C$ with the norm $\Vert(x_1,x_2)\Vert=\max\{|x_1|,\beta|x_2| \}$ and $Z=Y=\mathbb C$ with standard norms. Then $\tau:X \to Y:(x_1,x_2)\mapsto x_1+x_2$ is strictly coisometric.

Define $\varphi:Z\to Y:z\mapsto\alpha z$ and $\psi:Z\to X: z\to(0,\alpha z)$. Then $\varphi=\tau \circ \psi$, but we have $\Vert\varphi\Vert=\alpha$ and $\Vert\psi\Vert=\alpha\beta>\alpha=\Vert\varphi\Vert$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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