1
$\begingroup$

If I am not wrong to prove that a linear surjective map $\pi:(V_1,||.||_1)\to (V_2,||.||_2)$ is an isometric submersion between the normed vector spaces is equivalent proving that $\pi$ projects the closed unit ball to the closed unit ball. Why they are equivalent?

P.S. $\pi$ is said to be an isometric submersion if $\forall w\in V_2, ||w||_2=inf \{||v||_1: \pi(v)=w\}$

$\endgroup$
3
$\begingroup$

I prove first that $\pi$ is an isometric submersion iff $\pi$ projects the open unit ball in $V_1$ to the open unit ball in $V_2$.

Suppose that $\pi$ projects the open unit ball to the open unit ball. We need to prove that for arbitrary $w\in V_2$ $$||w||_2=\inf\{||v||_1:\pi(v)=w\}.$$ If $w=0$, this is trivial. The assumption implies that $||\pi||\leq1$ (operator norm). Therefore, for each $v\in V_1$ with $w=\pi(v)$ we have $$||w||_2\leq ||\pi||\cdot ||v||_1\leq ||v||_1.$$ Consequently, $$||w||_2\leq\inf\{||v||_1:\pi(v)=w\}.$$ If $||w||_2=1$, $(1-\epsilon)w$ is in the open unit ball for each $\epsilon>0$, so $(1-\epsilon)w=\pi(v_\epsilon)$ for a suitable $v_\epsilon$ in the open unit ball of $V_1$, i.e. $||v_\epsilon||_1<1$. In other words, $w=\pi\left(\frac{1}{1-\epsilon}v_\epsilon\right)$. Hence, $$\inf\{||v||_1:\pi(v)=w\}\leq ||\frac{1}{1-\epsilon}v_\epsilon||_1<\frac{1}{1-\epsilon}.$$ Since $\epsilon>0$ was arbitrary, we conclude $$\inf\{||v||_1:\pi(v)=w\}\leq 1=||w||_2,$$ so that the claim is proven for $||w||_2=1$. Let now $w\in V_2$ be arbitrary, as long as $w\neq 0$. Then $\frac{1}{||w||_2}w$ is a unit vector and, by what I have just shown, $$||\frac{1}{||w||_2}w||_2=\inf\{||v||_1:\pi(v)=\frac{1}{||w||_2}w\}.$$ Multiplying both sides by $||w||_2$ and noting that $\pi(v)=\frac{1}{||w||_2}w$ iff $\pi(||w||_2\cdot v)=w$ as well as $||||w||_2\cdot v||_1=||w||_2\cdot ||v||_1$, we have obtained the desired conclusion.

Suppose now, conversely, that $$||w||_2=\inf\{||v||_1:\pi(v)=w\}$$ and let $w$ be an element of the open unit ball in $V_2$. Then, there exists a $v\in V_1$ with $||v||<1$ and $\pi(v)=w$, so that the open unit ball in $V_2$ is contained in the image of the corresponding ball in $V_1$. On the other hand, for each $\tilde{v}\in V_1$ the assumption implies (set $w=\pi(\tilde{v})$) $$||\pi(\tilde{v})||_2=\inf\{||v||_1:\pi(v)=\pi(\tilde{v})\}\leq ||\tilde{v}||_1.$$ This shows the converse set inclusion, in other words that $\pi$ does indeed project the open unit ball to the open unit ball.

Finally, under the additional assumption that $V_1$ (and therefore $V_2$ since $\pi$ is surjective) is finite dimensional, I show the desired equivalence involving the closed unit balls. I accomplish this by showing that the conditions for the open and the closed balls are equivalent. Let $U_i$ and $C_i$ denote the open and closed unit balls in $V_i$, respectively. Firstly, $\pi(U_1)\subseteq U_2$ is clearly equivalent to $\pi(C_1)\subseteq C_2$, since both are equivalent to $||\pi||\leq 1$. If $\pi(C_1)\supseteq C_2$, then $\pi(U_1)\supseteq U_2$: for $y\in U_2$, there exists a $\rho>1$ with $\rho\cdot y\in U_2\subseteq C_2$, eq $\rho=\frac{2}{1+||y||}$. By assumption, $\rho\cdot y=\pi(x)$, $x\in C_1$. Hence, $y=\pi(\frac{1}{\rho}x)\in\pi(U_1)$. Conversely, suppose that $\pi(U_1)\supseteq U_2$. We have $$C_2\subseteq \rho U_2\subseteq\pi(\rho U_1)$$ for each $\rho>1$. Choose $y\in C_2$. For each $\rho\in(1,2]$, there exists $x_\rho\in\rho U_1$ with $y=\pi(x_\rho)$ - in other words, $||x_\rho||<\rho$. Consider the sequence $(x_{1+1/n})_{n\in\mathbb{N}}$. Since $2C_1$ is compact, there is a convergent subsequence $$(x_{1+1/n(k)})_{k\in\mathbb{N}}\to x.$$ By continuity of $\pi$, we have $\pi(x)=\lim_{k\to\infty}\pi(x_{1+1/n(k)})=y$. Additionally, $||x||=\lim_{k\to\infty}\underbrace{||x_{1+1/n(k)}||}_{<1+1/n(k)}\leq 1$, so $x\in C_1$. We conclude $y=\pi(x)\in\pi(C_1)$.


Note1: Originally, the question was asked differently, namely dropping the word "submersion" from the claim. For completeness' sake, here is my original answer:

Your claim is not correct: let $V_1=\mathbb{C}^2$ and $V_2=\mathbb{C}$, both normed by the appropriate Euclidean norms. Then the projection to the first component, $$(x,y)\mapsto x,$$ is surjective and maps the closed unit ball in $V_1$ to the one in $V_2$, but isn't isometric, since it isn't even one-to-one. I'm not certain right now, but perhaps if you require that $\pi$ is bijective, your claim could be correct.

Note2: It appears as though I never use the surjectivity of $\pi$. In fact, it is used to ensure the welldefinedness of $\inf\{||v||_1:\pi(v)=w\}$.

Note3: The fact that $V_1$ is finite dimensional is only used to show the implication $\pi(U_1)\supseteq U_2\Rightarrow\pi(C_1)\supseteq C_2$. The rest, especially the equivalence of "$\pi$ is an isometric submersion" and $\pi(U_1)=U_2$, is valid in the general case, too.

$\endgroup$
  • $\begingroup$ yes you are completely right. I should edit my question and ask is it is ai isometric submersion or not. $\endgroup$ – Majid Jun 17 '17 at 20:03
  • $\begingroup$ @Majid how do you define an isometric submersion? I was unfamiliar with the term; a quick google search revealed as the only definition the condition you wish to be equivalent to it... $\endgroup$ – haemi Jun 17 '17 at 21:16
  • $\begingroup$ I will add it to the question then. $\endgroup$ – Majid Jun 17 '17 at 21:27
  • $\begingroup$ @Majid I have updated my answer. $\endgroup$ – haemi Jun 18 '17 at 20:51
  • $\begingroup$ Your solution for this case sounds right. In fact I like it. Now let's see what is behind being closed! $\endgroup$ – Majid Jun 18 '17 at 23:50

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.