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 $\forall w\in V_2, ||w||_2=inf \{||v||_1: \pi(v)=w\}$


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.

  • $\begingroup$ 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

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