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This is a question from the chapter 5 of Pugh's Real Mathematical Analysis book:

The conorm of a linear transformation $T : \mathbb{R}^n \to \mathbb{R}^m$ is $$m(T) = \inf \left\{ \frac{|Tv|}{|v|}: v \neq 0 \right\}.$$ It is the minimum stretch that $T$ imparts to vectors in $\mathbb{R}^n.$ Let $U$ be the unit ball in $\mathbb{R}^n.$ Show that the conorm of $T$ is the radius of the largest ball contained in $TU.$

I guess the conorm of $T$ is the radius of the largest ball contained in $TU$ iff the boundary of $TU$ is $TS^{n-1}$, where $S^{n-1}$ is the unit sphere in $\mathbb{R}^n$. But how can I show that the boundary of $TU$ is $TS^{n-1}$? Do I need to assume that the norm on both $\mathbb{R}^n$ and $\mathbb{R}^m$ is the Euclidean norm, or I can proceed with a general norm?

At this point, I found two similar questions on this site, this and this, but I'm unable to proceed further with the answers provided. For instance, this says "A very similar argument can be made for the conorm." But I don't see how. (I believe this should be enough to not closing the question.)

Aleph-null
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  • $m(T) = \inf { |Tv|, , v \in \mathbb S^{n-1}}$ the proof is the same as the usual one. Also the boundary is not $TS^{n-1}$, it englobes it – julio_es_sui_glace Dec 20 '24 at 12:10
  • @julio_es_sui_glace I know that $m(T) = \inf { |Tv|, , v \in \mathbb S^{n-1}}$. But how can I proceed further from here? – Aleph-null Dec 20 '24 at 12:25
  • Also, it is not clear why the boundary of $TU$ is not $TS^{n-1}$, @julio_es_sui_glace. – Aleph-null Dec 20 '24 at 13:43
  • Because it is almost never a ball, it is rather a deformed ball, or the unit ball associated to the scalar product given by $S$ in the polar decomposition of $T$. You should draw it in dimension $2$, it makes an ellipse. Drawing in lower dimension should be your first reflex if you have no intuition – julio_es_sui_glace Dec 20 '24 at 13:48
  • @julio_es_sui_glace, for instance, let $T:\mathbb{R}^2 \to \mathbb{R}^2$ be defined as $ T((x, y)) = \begin{bmatrix} 2x\
    y \end{bmatrix}$. Then considering the Euclidean norm, the unit ball in $\mathbb{R}^2$ get transformed to the elliptic disc in $\mathbb{R}^2$ (centered at origin) with semi-major axis $2$ and semi-minor axis $1$. Clearly, the boundary of the linear transformation of the unit ball in $\mathbb{R}^2$ is the linear transformation of the unit sphere in $\mathbb{R}^2$ here.     – Aleph-null Dec 20 '24 at 14:56
  • You are right, I read (and did not write) $S^{n-1}$ instead of $TS^{n-1}$ – julio_es_sui_glace Dec 20 '24 at 15:00
  • @julio_es_sui_glace, I still couldn't figure out an way to show that the conorm of $T$ is the radius of the largest ball contained in $TU$. – Aleph-null Dec 20 '24 at 15:07
  • The solution is provided in the link of the second commentary... – julio_es_sui_glace Dec 20 '24 at 15:10
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    @julio_es_sui_glace, it says "A very similar argument can be made for the conorm." I don't know how. – Aleph-null Dec 20 '24 at 15:16
  • replace min by max, and $\le$ by $\ge$ or the converse – julio_es_sui_glace Dec 20 '24 at 15:32
  • @julio_es_sui_glace, I'm sorry, still not clear to me. – Aleph-null Dec 20 '24 at 15:40
  • Differentiate the square of the quotient to find the critical points. Or use Lagrange multipliers. Then do some linear algebra. – Deane Dec 20 '24 at 15:43

1 Answers1

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You need $n \ge m$, otherwise you will have some $y \in \mathbb R^m$ such that $T^{-1} (\mathbb R y) = \{0\}$, thus for any injective $T$ the result will be false.

You can also see that you need $n \le m$ otherwise you can define $$T = \begin{pmatrix} I_m & 0_{n-m} \end{pmatrix}$$ And you will have $m(T) = 0$ since it is not injective but $TU_n = U_m$.

So we suppose $n=m$.

First we suppose $T$ invertible.

First show that the ball $B$ of radius $m(T)$ is contained in the image of the unity ball $TU$:

For $y \in B, x\neq 0 \, \|y \| \le m(T)$, now we set $x = T^{-1} y$, $\|Tx\| = m(T)$ but $\|Tx\| \ge m(T) \cdot \|x\|$ which means $\|x\| \le 1$, thus $y \in TU$.

Now let us show that any bigger ball fails to lie in $TU$: take $B'$ the ball of radius $r > m(T)$. By definition of $m(T)$, there exists $x \in V$ such that $\frac{\|Tx\|}{\|x\|} < r$ so by setting $y = x / \|x\|$ we have $\|Ty\| < r$ which means $z = \frac{r}{\|Ty\|} Ty \in B’ $ but $\|T^{-1} z\| = \frac{r}{\|Ty\|} \|y\| > 1$ thus $z \notin TU$.

Now if $T$ is not invertible it is not surjective and not injective, thus $m(T) = 0$, so if we take any ball $B$ with positive radius $\epsilon$, it will contain $\frac{\epsilon}{\|y\|} y$ for some $y \notin T \mathbb R^n$ which is impossible, which proves the case.

julio_es_sui_glace
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    But if the noninvertible linear transformation is injective, then conorm is not $0$. How would you argue in that situation (to show that the ball $B$ of radius $m(T)$ is contained in $TU$)? Also, you didn't show why the boundary of $TU$ is $TS^{n-1}$, but used in your proof (I was unable to show it with respect to the Euclidean norm, let alone for a general norm). – Aleph-null Dec 21 '24 at 11:54
  • I read $n = m$ for the injective part. I believe it is false when $n<m$ since you will always have elements in the neighbourhood of $0$ that don’t have a pre image, thus the biggest ball will be $0$ – julio_es_sui_glace Dec 22 '24 at 16:53