0
$\begingroup$

I am trying to understand definition of Euclidean norm on a finite dimensional space, $\mathbb{R}^{n}$, denoted as $\mathbb{E}$. The dual space is denoted by $\mathbb{E}^{*}$.

In the attached screenshot from this paper (see the highlighted section), the self-adjoint operator $B$ is used to define Euclidean norm on $\mathbb{E}$, and $\mathbb{E}^{*}$. In my knowledge, a self-adjoint operator maps one space to itself. for example $ X\rightarrow X.$

Defining Euclidean norm

Therefore, my question is how we can use $B$ to define the norm.

Check for triangle inequality:

\begin{align} \langle B(x+y), x+y \rangle &= \langle Bx+By, x+y \rangle \\ &= \langle Bx,x\rangle + \langle Bx, y \rangle + \langle By, x\rangle + \langle By, y\rangle \\ &= \langle Bx,x\rangle + \langle y, Bx \rangle + \langle y, Bx\rangle + \langle By, y\rangle\; (B \text{ is self-adjoint}) \\ &\leq \|x\|^{2} + \|y\|^{2} +2 |\langle y, Bx \rangle| . \end{align}

Next, we have

\begin{align} |\langle y, Bx \rangle|&= |\langle \sqrt{B}y, \sqrt{B}x \rangle| \; ( \because B>0, \sqrt{B} \; \text{exists}) \\ &\leq \sqrt{\langle Bx, x \rangle} \sqrt{\langle By, y \rangle} \; (\text{Cauchy schwartz})\\ &= \|x\|\|y\| . \end{align}

Therefore, we end up with

$$ \|x+y\|^{2} \leq (\|x\|+\|y\|)^{2} .$$

I have given my proof for triangle inequality for the inner product defined in the screenshot, as advised by the other members.

$\endgroup$
8
  • 2
    $\begingroup$ I don't understand what the issue is here. They have explained exactly how to take the norm in the image. $Bx$ is in $\mathbb{E}^*$, so $\langle Bx, x\rangle$ makes sense. If you read up in the screenshot, you will see that they defined for $s\in \mathbb{E}^*,x\in\mathbb{E}$, $\langle s, x\rangle = s (x)$. Similarly $\langle g, B^{-1}g\rangle$ makes sense as well. Ofcourse, you must prove that these definitions do give you inner products but there is no ambiguity in regards to how they are defined. $\endgroup$ Commented Nov 25 at 21:58
  • $\begingroup$ I mean norms, not inner products in my message above. $\endgroup$ Commented Nov 25 at 22:06
  • 1
    $\begingroup$ Is your question that a self-adjoint operator maps a space to itself but $B$ maps $\mathbb E$ to $\mathbb E^*$? $\endgroup$ Commented Nov 25 at 22:16
  • $\begingroup$ @J.W.Tanner Thanks for the response. Yes, I am primarily concerned about how B being a self adjoint operator has been used to define this norm. $\endgroup$ Commented Nov 25 at 23:00
  • 1
    $\begingroup$ @DeanMiller It is this open access paper by Nesterov, I am using; link.springer.com/article/10.1007/s10957-023-02163-x titled: Set-Limited Functions and Polynomial-Time Interior-Point Methods $\endgroup$ Commented Nov 26 at 16:52

1 Answer 1

Reset to default
1
$\begingroup$

There is a problem here because the standard notion of a self-adjoint operator only makes sense for a linear operator going from $\mathbb{E}$ into $\mathbb{E}$, whereas the relevant article is trying to talk about a self-adjoint operator from one finite-dimensional vector space into a completely different finite-dimensional vector space. The author does not clarify what is meant by a self-adjoint operator in this situation, but I would advise against this terminology in this situation due to the fact that any such definition would go against the common understanding of a self-adjoint operator.

Also note that writing $\langle x, By\rangle$ for $x, y\in \mathbb{E}$ does not make sense because the map $\langle \cdot , \cdot \rangle$ is defined from $\mathbb{E}^{*} \times \mathbb{E}$ into $\mathbb{R}$.

The best thing I can think of is to attempt to come up with some conditions on a linear operator $B\colon \mathbb{E} \to \mathbb{E}^{*}$ that imply the map $u\colon \mathbb{E}\times\mathbb{E} \to \mathbb{R}$ defined by $u(x, y) = \langle Bx, y\rangle$ is an inner product on $\mathbb{E}$. Then the norm $\|\cdot\|$ induced by $u$ will be given by $\|x\| = \langle Bx, x\rangle^{\frac{1}{2}}$ for each $x\in \mathbb{E}$. Let $B\colon \mathbb{E} \to \mathbb{E}^{*}$ be a linear operator satisfying the following conditions.

  1. For all $x,y \in \mathbb{E}$ we have $\langle By, x\rangle = \langle Bx, y\rangle$.

  2. For all $x\in \mathbb{E}$ we have $\langle Bx, x\rangle \geq 0$.

  3. For each $x\in \mathbb{E}\setminus\{0\}$ we have $\langle Bx, x\rangle > 0$.

Then it can be shown that the map $u\colon \mathbb{E} \times \mathbb{E} \to \mathbb{R}$ defined by $u(x, y) = \langle Bx, y\rangle$ is an inner product on $\mathbb{E}$. Here is an outline of the steps.

  • That the map $u$ is bilinear follows from both the map $\langle \cdot , \cdot \rangle$ being bilinear and the map $B$ being linear.

  • Condition 1. implies that the bilinear form $u$ is symmetric.

  • Conditions 2. and 3. imply that $u(x, x) \geq 0$ for all $x\in \mathbb{E}$ and that for each $x\in \mathbb{E}$ we have $u(x, x) = 0$ if and only if $x = 0$.

Unfortunately, I am not aware of a nice condition to check whether a given linear operator $B\colon \mathbb{E} \to \mathbb{E}^{*}$ satisfies conditions 1. to 3. above.

$\endgroup$
4
  • $\begingroup$ It makes perfect sense to talk about a self-adjoint here. The natural definition of an adjoint in this context is the unique $B'$ such that $\langle Bx,y\rangle=\langle B'y,x\rangle$. This $B'$ has the same domain and codomain as $B$, so we can ask for a $B$ such that $B'=B$. We can also define adjoints for $\mathbb E\to\mathbb E$ and $\mathbb E^*\to\mathbb E^*$, but taking adjoints swaps between these two spaces so they cannot be "self-adjoint". $\endgroup$ Commented Nov 27 at 17:13
  • $\begingroup$ @NicholasTodoroff Thanks for the comment. The issue is that the definition for an adjoint operator you are proposing in this setting conflicts with the usual definition of an adjoint operator, which should map $\mathbb{E}^{*}$ into $\mathbb{E}$, and consequently does not appear to be that natural. $\endgroup$ Commented 2 days ago
  • $\begingroup$ You're thinking of a situation where we have both domain and codomain equipped with bilinear forms. That's not what we have, we just have a single bilinear pairing. Anyway, there is no conflict whatsoever. Let $\mathfrak E=\mathbb E^*\oplus\mathbb E$, and on this space define a bilinear form $\langle\phi+x,\psi+y\rangle=\phi(y)+\psi(x)$ where $\phi,\psi\in\mathbb E^*$ and $x,y\in\mathbb E$. Note that this form comes from the natural isomorphism $\mathfrak E^*\cong\mathfrak E$. $\endgroup$ Commented 2 days ago
  • $\begingroup$ The map $B:\mathbb E\to\mathbb E^*$ naturally extends to a map $\mathfrak E\to\mathfrak E$, and it's adjoint $B':\mathfrak E\to\mathfrak E$ defined by $$\langle B(\mathfrak x),\mathfrak y\rangle=\langle\mathfrak x,B'(\mathfrak y)\rangle$$for all $\mathfrak x,\mathfrak y\in\mathfrak E$ satisfies $B'|_{\mathbb E^*}=0$ and $\operatorname{Im}(B')\in\mathbb E^*$, and so gives us precisely the map $B'|_{\mathbb E}:\mathbb E\to\mathbb E^*$ from before. $\endgroup$ Commented 2 days ago

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.