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Operator norm --- Remarks by Browder After Lemma 8.4 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,911
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...

I need some help in fully understanding some remarks by Browder after Lemma 8.4 pertaining to the "operator norm" Lemma 8.4 ...


The relevant text including Lemma 8.4 reads as follows:




Browder - 1 - Lemma 8.4 ... PART 1 .. .png
Browder - 2 - Lemma 8.4 ... PART 2 ... .png






Near the end of the above text we read the following:

" ... ... We leave it to the reader to show that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ... "



Could someone please demonstrate rigorously that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ...




Help will be much appreciated ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,694
Could someone please demonstrate rigorously that the sup and inf in (8.2) and (8.3) are actually max and min, i.e. that they are actually attained ... ...
I think that you are going to have to quote a theorem from earlier in the Analysis course in order to prove these results.

The function given by $f(\mathbf{v}) = |T\mathbf{v}|$ is a continuous function from $\Bbb{R}^n$ to $\Bbb{R}$. The set $\{\mathbf{v}\in \Bbb{R}^n:|\mathbf{v}|\leqslant1\}$ is a closed, bounded subset of $\Bbb{R}^n$. There is a theorem that a continuous function on a closed, bounded set is bounded and attains its bounds. That explains why the sup in (8.2) is attained.

As for (8.3), you can check that the inf in (8.3) is equal to the sup in (8.2). So if the sup in (8.2) is attained at some vector $\mathbf{v}$, then the inf in (8.3) is attained at that same vector..
 

steep

Member
Dec 17, 2018
51
the above approach using Compactness probably is the right answer for an analysis text

- - - - -
However, there is an alternative algebraic approach to (8.2) that I hope is of considerable interest to OP. First focus on $\big \Vert \mathbf v \big \Vert_2 = 1$ and consider
$0 \leq \sigma_n \leq \sigma_{n-1} \leq ... \leq \sigma_2 \leq \sigma_1$
then for $j\in\{1,2,..., n\}$ we have the point-wise bound

$\sigma_j^2 \leq \sigma_1^2$
which is preserved under rescaling by real non-negative numbers $w_j$ (in particular for convex combinations, where $\sum_{j=1}^n w_j = 1$)

so we have
$w_j \sigma_j^2 \leq w_j \sigma_1^2$
and summing over the bound
$\sum_{j=1}^n w_j \sigma_j^2 \leq \sum_{j=1}^n w_j \sigma_1^2 =\sigma_1^2 \cdot \sum_{j=1}^n w_j = \sigma_1^2$


(now to accommodate the 'more general' case of $\big \Vert \mathbf v \big \Vert_2 \leq 1$, insert a slack parameter $\sigma_{n+1} = 0 $ and $w_{n+1}\geq 0 $, where again $\sum_{j}w_j = 1$ and re-run the above argument)

- - - - -
what are these sigmas? They are the singular values of $\mathbf A$ and recall that one way of constructing them is to look at the eigenvalues of $\big(\mathbf A^* \mathbf A\big)$ (and then take square roots). Why should the eigenvalues of $\mathbf A^* \mathbf A$ come up? Because $\big \Vert \mathbf A\mathbf x \big \Vert_2^2 = \mathbf x^* \mathbf A^*\mathbf A \mathbf x$ and hopefully one recalls how to examine quadratic form problems....

There are a few more details that need filled in here, and I'd like to leave them as an exercise for the reader Peter

incidentally the trace norm is better known as a Frobenius norm (or Schatten 2 norm, in which case the operator norm is the Schatten $\infty$ norm)