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10

Here's the way I like to think about it. I'll start with the finite dimensional space $\Bbb{R}^n$ because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity $z^Tx$ represents a linear functional on $\Bbb{R}^n$, that is a linear function which eats a vector and spits out a real number: $$ ...


4

The dual space is a space of linear functionals. If we want to define a norm on the dual space, we do what we always do to measure the "size" of a linear transformation: we use an operator norm. Alternatively, the dual norm of $z$ is the matrix norm of the matrix $z^T$.


4

Take a real matrix $A \in \mathbb{R}^{n \times m}$ for example. Now let this matrix represent an input-output relation like $y = Au$. The $p$-norm of $A$ is $\sup\limits_{||u||_p \neq 0}\frac{||y||_p}{||u||_p}$ where $||v||_p$ denotes the $p$-norm of vector $v$. In other words, it is a measure of amplification of the input.


4

There are actually multiple ways to assign a norm to a matrix, in fact there are multiple ways to give a norm to a vector. With vectors in $\mathbb{R}^n$ the choice that is most "geometrically appealing" is the Euclidean one $$\|(x_1,...,x_n)\| = \sqrt{x_1^2 + ... + x_n^2}$$ However, there are others, I'd advise looking up $\ell^p$-norms. One "obvious" ...


2

If $k_1,\dots,k_p$ are kernels with RKHSs $\mathcal{H}_i$, then $k=\sum_i k_i$ is indeed a positive definite kernel again without further assumptions. The relation $\lVert f\rVert_{\mathcal{H}}^2 = \sum_{i=1}^{p}\lVert f\rVert_{\mathcal{H}_i}^2$ holds automatically for all functions that are in the intersection of all participating RKHSs. There is no ...


2

Dual norm is a particular case of the support function, specifically it is the support function of the unit ball of the original norm. When the unit ball is smooth enough $\|z\|_*$ is the Euclidean distance from the origin to the hyperplane with the normal vector $z$ (of unit Euclidean length) tangent to the ball. The equation of this hyperplane is ...


2

A common use is in topologizing certain function spaces. For example, let $\Omega \subset \mathbb{R}$ be open. We want a topology on $C^\infty(\Omega)$ the collection of smooth functions. Intuitively we'd like convergence $f_n \to f$ to imply the convergence locally of all derivatives. So what we do is we use a family of semi-norms to do it. $\|f\|_{K,n}$ is ...


2

Hint: If you are given a proof for those estimates, look at them. The proofs contain some other inequalities that you may be more familiar with. When does equality hold in them? A different kind of hint: What special kind of vectors can you come up with in $\mathbb R^n$? What are some simple nonzero vectors? The first inequality is not optimal and thus ...


1

To simplify my answer, I'll ignore the "continuous" requirement and assume there is an appropriate inner product for that norm. Let $b$ be a real number, $$f(x) = \left\{ {\begin{array}{*{20}{c}} {1,}&{x = 0} \\ {0,}&{x \ne 0} \end{array}} \right.$$ and $$g(x) = \left\{ {\begin{array}{*{20}{c}} {1,}&{x = 2} \\ {0,}&{x \ne 2} ...


1

The picture is particularly simple if you consider a positive self-adjoint matrix. In this case the operator norm is the largest eigenvalue. From the point of view of quantum mechanics, this is the largest possible outcome for the corresponding variable in any given experiment.


1

Use the definition. If $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n x_k^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n x_k^2 ,$$ then $$\frac{\partial}{\partial x_j}f(x) =\frac{\partial}{\partial x_j}\sum_{k=1}^n x_k^2=\sum_{k=1}^n \underbrace{\frac{\partial}{\partial x_j}x_k^2}_{\substack{=0, \ \text{ if } j \neq k,\\=2x_j, \ \text{ else }}}= 2x_j.$$ It follows ...


1

It looks fine. I think the denseness is used implicitly when you want to show that $\|\cdot\|_0$ is a norm.


1

Some things to check: Is it zero at the origin? Is it positive everywhere else? Is it homogeneous, meaning that $\|(tx,ty)\| = t\|(x,y)\|$ for all $t>0$ and all $x,y\in\mathbb R$? If the answers to those questions are yes (and they are) we have to deal with the hard part: the triangle inequality. Luckily, the homogeneity makes it easier: The ...


1

We have to maximize $$\|A{\bf x}\|^2=\langle A{\bf x}, A{\bf x}\rangle=\langle A^*A\,{\bf x},{\bf x}\rangle$$ subject to the condition $\|x\|^2=1$. The map $A^*A$ , resp. its matrix $[A^*A]$, is symmetric positive definite. Therefore there exists an orthonormal basis of ${\mathbb R}^n$ with $$[A^*A]={\rm diag}(\lambda_1,\lambda_2,\ldots,\lambda_n),\qquad ...



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