Linked Questions

3
votes
1answer
596 views

Operator norm. Alternative definition [duplicate]

Let $T\colon X\to Y$ be a linear operator with norm $$\|T\|=\sup_{\|x\|=1}\|Tx\|.$$ Prove that $$\|T\|=\sup_{\|x\|\leq 1}\|Tx\|.$$
0
votes
2answers
198 views

Show that two definitions of matrix norm are in fact equivalent $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} = \max\limits_{\|x\| = 1} \|Ax\|$ [duplicate]

I found a proof in an online course note which purports to show that two definitions of matrix norms are equivalent, however, I have some doubts regarding the proof, I would like a second pair of eyes ...
0
votes
1answer
431 views

Equivalent definition operator norm [duplicate]

Let $T: X \to Y$ be a bounded linear map between normed spaces. The operator norm is defined by $$\sup_{\|x\| = 1} \|T(x)\|$$ Is this equivalent to $$\sup_{x \in B(0, 1)} \|T(x)\|$$ where $B(0, 1)$...
1
vote
1answer
164 views

Why $\|S\|=\sup_{\|x\| \leq 1}\|Sx\|?$ [duplicate]

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. It is well known that if $S\in \mathcal{B}(F)$, then $$\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\...
2
votes
1answer
124 views

Why $\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?$ [duplicate]

Let $E$ be a complex Hilbert space. For $A\in\mathcal{L}(E)$, I want to show that $$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|=\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.$$ Clearly, since $$\{\|Ax\|...
0
votes
1answer
165 views

Matrix norm, equivalent definitions [duplicate]

I find two main definitions for the norm of a matrix or linear operator. First definition: $\lVert A\rVert := sup \{|A(\frac{x}{|x|})|: x\neq 0 \} $ Second definition: $\lVert A\rVert := sup \{|A(\...
0
votes
1answer
44 views

Understanding Matrix Norms [duplicate]

I'm trying to gain intuition for matrix norms. I'd like to know why if $A\in\mathbb{R}^{n\times n}$ that $||A||$ is equal to both $\max _{x\in\mathbb{R}^n}\frac{||Ax||}{||x||}$ and $\max_{||x||=1}||Ax|...
1
vote
0answers
27 views

Norms on vector spaces [duplicate]

I would appreciate it if you could give me a few hints as to how I should go about solving this problem. Suppose $T:\mathfrak{X}\rightarrow \mathfrak{Y}$ is bounded. Show that $\left\|T\right\|=\inf\{...
4
votes
2answers
5k views

Proving that the closure of a set contains the $\inf$ and $\sup$

I came across the following problem about closures: If $A$ is a bounded nonempty subset of $\mathbb{R}$, prove that $\sup A \in \overline{A}$ and $\inf A \in \overline{A}$. Proof. By hypothesis, $...
6
votes
2answers
447 views

Equivalent characterizations of the dual norm on finite dimensional vector spaces

In their book on Convex Optimization, Boyd and Vandenberghe state that given a norm, $||\cdot||$, defined on $\mathbb{R}^n$, the dual norm is defined as $$||z||_*= \sup \{ z^Tx : ||x|| \leq 1 \}$$ ...
4
votes
1answer
785 views

Infimum of a set is its limit point

Lemma: $A\subset\mathbb{R}, z=\inf(A)$, and $z \notin A$, then $z$ is an accumulation point of $A$. Here is my proof: Let $e>0$ be given. Consider $N(z,e)$. Note that $N(z,e)=(z-e,z+e)$. Since $...
2
votes
2answers
69 views

Do contraction semigroups admit exponential representation?

Given a Banach space $\mathcal N$, as contraction semigroup is defined as a set of bounded operators $P^t$, $0\le t\le+\infty$ defined everywhere in $\mathcal N$, such that \begin{equation*} P^0=1, \...
2
votes
1answer
113 views

Operator theory : Matrix Norm [duplicate]

Definition of Norm : Suppose $A$ be an $m\times n$ matrix. Then $\|A\|$ is defined by : \begin{align} \|A\|= &\sup\{\|Ax\| : \|x\|=1\}\\ = &\sup\{\|Ax\| : \|x\|\leq 1\} \end{align} These ...
1
vote
3answers
86 views

Why is operator norm defined the way it is?

Is there an intuition for the infimum definition of ${\| A \|}_{\mathrm{op}}$ without using a different, equivalent definition? I am referring to the definition, given an operator $A: W \rightarrow V$...
2
votes
2answers
40 views

Equivalent operator norm as $|\langle Au,v\rangle|$

Suppose $A$ is a bounded linear operator on Hilbert space $H$. We know that $\|A\|_{op} := \sup \{\|Au\| : u \in H,\ \|u\| = 1\}$. Curious to know if we can write this: $$\|A\|_{op} = \sup \{|\langle ...

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