# Tagged Questions

A vector space $E$, generally over the field $\mathbb R$ or $\mathbb C$ with a map $\lVert \cdot\rVert\colon E\to \mathbb R_+$ satisfying some conditions.

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### How do I link dimension of a normed vector space with closedness?

Let $X$ be a Normed Vector Space, for any $x\in X$ and $r>0$. Let $W:=\{y\in X : \|y-x\|\leq r\}$ and $S:=\{y\in X : \|y-x\|<r\}$ Prove: $W$ is closed if $\dim(X)<\infty$ I can't think of a ...
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### Show two norms are equivalent

Let $$N(z)=\Bigg(\sum_{n=1}^{\infty} \bigg|\frac{c_n}{n}\bigg|^3\Bigg)^{1/3}$$ be a norm on $\ell^3$ where $z=(c_n)_{n\geq1} \in \ell^3.$ Are the norms $N(\cdot)$ and $\|\cdot\|_3$ equivalent? I ...
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### For a normal operator is it true that $\|T^*T^2\| = \|T^3\|$?

For a normal operator is it always true that $\|T^*T^2\| = \|T^3\|$? See the accepted answer for the case in a Hilbert space Update: how about $\|T^*T^2\| = \|T\|^3$ in a Hilbert space
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### Can we detect smoothness of a norm by its behavior along paths?

We say a norm $\| \cdot \|$ on $\mathbb{R}^n$ is smooth if it is smooth as a function $\mathbb{R}^n\setminus \{0\} \to \mathbb{R}$. (i.e, after restricting the domain). We say a norm is smooth along ...
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### $X$ be Banach space with a proper dense subspace $Y$. Can the identity operator on $Y$ be extended to a continuous function from $X$ into $Y$?

Let $X$ be any Banach space with a proper dense subspace $Y$. Can the identity operator on $Y$ be extended to a continuous function from $X$ into $Y$ ?
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### Is it possible to extended finite rank continuous linear transformation to a continuous linear transformation with same range?

Let $X,Y$ be normed linear spaces , $W$ be a linear subspace of $X$ , let $T:W \to Y$ be a continuous linear tranformation with finite rank i.e. $T(W)$ is finite dimensional ; then can we extend $T$ ...
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### Proving linear operator is bounded

Prove that the formula $T(b_1,b_2,b_3,...,b_n,...) = (b_1, b_2/2 ,..., b_n/n ,...)$ deﬁnes a bounded linear operator $T : (ℓ^∞,∥·∥_∞)→(ℓ^∞,∥·∥_∞)$. Proving that it is linear is easy. Need help with ...
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### Weak convergence $\iff$ strong convergence in finite dimensional space

I am seeking a proof of the following claim. Weak convergence $\implies$ strong convergence in a finite-dimensional normed linear space. Thank you.
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### $X,Y$ be Banach , $T \in \mathcal B(X,Y)$ be onto ; then , for every sequence $y_n \to y \in Y$ , $\exists x_n \to x\in X$ s.t. $T(x_n)=y_n , T(x)=y$?

Let $X,Y$ be Banach spaces , $T:X \to Y$ be a surjective continuous linear transformation , then is it true that for every convergent sequence $\{y_n\}$ in $Y$ , converging to $y \in Y$ , there exist ...
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### What are the differences between $l^p$ space and $L^p$ space?

I always thought small $l^p$ space, or so-called Banach space, is the space equipped with a norm similar to vector norm as $$||x||_p = \left( \sum_{i\in\mathbb{N}} |x_i|^p \right)^{\frac{1}{p}}$$ ...
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### Isometry and isomorphism normed spaces

Problem. Let $X$, $Y$ be real normed vector spaces and $f$ isometry space $X$ in the space $Y$. Show that there is isomorphism $A$ spaces $X$ on the space $Y$ and vector $c \in Y$ such ...
Let $-\infty\lt a\lt b \lt \infty$. Let $C[a,b]$ is space of continuous functions and a function $\|.\|:C[a,b]\to R$ is given by $\|f\|:=\int_a^b |f(x)|d(x)$ a). Show that $\|\cdot\|$ is a norm in $C[... 1answer 60 views ### Characterizing orthogonally-invariant norms on the space of matrices Denote by$M_n$the space of$n \times n$real matrices. We say a norm on$M_n$is orthogonal invariant if: $$\|OX \|=\| XO\|=\|X \| \, \, \forall O \in O_n,X \in M_n$$ I am trying to characterize ... 0answers 26 views ### Find$\|B\|_{\ell^1 \rightarrow \ell^1} B(a_1,a_2,...)=(a_1/1,a_2/2,\ldots, a_n/n,\ldots) $This is a linear operator defined on$\ell^1$. Notice that$\|Bx\|=\sum |a_n/n| \le \sum |a_n| =\|x\|$. So$\|B\|=\sup_{\|x\|_1 \le 1} \|Bx\|...
I am trying to show that there exists a normed vector space which is isometric to a proper subspace of itself. I have been playing around with the $l^\infty$ norm on $\mathbb{N}$, but am struggling to ...