Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{L}(\mathbb{R}^m,\;\mathbb{R}^k):=\{T:\mathbb{R}^m \longrightarrow \mathbb{R}^k : T$ is a linear transformation$\}$ the Vector Space of Linear Transformations with norm $\|T||=\sup\{\|Tx\|:\|x\|=1\}$.

Let $(T_n)_{n\ge1}$ a sequence in $\mathcal{L}(\mathbb{R}^m,\;\mathbb{R}^k)$ and $(u_n)_{n \ge 1}$ $\subset$ $\mathbb{R}^m$ such that :

  • $\forall n\ge 1, \|u_n\|=1$ and $u_n \to u$ (convergent)

  • $\forall n\ge 1, \|T_n\|\ge 1$

Then we can conclude that: $\liminf_{n \to \infty}\|T_n(u_n)\|=L>0$.

I think the proof is by contradiction, suppose $\liminf_{n \to \infty}\|T_n(u_n)\|=0$, but how to use the condition $\|T_n\|\ge 1$?

Any hints would be appreciated.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

It's not necessarily true: take $T_n:=\pmatrix{1/n&0\\0&n}$ and $u_n=u=\pmatrix{1\\ 0}$. The norm of $T_n$ is $n$ but $T_n(u_n)=\frac 1n u$ converges to $0$.

share|improve this answer

I think that what you stated above is not true:

  • Take $T\colon \mathbb R^2 \to \mathbb R$, $T(x,y)=y$; clearly $\lVert T \rVert=1$.
  • Put $T_n=T$ for each $n$.
  • Put $u_n=u=(1,0)$ for each $n$.

You have $u_n\to u$ and $T_n(u_n)=0$ for each $n$.

Are you sure that you formulated your questions correctly?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.