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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.

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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 $0$.

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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?

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