Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a finite-dimensional normed space and $T_n : X \to X$ a sequence of linear operators such that $\lim_nT_nx = 0$ for all $x$ in $X$. Prove that $\lim_n\|T_n\|=0$.

share|cite|improve this question
up vote 2 down vote accepted

First, we must understand that the sequence $(\|T_n\|)$ is bounded by, say, $C$, since this sequence of operators is pointwise bounded (Banach-Steinhaus theorem, here we use the fact that finite dimensional normed spaces are necessarily complete). Now, assume that the claim $\lim_{n\to \infty} \|T_n\|=0$ is not true, then for some subsequence (which I still denote by $(T_n)$) we have $\|T_n x_n\| \geq \varepsilon$ for some $\varepsilon$ (each $x_n$ is of norm $1$). By compactness (here we use the fact that the dimension is finite once again), we may assume that $(x_n)$ converges to some $x$ (after passing to further subsequence). We can now use triangle inequality to conclude $$ \|T_n x_n\| \leq \|T_n (x-x_n)\| + \|T_n x\| \leq C \|x_n -x\| + \|T_n x\|$$ and both terms converge to $0$, so we are done.

EDIT Here comes much better solution. Let $(e_1, \dots, e_n)$ be a basis of $X$. We have $T_n x = \sum_{k=1}^{n} x_k T_n e_k \leq \|(x_k)\|_2 \cdot \|(T_n e_k)\|_2$. Since all norms on finite dimesional space are equivalent this is less than $M \|x\|_X \cdot \|(T_n e_k)\|_2$ for some constant $M$ (dependent on dimension). This means that $\|T_n\| \leq M \|(T_n e_k)\|_2$ and this tends to $0$ from assumptions. Essentially, we just say that a sequence in $\mathbb{R}^{n}$ with any norm converges iff its coordinates converge in $\mathbb{R}$.

share|cite|improve this answer
By "compactness" you mean "locally compact" right? – ctlaltdefeat Mar 10 '13 at 20:03
I meant compactness of the unit ball. I also added another solution. – Mateusz Wasilewski Mar 10 '13 at 20:15
It seems that my edit coincides with Davide Giraudo's answer. – Mateusz Wasilewski Mar 10 '13 at 20:24
Thanks, nice! Also, I hope I'm not rude but I've got another question if you want to see :P… – ctlaltdefeat Mar 10 '13 at 20:31

Hint: given $N$ the dimension and $\{v_1,\dots,v_N\}$ a basis, we can assume that the norm is given by $\left\lVert \sum_{j=1}^Na_jv_j\right\rVert:=\max_{1\leqslant j\leqslant N}|a_j|$. Then if $x=\sum_{j=1}^Na_jv_j$ is in the unit ball, then $$\lVert T_n(x)\rVert\leqslant \max_{1\leqslant j\leqslant N}\lVert T_n(v_j)\rVert.$$

share|cite|improve this answer

Your Answer


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.