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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 $\lim_n\|T_n\|=0$.

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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 $\mathbb{R}$.

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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
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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 math.stackexchange.com/questions/325702/… –  ctlaltdefeat Mar 10 '13 at 20:31
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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 $$\lVert T_n(x)\rVert\leqslant \max_{1\leqslant j\leqslant N}\lVert T_n(v_j)\rVert.$$

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