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I'm trying to prove the theorem, that states, that if I have a normed vector space with a finite dimension (so that each vector I can express as a linear combination $$ \vec{v}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n $$ where the vectors $\{\vec{e_k}\}_{k=1}^{n} $ are linearly independent), and I have some norm defined there, and I have a sequence of vectors $$ \{\vec{v_k}\}_{k=1}^{\infty} $$ that converges in norm to some vector $\vec{V}$, then it also converges to it Point-wisely, that is, if $$\vec{v}_k=a_k\vec{e}_1+b_k\vec{e}_2+c_k\vec{e}_3+\cdots+q_k\vec{e}_n, $$ and $$\vec{V}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n, $$ and $$\lim_{k\rightarrow\infty} \lVert\vec{v}_k-\vec{V}\rVert=0,$$ then also $$\begin{align*} \lim_{k\rightarrow\infty} |a_k-a|&=0,\\ \lim_{k\rightarrow\infty} |b_k-b|&=0,\\ &\,\vdots\\ \lim_{k\rightarrow\infty} |q_k-q|&=0. \end{align*}$$


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up vote 4 down vote accepted

Apart from the given norm $\|\cdot\|$, we can define another norm by $$\|\sum_{j=1}^k \lambda_je_j\|_{\infty}=\max_{1\le j\le k} |\lambda_j|$$ where the $\lambda_j$ are scalars. (You should check that this is a well-defined norm.) Any two norms on a finite-dimensional vector space are equivalent, so in particular there is a constant $C>0$ so that $\|x\|_{\infty}\le C\|x\|$ for all vectors $x$. This inequality shows that if $\|v_k-V\|\to 0$, then we must have $\|v_k-V\|_{\infty}\to 0$ as $k\to\infty$, which implies that each component of $v_k-V$ converges to $0$, or that each component of $v_k$ converges to the corresponding component of $V$.

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Luckily, there is quite a nice answer (with links to some sites containing proofs) proving the equivalence of any norm on a finite-dimensional space with the sup-norm. Also, the constant $C$ is missing in your inequality $\|x\|_{\infty} \leq \|x\|$ which should read $\|x\|_{\infty} \leq C\|x\|$. – t.b. May 17 '11 at 17:17
@Theo: thanks, I've inserted the missing $C$ and a link to a proof. – mac May 17 '11 at 17:24
Thanks to Theo Buehler, the link you have given, had the answer. – Maxim May 17 '11 at 19:41
Seems link is broken, should probably be… – Matifou Jan 24 '15 at 23:31

You have $$ \| \vec{v_k} - \vec V \|^2 = |a_k - a|^2 + |b_k - b|^2 + ... + |q_k - q|^2 \ge |a_k - a|^2 $$ for instance, so that $0 \le |a_k - a| \le \| \vec{v_k} - V \| \to 0$, so that you can use a so-called sandwich theorem to conclude that your limit is zero. With a better notation for the components of $v_k$, something like $v_{k_i}, 1 \le i \le n$, the proof would be more elegant though.

Note that this statement is not just an implication, it's an equivalence : a vector sequence in a finite dimensional space whose components converge is always converging to the vector with components being the limits of the components of the sequence, i.e. $$ a_k \to a, b_k \to b, \dots, q_k \to q \quad \Longleftrightarrow \quad v_k \to V. $$

EDIT : Oh well. This isn't much general, I admit, 'cause I tend to spare the details when it's getting late, so I didn't notice you were in a context of a general norm. The only way to get an upper bound function of $\|v_k - V\|$ is to use norm equivalence on finite-dimension space, so I guess I can't make any better than what's been said.

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The first equation doesn't hold in general. For example, take the normed space $\mathbb{R}^2$ with the sup norm $\|(x,y)\|=\max\{|x|,|y|\}$, and the standard basis $(1,0),(0,1)$. – mac May 17 '11 at 17:07
Oh, didn't notice the norm wasn't euclidian. I'll edit my proof. – Patrick Da Silva May 18 '11 at 1:53
Thanks anyway Patrick :) – Maxim May 21 '11 at 21:20

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