4
$\begingroup$

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 $$\begin{align*} \lim_{k\rightarrow\infty} |a_k-a|&=0,\\ \lim_{k\rightarrow\infty} |b_k-b|&=0,\\ &\,\vdots\\ \lim_{k\rightarrow\infty} |q_k-q|&=0. \end{align*}$$

Thanks.

$\endgroup$
6
$\begingroup$

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 $V$.

$\endgroup$
4
  • 1
    $\begingroup$ 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 $\|x\|_{\infty} \leq C\|x\|$. $\endgroup$ – t.b. May 17 '11 at 17:17
  • $\begingroup$ @Theo: thanks, I've inserted the missing $C$ and a link to a proof. $\endgroup$ – mac May 17 '11 at 17:24
  • $\begingroup$ Thanks to Theo Buehler, the link you have given, had the answer. $\endgroup$ – Maxim May 17 '11 at 19:41
  • 1
    $\begingroup$ Seems link is broken, should probably be planetmath.org/… $\endgroup$ – Matifou Jan 24 '15 at 23:31
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $(1,0),(0,1)$. $\endgroup$ – mac May 17 '11 at 17:07
  • $\begingroup$ Oh, didn't notice the norm wasn't euclidian. I'll edit my proof. $\endgroup$ – Patrick Da Silva May 18 '11 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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