Does a sequence of vectors whose norms converge to zero necessarily converge to the zero vector?

Question

Let $V$ be an $N$-dimensional normed vector space, and let $(v_n)$ be a sequence of vectors such that $\lim (\left \| v_n \right \|)=0$. Given some basis $\{e^1,e^2,..., e^N \}$, we can write each $v_n$ as $a_1^{(n)}e^1+a_2^{(n)}e^2 +\dots + a_N^{(n)}e^N$.

Is it necessarily true that for each fixed $k \in \{1, 2, \dots, N\}$, $\lim_{n \to \infty} a_k^{(n)}=0$ ? If so, how can we prove this using only the axioms for a generic norm?

Answer 1

I thought the simplest way is that $\lvert\lvert v_n \rvert\rvert \rightarrow 0$ is the same as $\lvert \lvert v_n - 0 \rvert\rvert \rightarrow 0$, i.e. $v_n \rightarrow 0$.

EDIT 1: Let $v_n = a_1^ne_1 + \cdots + a_N^ne_N$. Then $\lvert\lvert v_n \rvert\rvert = \lvert\lvert \sum_{i=1}^N a_i^n e_i \rvert\rvert \leq \sum_{i=1}^N \lvert \lvert a_i^ne_i \rvert\rvert$ by the triangle inequality. Thus if $v_n \rightarrow 0$ component-wise, i.e. $a_i^n \rightarrow 0$, then $\lvert\lvert a_i^ne_i \rvert\rvert = \lvert a_i^n \rvert \lvert\lvert e_i \rvert\rvert \rightarrow 0$ and hence $\lvert \lvert v_n \rvert \rvert \rightarrow 0$, i.e. $v_n \rightarrow 0$ in norm.

On the other hand, if $v_n \rightarrow 0$ in norm but not all $a_i^n \rightarrow 0$, let $v = a_1e_1 + \cdots + a_Ne_N$ where $a_i = \lim_{n\to\infty} a_i^n$. Then $v$ is nonzero, since $(e_j)_{j=1}^N$ is a basis and at least one $a_i \neq 0$, and $\lvert \lvert v_n - v \rvert \rvert \leq \sum_{i=1}^N \lvert \lvert (a_i^n - a_i)e_i \rvert \rvert \rightarrow 0$ so $v_n \rightarrow v$, a contradiction since by continuity of the norm, $\lvert \lvert v_n \rvert \rvert \rightarrow \lvert \lvert v \rvert \rvert \neq 0$.

EDIT 2: The above assumes the limit exists, which may not be true. I'm not sure how to show the limit exists.

Answer 2

Here is a proof that uses no theorem on normed linear spaces: Let $x_n=(a_1^{(n)},a_2^{(n)},\cdots,a_N^{(n)})$ and $r_n$ be its (usual) norm in $\mathbb R^{N}$. If $\{r_n\}$ is unbounded there is a subsequence $\{r_n'\}$ that tends to $\infty$. Now $\frac {v_n'} {r_n'} \to 0$ and we get unit vectors $(b_1^{(n')},b_2^{(n')},\cdots,b_N^{(n')})$ (obtained by normalizing $(a_1^{(n')},a_2^{(n')},\cdots,a_N^{(n')})$) such that $\sum b_i^{(n')}e_i \to 0$. But any sequence of unit vectors in $\mathbb R^{N}$ has a subsequence which tends to unit vector. We end up with $\sum c_i e_i=0$ where $(c_1,c_2,\cdots,c_n)$ is a unit vector contradicting linear independence of $e_i$'s. Hence $\{r_n\}$ is bounded. Now extract a convergent subsequence and take the limit in the equation $v_n=\sum a_1^{(n)}e_i$ to see that the limit of the subsquence must be $(0,0,\cdots,0)$. Now you can apply this argument to subsequences of $(a_1^{(n)},a_2^{(n)},\cdots,a_N^{(n)})$ to conclude that every subsequence of this has a further subsequence converging to the zero vector. Hence $a_i^{(n)} \to 0$ for each $i$.

PS This approach can be used prove many of the basic theorem about finite dimensional normed linear spaces from scratch. Unfortunately it is not seen in text books. By this approach you can prove that finite dimensional subspaces are closed, that linear maps on finite dimensional spaces are continuous and all norms on finite dimensional spaces are equivalent.

Answer 3

To answer the question in the title (which is a little bit different than the question in the body of the OP):

Let's get back to the definitions! The benefit is that we don't need component, and the answer would work in infinite dimension.

In a normed vector space, we have a natural distance between vectors $v$ and $w$ given by $||v-w||$. Therefore, convergence in a normed vector space is defined as the convergence in the metric space induced by the norm, that

Definition $1$: Let $v_n$ be a sequence of vectors in a normed space $(V,||\cdot||)$. We say that $v_n$ converges to a vector $v$ in $V$ if, for any $\varepsilon >0$, there exists $N\in\mathbb N$ such that, if $n\geqslant N$, then $||v_n-v||<\varepsilon$.

In plain English, we can make $v_n$ as close as we want to $v$ (that is, we can make the distance between $v_n$ and $v$ as small as we want), provided that $n$ is large enough.

Now, consider the sequence $x_n=||v_n||$ of the norms of the vectors $v_n$. That's a sequence of real numbers so its convergence is defined by the "classic" (Calculus 1 or 2 course) convergence of sequences:

Definition $2$: Let $x_n$ be a sequence of real numbers. We say that $x_n$ converges to a real number $\ell$ if, for any $\varepsilon >0$, there exists $N\in\mathbb N$ such that, if $n\geqslant N$, then $|x_n-\ell|<\varepsilon$.

This time, we are using the distance between real numbers, that is the absolute value of their difference.

We are finally in position to answer the question in the title (and a little bit more):

Theorem: Let $v_n$ be a sequence of vectors in a normed space $(V,||\cdot||)$. Then $v_n$ converges to the zero vector $0$ if and only if its sequence of norms $x_n=||v_n||$ converges to $0$.

The proof is essentially that $|x_n-0|=|x_n|=|\;||v_n||\;|=||v_n||=||v_n-0||$. Indeed, assume that $x_n=||v_n||$ converges to $0$. We will show that $v_n$ has to converge to the zero vector. For, let $\varepsilon>0$. Since $x_n$ converges to $0$, there exists $N$ in $\mathbb N$ such that, if $n\geqslant N_1$, we have $|x_n|<\varepsilon$. Therefore, if $n\geqslant N$, we have also $||v_n||<\epsilon$, so the claim is proved.

It should be obvious that the proof of the converse is almost identical. I think that a source of confusion is that it seems "too easy".

Answer 4

Going off this question, is it possible for there to be an infinite sequence whose L2 norm goes to zero as N goes to infinity, but for every t in it's duration, the signal at t does NOT tend to zero?