Bounded Coordinates and Convergent Subsequences

Let \(V\) be a finite-dimensional normed space with basis \(\{e_1,\ldots,e_n\}\). Suppose \((v_j)\) is a bounded sequence in \(V\), and write

$$ v_j=x_{j,1}e_1+\cdots+x_{j,n}e_n. $$

We want to choose one subsequence along which every coordinate converges.

1. Bounded vectors have bounded coordinates.

Recall the coordinate norm

$$ \left\|\sum_{i=1}^{n}x_i e_i\right\|_0 =\sum_{i=1}^{n}|x_i|. $$

Since all norms on a finite-dimensional space are equivalent, there exists \(K>0\) such that \(\|v\|_0\le K\|v\|\) for every \(v\in V\).

Because \((v_j)\) is bounded, there is a constant \(M>0\) such that \(\|v_j\|\le M\) for every \(j\). Consequently,

$$ |x_{j,i}| \le\sum_{r=1}^{n}|x_{j,r}| =\|v_j\|_0 \le K\|v_j\| \le KM. $$

Thus each coordinate sequence \((x_{j,i})_{j=1}^{\infty}\) is bounded.

2. Choose a subsequence for which every coordinate converges.

By Bolzano–Weierstrass in \(\mathbb{R}\) or \(\mathbb{C}\), the first coordinate sequence has a convergent subsequence. Choose increasing indices \(j_k^{(1)}\) such that

$$ x_{j_k^{(1)},1}\to x_1. $$

Along these indices, the second coordinate sequence remains bounded. Choose a further subsequence, with indices \(j_k^{(2)}\), such that

$$ x_{j_k^{(2)},2}\to x_2. $$

The first coordinate still converges to \(x_1\) along this further subsequence, because every subsequence of a convergent sequence has the same limit.

Continue this process through the last coordinate. Since there are only finitely many coordinates, we obtain one sequence of indices \(j_k\) such that

$$ x_{j_k,i}\to x_i \qquad\text{for every }i=1,\ldots,n. $$

3. Convert coordinate convergence into norm convergence.

Define \(v=x_1e_1+\cdots+x_ne_n\). Then

$$ \begin{aligned} \|v_{j_k}-v\| &=\left\|\sum_{i=1}^{n}(x_{j_k,i}-x_i)e_i\right\|\\ &\le\sum_{i=1}^{n}|x_{j_k,i}-x_i|\,\|e_i\| \longrightarrow 0. \end{aligned} $$

The sum has finitely many terms, each tending to zero. Hence \(v_{j_k}\to v\) in the norm of \(V\).

Sequential Characterization of Compactness
A subset \(E\) of a metric space is compact if and only if it is sequentially compact. This means that every sequence in \(E\) has a subsequence converging to a point of \(E\).

In particular, a sequence in the closed unit ball has a convergent subsequence by the preceding argument. Its limit remains in the ball because the ball is closed. This proves compactness of the closed unit ball in finite dimensions.

Proof of Riesz’s Lemma

Riesz’s Lemma
Let \(V\) be a normed space, and let \(Y\subsetneq Z\subseteq V\) be linear subspaces with \(Y\) closed in \(Z\). For every \(\theta\in(0,1)\), there exists \(v\in Z\) such that \(\|v\|=1\) and \(\displaystyle\inf_{y\in Y}\|v-y\|\ge\theta\).

1. Choose a vector outside the closed subspace.

Since \(Y\) is a proper subspace of \(Z\), choose \(v_0\in Z\setminus Y\). Define its distance from \(Y\) by

$$ a=\inf_{y\in Y}\|v_0-y\|. $$

We claim that \(a>0\). If \(a=0\), then for every positive integer \(j\) there would exist \(y_j\in Y\) such that \(\|v_0-y_j\|\lt 1/j\). Thus \(y_j\to v_0\). Since \(Y\) is closed in \(Z\) and \(v_0\in Z\), this would imply \(v_0\in Y\), contradicting the choice of \(v_0\).

Also, \(a\) is finite because \(0\in Y\) and \(a\le\|v_0-0\|=\|v_0\|\). Therefore \(0\lt a\lt\infty\).

2. Choose a point sufficiently close to the infimum.

Fix \(\theta\in(0,1)\). Since \(a>0\) and \(\theta\lt 1\), we have \(a/\theta>a\). By the definition of the infimum, there exists \(y_0\in Y\) such that

$$ a\le\|v_0-y_0\|\lt\frac{a}{\theta}. $$

Indeed, if every \(y\in Y\) satisfied \(\|v_0-y\|\ge a/\theta\), then \(a/\theta\) would be a lower bound strictly greater than the infimum \(a\).

Set \(d=\|v_0-y_0\|\). We have \(d\ge a>0\), so we can define

$$ v=\frac{v_0-y_0}{d} =\frac{v_0-y_0}{\|v_0-y_0\|}. $$

Since \(v_0,y_0\in Z\) and \(Z\) is a linear subspace, \(v\in Z\). By absolute homogeneity,

$$ \|v\|=\frac{\|v_0-y_0\|}{d}=1. $$

3. Bound the distance from the normalized vector to every point of the subspace.

Let \(y\in Y\) be arbitrary. Then

$$ \begin{aligned} \|v-y\| &=\left\|\frac{v_0-y_0}{d}-y\right\|\\ &=\frac{1}{d}\|v_0-y_0-dy\|\\ &=\frac{1}{d}\|v_0-(y_0+dy)\|. \end{aligned} $$

Because \(Y\) is a linear subspace and \(y_0,y\in Y\), the vector \(y_0+dy\) belongs to \(Y\). By the definition of \(a\),

$$ \|v_0-(y_0+dy)\|\ge a. $$

Consequently,

$$ \|v-y\|\ge\frac{a}{d}>\theta, $$

where the last inequality follows from \(d\lt a/\theta\).

The same lower bound \(a/d\) holds for every \(y\in Y\). Taking the infimum therefore gives

$$ \inf_{y\in Y}\|v-y\| \ge\frac{a}{d} >\theta. $$

In particular, the distance is at least \(\theta\), as required.

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi