📂Banach Space

Proof of Riesz's Theorem

Theorem¹

Let the scalar field of a normed space $(X , \left\| \cdot \right\|)$ be $\mathbb{C}$. Then

$X$ is finite-dimensional. $\iff$ $\overline{ B ( 0 ; 1 ) }$ is compact.

Explanation

$\overline{ B ( 0 ; 1 ) } := \left\{ x \in X : \| x \| \le 1 \right\}$ denotes the closed unit ball. According to Riesz’s theorem, to determine whether the entire space is finite-dimensional, it suffices to check only a very small region. Usually one merely thinks of examples of finite-dimensional normed spaces without ever pondering a necessary and sufficient condition for them, which makes this a truly mathematical theorem.

Proof

Strategy: We give a simple homeomorphism from the tractable $\mathbb{C}^{n}$ to $X$, carrying the compactness of $\mathbb{C}^{n}$ over to $X$. For the converse direction, based on the compactness of $\overline{ B ( 0 ; 1 ) }$, we construct a certain finite-dimensional vector space and then show that it actually contains $X$.

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• $(\implies)$

  If $\dim X = n$, then there exists a basis $\left\{ e_{1} , \cdots , e_{n} \right\}$ of $X$. With respect to it, define the function $f : ( \mathbb{C}^{n} , \| \cdot \|_{1} ) \to (X , \| \cdot \| )$ by $f(\lambda_{1} , \cdots , \lambda_{n} ) : = \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n}$; then $f$ is a continuous bijection.

  The closed unit ball in $\mathbb{C}^{n}$, $\overline{ B_{ \| \cdot \|_{1} } ( 0 ; 1 ) } = \left\{ (\lambda_{1} , \cdots , \lambda_{n}) \in \mathbb{C}^{n} \ | \ | \lambda_{1} | + \cdots + | \lambda_{n} | \le 1 \right\}$, is compact by the Heine-Borel theorem. Since $f$ is continuous, $f \left( \overline{ B_{ \| \cdot \|_{1} } ( 0 ; 1 ) } \right)$ is also compact.

  Meanwhile, since $\| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \| \le | \lambda_{1} | + \cdots + | \lambda_{n} |$, we have $\overline{ B ( 0 ; 1 ) } \subset f \left( \overline{ B_{ \| \cdot \|_{1} } ( 0 ; 1 ) } \right)$. Since $\overline{ B ( 0 ; 1 ) }$ is a closed subset of the compact set $f \left( \overline{ B_{ \| \cdot \|_{1} } ( 0 ; 1 ) } \right)$, it is compact.

• $(\impliedby)$

  Let $0 < \varepsilon < 1$.

Since $\overline{ B ( 0 ; 1 ) }$ is compact, for the open cover $\displaystyle \bigcup_{x \in \overline{ B ( 0 ; 1 ) } } { B \left( x ; \varepsilon \right) }$ there exists a finite subcover satisfying $\displaystyle \overline{ B ( 0 ; 1 ) } \subset \bigcup_{i=1}^{m} B \left( x_{i} ; \varepsilon \right)$. With respect to it, let $M := \text{span} \left\{ x_{1} , \cdots , x_{n} \right\}$.

$\displaystyle \overline{ B ( 0 ; 1 ) } \subset \bigcup_{i=1}^{m} B \left( x_{i} ; \varepsilon \right)$ means, in other words, that $\displaystyle \overline{ B ( 0 ; 1 ) } \subset \bigcup_{ m \in M } B \left( m ; \varepsilon \right)$ holds. Since $ m \in \text{span} \left\{ x_{1} , \cdots , x_{n} \right\}$ from the very beginning, no matter how small the diameter $\varepsilon$ of the balls is taken, the above inclusion continues to hold. Therefore, for $k \in \mathbb{N}$,

$$ \overline{ B ( 0 ; 1 ) } \subset \bigcup_{ m \in M } B \left( m ; \varepsilon \right) \subset \bigcup_{ m \in M } B \left( m ; \varepsilon^2 \right) \subset \cdots \subset \bigcup_{ m \in M } B \left( m ; \varepsilon^k \right) $$

Now, considering an arbitrary nonzero vector $x \in X$, for some $y_{k} \in M$, $\displaystyle z_{k} : = B ( 0 ; \varepsilon^k )$,

$$ {{ x } \over { \| x \| }} = y_{k} + z_{k} $$

As $k \to \infty$, $z_{k} \to 0$, so

$$ y_{k} = {{ x } \over { \| x \| }} - z_{k} \to {{ x } \over { \| x \| }} \in \overline{ M } = M $$

That is, $x \in M$, so $X \subset M$; but since $M \subset X$,

$$ X = \text{span} \left\{ x_{1} , \cdots , x_{n} \right\} $$

Therefore, $X$ is a finite-dimensional vector space.

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See Also

Generalization from Euclidean Space

Riesz’s theorem points out the compactness of the closed unit ball $\overline{B (0;1)}$ in a normed space as an equivalent condition for being finite-dimensional. In Euclidean space, the $k$-cell $[0,1]^{k}$ is compact, and a homeomorphism with the closed unit ball exists, so Riesz’s theorem can be viewed as a generalization of the compactness of the $k$-cell.