Topological Equivalence of Norms

A single vector space can carry many different norms. Two norms are equivalent when they induce the same topology — the same open sets, the same convergent sequences, the same continuous functions. In finite dimensions this never matters: all norms on $\mathbb{R}^n$ are equivalent. In infinite dimensions it matters enormously, and the choice of norm determines which operators are bounded and which sequences converge.

Definition 1 (Equivalent Norms)

Two norms $\|\cdot\|^1$ and $\|\cdot\|^2$ on a vector space $X$ are equivalent if there exist constants $a, b > 0$ such that

a \|x\|^1 \leq \|x\|^2 \leq b \|x\|^1 \quad \forall x \in X.

(1)

Equivalent norms induce the same underlying topology on $X$ : they generate the same open sets and hence the same notion of “close.”

Geometric intuition. The condition $a\|x\|^1 \leq \|x\|^2 \leq b\|x\|^1$ says the unit ball of one norm can be scaled to sit inside the unit ball of the other, and vice versa. In $\mathbb{R}^2$ the $\ell^1$ ball (a diamond), the $\ell^2$ ball (a circle), and the $\ell^\infty$ ball (a square) all have different shapes — yet each can be scaled to contain, or be contained in, any of the others. This mutual containment of balls is exactly equivalence of topologies.

Example 1 (Unit balls in $\mathbb{R}^n$ fit inside each other)

For $x \in \mathbb{R}^n$ the $\ell^p$ norms satisfy the pointwise inequalities

\|x\|_\infty \leq \|x\|_2 \leq \|x\|_1 \leq n\,\|x\|_\infty, \qquad \|x\|_2 \leq \sqrt{n}\,\|x\|_\infty, \qquad \|x\|_1 \leq \sqrt{n}\,\|x\|_2.

(2)

Writing $B_p = \{x : \|x\|_p \leq 1\}$ for the closed unit ball of $\|\cdot\|_p$ , the smaller the norm, the larger the ball. The inequalities above translate directly into the nested inclusions

B_1 \subset B_2 \subset B_\infty \subset \sqrt{n}\, B_2 \subset n\, B_1.

(3)

So the diamond, the disk, and the square each fit inside a rescaled copy of the others. Pictorially in $\mathbb{R}^2$ : the diamond $B_1$ is inscribed in the disk $B_2$ , which is inscribed in the square $B_\infty$ ; conversely, the square shrunk by $\sqrt{2}$ fits back inside the disk, and the disk shrunk by 2 fits back inside the diamond. Because the scaling factors $1, \sqrt{n}, n$ are finite, each pair of $\ell^p$ norms on $\mathbb{R}^n$ is equivalent.

Theorem 1 (Equivalence of Norms in $\mathbb{R}^n$ )

On $\mathbb{R}^n$ all norms are equivalent.

Proof 1

Remark 1 (Failure in Infinite Dimensions)

This theorem fails in infinite dimensions. For instance, on $\mathcal{C}^0([0,1])$ the norms $\|\cdot\|_\infty$ and $\|\cdot\|_1$ are not equivalent: a sequence of tall, narrow spikes can converge in $L^1$ without converging uniformly. The choice of norm — and hence the topology — matters fundamentally in infinite-dimensional analysis and determines which operators are bounded, which functionals are continuous, and which sequences converge.

The proof above breaks down exactly because the unit sphere of one norm need not be compact in the topology of another. Equivalence of norms on $\mathbb{R}^n$ and compactness of the unit ball are two sides of the same finite-dimensional coin.

Definition 1 (Equivalent Norms)

Example 1 (Unit balls in Rn\mathbb{R}^nRn fit inside each other)

Theorem 1 (Equivalence of Norms in Rn\mathbb{R}^nRn)

Proof 1

Remark 1 (Failure in Infinite Dimensions)

Example 1 (Unit balls in $\mathbb{R}^n$ fit inside each other)

Theorem 1 (Equivalence of Norms in $\mathbb{R}^n$ )