📂Linear Algebra

Homogeneity of Norms

Definition

On a vector space $V$, if for two norms $\left\| \cdot \right\|_{\alpha}, \left\| \cdot \right\|_{\beta}$ defined and for any vector $\mathbf{v} \in V$ the following $$c \left\| \mathbf{v} \right\|_{\alpha} \le \left\| \mathbf{v} \right\|_{\beta} \le C \left\| \mathbf{v} \right\|_{\alpha}$$ is satisfied with some constant $c , C >0$, the two norms are equivalent.

Theorem

Preservation of inequalities

• [1]: If norms $\left\| \cdot\right\|_{\alpha}$ and $\left\| \cdot \right\|_{\beta}$ defined on vector space $V$ are equivalent, then for all $\mathbf{x}, \mathbf{y} \in V$, the following holds. $$\left\| \mathbf{x} \right\|_{\alpha} \le \left\| \mathbf{y} \right\|_{\alpha} \implies \left\| \mathbf{x} \right\|_{\beta} \le \left\| \mathbf{y} \right\|_{\beta}$$

Equivalence of norms defined in a complex space

• [2]: All norms defined on the vector space $\mathbb{C}^n$ are equivalent.

Explanation

For two norms to be equivalent means that it is possible to use different norms without issue when dealing with inequalities using norms. Naturally, one can think of utilizing this by converting a hard-to-use norm into an easier one.

Explained in mathematical terms, the equivalence occurs when one norm can be expanded or contracted to be bigger or smaller than the other. This definition can be considered quite intuitive, recalling that the concept of norms comes from length. It would be convenient to think that regardless of the original unit of measurement for length, it can be utilized in comparisons by expanding or contracting it.

Proof

[1]

Given equivalent norms $\left\| \cdot\right\|_{\alpha}$ and $\left\| \cdot \right\|_{\beta}$, let’s say $\left\| \mathbf{x} \right\|_{\alpha} \le \left\| \mathbf{y} \right\|_{\alpha}$.

Assuming $\left\| \mathbf{x} \right\|_{\beta} < \left\| \mathbf{y} \right\|_{\beta}$, according to the definition of equivalence, there must exist $C_{\mathbf{y}} , C_{\mathbf{x}}$ satisfying the following for all $\mathbf{x}, \mathbf{y} \in V$. $$\begin{align*} \left\| \mathbf{x} \right\|_{\alpha} \le & \left\| \mathbf{y} \right\|_{\alpha} \\ =& C_{\mathbf{y}} \left\| \mathbf{y} \right\|_{\beta} \\ < & C_{\mathbf{y}} \left\| \mathbf{x} \right\|_{\beta} \\ \le & C_{\mathbf{y}} C_{\mathbf{x}} \left\| \mathbf{x} \right\|_{\alpha} \end{align*}$$ However, if $\mathbf{y} = \mathbf{0}$ then $\mathbf{x} = \mathbf{0}$, therefore $$0 = \left\| \mathbf{x} \right\|_{\alpha} < C_{y} C_{x} \left\| \mathbf{x} \right\|_{\alpha} = 0$$ and there does not exist $C_{\mathbf{y}}$, $C_{\mathbf{x}}$ satisfying the above equation, contradicting the premise that $\left\| \cdot\right\|_{\alpha}$ and $\left\| \cdot \right\|_{\beta}$ are equivalent. Thus, we obtain the following result. $$\left\| \mathbf{x} \right\|_{\alpha} \le \left\| \mathbf{y} \right\|_{\alpha} \implies \left\| \mathbf{x} \right\|_{\beta} \le \left\| \mathbf{y} \right\|_{\beta}$$

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[2]

Strategy: Use the extreme value theorem to show both the existence of $c$ and $C$ at once.

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For $S = \left\{ z : \left\| z \right\|_{2} = 1 \right\}$, define the function $h : S \to \mathbb{R}$ as follows: $\displaystyle h(z) := {{ \left\| z \right\|_\beta } \over { \left\| z \right\|_\alpha }}$. Since $h$ is continuous on the compact set $S$, by the extreme value theorem, there exists $c, C$ satisfying $c \le h(z) \le C$, thus the following holds.

$$c \left\| z \right\|_{\alpha} \le \left\| z \right\|_{\beta} \le C \left\| z \right\|_{\alpha}$$

Meanwhile, since $h$ is a function defined by the ratio of norms, $h(z) > 0$, and its minimum value must also be $c>0$. Therefore, norms $\left\| \cdot \right\|_{\alpha}$ and $\left\| \cdot \right\|_{\beta}$ are equivalent.

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The Euclidean space $\mathbb{R}^n$ being a subspace of $\mathbb{C}^n$, its high utility goes without saying. Moreover, as a more generalized fact, norms defined in a finite-dimensional vector space are all equivalent.