Equivalence of Norms in a Finite-Dimensional Linear Space

Equivalence of Norms in a Finite-Dimensional Linear Space

Definition: Let $X$ be a linear space and let $\| \cdot \|_1$ and $\| \cdot \|_2$ be norms defined on $X$. Then the norm $\| \cdot \|_1$ is said to be Equivalent to $\| \cdot \|_2$ if there exists real numbers $C, D \in \mathbb{R}$, $C, D > 0$ such that $C \| x \|_1 \leq \| x \|_2 \leq D \| x \|_1$ for all $x \in X$.

Note that if $\| \cdot \|_1$ is equivalent to $\| \cdot \|_2$ then:

(1)
\begin{align} \quad C \| x \|_1 \leq \| x \|_2 \leq D \| x \|_1 \quad \forall x \in X \end{align}

Therefore:

(2)
\begin{align} \quad \frac{1}{D} \| x \|_2 \leq \| x \|_1 \leq \frac{1}{C} \| x \|_2 \quad \forall x \in X \end{align}

So $\| \cdot \|_2$ is also equivalent to $\| \cdot \|_1$.

We will now prove a very important result which says that if $X$ is a finite-dimensional linear space then any two norms defined on $X$ are equivalent.

Theorem 1: Let $X$ be a finite-dimensional linear space with $\mathrm{dim} (X) = n$. Then any two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on $X$ are equivalent.
  • Proof: Let $\{ e_1, e_2, ..., e_n \} \subseteq X$ be a basis of $X$. Then $\mathrm{span} (e_1, e_2, ..., e_n) = X$. So for every $x \in X$ there exists $a_1, a_2, ..., a_n \in \mathbb{C}$ such that:
(3)
\begin{align} \quad x = a_1e_1 + a_2e_2 + ... + a_ne_n = \sum_{k=1}^{n} a_ke_k \end{align}
  • Define a norm $\| \cdot \|_{*} : X \to [0, \infty)$ for all $x \in X$ by:
(4)
\begin{align} \quad \| x \|_{*} = \left ( \sum_{k=1}^{n} | a_k |^2 \right )^{1/2} \end{align}
  • We first show that $\| \cdot \|_{*}$ is indeed a norm on $X$. Suppose that $x = 0$. Then since $\{ e_1, e_2, ..., e_n \}$ is linearly independent in $X$ we must have that the equation $x = 0 = a_1e_1 + a_2e_2 + ... + a_ne_n$ implies that $a_1 = a_2 = ... = a_n = 0$. So $\| 0 \|_{*} = 0$. Now suppose that $\| x \|_{*} = 0$. Then by definition, $\displaystyle{\sum_{k=1}^{n} | a_k|^2 = 0}$ which implies that $|a_k| = 0$ for each $k \in \{ 1, 2, ..., n \}$. So $a_1 = a_2 = ... = a_n = 0$. Thus $x = 0$. So $\| x \|_{*} = 0$ if and only if $x = 0$.
  • Now let $x \in X$ and let $\lambda \in \mathbb{C}$. If $\displaystyle{x = \sum_{k=1}^{n} a_ke_k}$ then $\displaystyle{\lambda x = \sum_{k=1}^{n} \lambda a_ke_k}$. So:
(5)
\begin{align} \| \lambda x \| = \left ( \sum_{k=1}^{n} | \lambda a_k|^2 \right )^{1/2} = \left ( \sum_{k=1}^{n} |\lambda|^2 |a_k|^2 \right )^{1/2} = \left ( |\lambda|^2 \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} = |\lambda| \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} = |\lambda| \| x \|_{*} \end{align}
  • Lastly, let $x, y \in X$ with $\displaystyle{x = \sum_{k=1}^{n} a_ke_k}$ and $\displaystyle{y = \sum_{k=1}^{n} b_ke_k}$. Then:
(6)
\begin{align} \quad \| x + y \| = \left ( \sum_{k=1}^{n} |a_k + b_k|^2 \right )^{1/2} \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} + \left ( \sum_{k=1}^{n} |b_k|^2 \right )^{1/2} = \| x \| + \| y \| \end{align}
  • Now, let $\| \cdot \|$ be any norm on $X$. We will show that $\| \cdot \|$ and $\| \cdot \|_{*}$ are equivalent.
  • Showing that there exists a $D \in \mathbb{R}$, $D > 0$ such that $\| x \| \leq D \| x \|_{*}$ for all $x \in X$: We let:
(7)
\begin{align} \quad M = \max_{1 \leq k \leq n} \| e_k \| \end{align}
  • Then for $\displaystyle{x = \sum_{k=1}^{n} a_ke_k}$ we have that:
(8)
\begin{align} \quad \| x \| &= \biggr \| \sum_{k=1}^{n} a_ke_k \biggr \| \\ & \leq \sum_{k=1}^{n} |a_k| \| e_k \| \\ & \leq \sum_{k=1}^{n} |a_k| M \\ & \leq M \sum_{k=1}^{n} |a_k| \quad (*) \end{align}
  • Recall the Cauchy-Schwarz inequality:
(9)
\begin{align} \quad \sum_{k=1}^{n} s_k \overline{t_k} \leq \left ( \sum_{k=1}^{n} |s_k|^2 \right ) \left ( \sum_{k=1}^{n} |t_k|^2 \right ) \end{align}
  • By setting $s_k = a_k$ and $t_k = 1$ we have that:
(10)
\begin{align} \quad \sum_{k=1}^{n} a_k \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \left ( \sum_{k=1}^{n} 1 \right )^{1/2} = \sqrt{n} \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \end{align}
  • So from $(*)$ we conclude that:
(11)
\begin{align} \quad \| x \| \leq M \sqrt{n} \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} = M \sqrt{n} \| x \|_{*} \end{align}
  • Showing that there exists a $C \in \mathbb{R}$, $C > 0$ such that $C \| x \|_{*} \leq \| x \|$ for all $x \in X$: We define a function $f : \mathbb{C}^n \to [0, \infty)$ for all $(x_1, x_2, ..., x_n) \in \mathbb{C}^n$ by:
(12)
\begin{align} \quad f(x_1, x_2, ..., x_n) =\biggr \| \sum_{k=1}^{n} a_ke_k \biggr \| \end{align}
  • Observe that $f(x_1, x_2, ..., x_n) \geq 0$ for all $(x_1, x_2, ..., x_n) \in \mathbb{C}^n$ and that $f(x_1, x_2, ..., x_n) = 0$ if and only if $(x_1, x_2, ..., x_n) = (0, 0, ..., 0)$. Also, $f$ is a continuous function since if $\epsilon > 0$ is given then choose $\displaystyle{\delta = \frac{\epsilon}{M \sqrt{n}}}$. Then if $\| \xi - \eta \| < \delta$ and $\xi = (a_1, a_2, ..., a_n), \eta = (b_1, b_2, ..., b_n) \in \mathbb{C}^n$ we have that:
(13)
\begin{align} \quad |f(\xi) - f(\eta)| &= \biggr \lvert \biggr \| \sum_{k=1}^{n} a_ke_k \biggr \| - \biggr \| \sum_{k=1}^{n} b_ke_k \biggr \| \biggr \rvert \\ & \leq \biggr \| \sum_{k=1}^{n} (a_k - b_k) e_k \biggr \| \\ & \leq M \sqrt{n} \| \xi - \eta \| \\ & < M \sqrt{n} \delta \\ & < \epsilon \end{align}
  • Now since the unit sphere in $\mathbb{C}^n$ is compact and $f$ is continuous, we have that there exists a $\delta > 0$ such that if $\| \xi \| = 1$ then $f(\xi) \geq \delta$, and so:
(14)
\begin{align} \quad \biggr \| \sum_{k=1}^{n} a_ke_k \biggr \| \geq \delta \end{align}
  • If $\xi \neq 0$ then $\| \xi \| = \| x \|_{*}$, and so:
(15)
\begin{align} \quad \frac{\biggr \| \sum_{k=1}^{n} a_ke_k \biggr \|}{\| x \|_*} \geq \delta \quad \Leftrightarrow \quad \biggr \| \sum_{k=1}^{n} a_ke_k \biggr \| \geq \delta \| x \|_* \quad \Leftrightarrow \quad \delta \| x \|_* \leq \| x \| \end{align}
  • So $\| \cdot \|$ is equivalent to $\| \cdot \|_*$. So any two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent. $\blacksquare$
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