Continuity of the Norm on Normed Linear Spaces

Recall from the Normed Linear Spaces page that a normed linear space is a linear space $X$ with a norm $\| \cdot \|$ defined on $X$, where $\| \cdot \| : X \to [0, \infty)$ satisfies the following properties:

• $\| x \| = 0$ if and only if $x = 0$.

• $\| \alpha x \| = | \alpha | \| x \|$ for all $\alpha \in \mathbb{R}$ (or $\mathbb{C}$) and for all $x \in X$.

• $\| x + y \| \leq \| x \| + \| y \|$ for all $x, y \in X$ (Triangle Inequality).

We will now prove a nice result which says that every norm on a normed linear space is a continuous function on $X$. We first prove the following property of norms on a normed linear space.

Lemma 1:Let $X$ be a normed linear space with norm $\| \cdot \|$. Then for all $x, y \in X$ we have that $| \| x \| - \| y \| | \leq \| x - y \|$.

• Proof: Let $x, y \in X$. Then by the triangle inequality for norms we have that:

(1)

\begin{align} \quad \| x \| = \| x - y + y \| \leq \| x - y \| + \| y \| \end{align}

• Therefore:

(2)

\begin{align} \quad \| x \| - \| y \| \leq \| x - y \| \quad (*) \end{align}

• Similarly we have that:

(3)

\begin{align} \quad \| y \| = \| y - x + x \| \leq \| y - x \| + \| x \| = - \| x - y \| + \| x \| \end{align}

• Therefore:

(4)

\begin{align} \quad \| x \| - \| y \| \geq - \| x - y \| \quad (**) \end{align}

• From $(*)$ and $(**)$ we conclude that:

(5)

\begin{align} \quad | \| x \| - \| y \| | \leq \| x \| - \| y \| \quad \blacksquare \end{align}

Theorem 2: Let $X$ be a normed linear space with norm $\| \cdot \|$. Then $\| \cdot \|$ is continuous on $X$.

• Proof: Let $\epsilon > 0$ be given. Let $\delta = \epsilon$. Then for all $x, y \in X$ with $\| x - y \| < \delta$ we have that:

(6)

\begin{align} \quad | \| x \| - \| y \| | \leq \| x - y \| < \delta = \epsilon \end{align}

• Therefore $\| \cdot \|$ is continuous on $X$. $\blacksquare$