The C(K) Normed Linear Spaces

# The C(K) Normed Linear Spaces

Definition: Let $K$ be a compact metric space. We denote the Continuous Functions on $K$ Normed Linear Space by $C(K)$, which we define to be the set of all continuous real-valued (or complex-valued) functions defined on $K$ with the norm $\| \cdot \|_{\sup} : C(K) \to [0, \infty)$ defined for all $f \in C(K)$ by $\displaystyle{\| f \|_{\sup} = \sup_{x \in K} \{ |f(x)| \}}$. |

Observe that for each $f \in C(K)$, since $f$ is continuous and $K$ is compact then $f$ attains a maximum on $K$.

Proposition 1: If $K$ is a compact metric space then $(C(K), \| \cdot \|_{\sup})$ is a normed linear space. |

**Proof:**The set of all functions defined on $K$ is itself a linear space and $C(K)$ is a subset of this space, so it suffices to show that $C(K)$ is closed under addition, closed under scalar multiplication, and contains the zero function.

- Let $f, g \in C(K)$. Then $f + g$ is also continuous on $K$. Similarly, if $\alpha \in \mathbb{R}$ then $\alpha f$ is continuous on $K$. Lastly, the zero function is trivially continuous on $K$. Hence $C(K)$ is a linear space.

- We will now shows that $\| \cdot \|_{\sup}$ is a norm on $C(K)$.

**Showing that $\| f \|_{\sup} = 0$ if and only if $f = 0$:**Suppose that $\| f \|_{\sup} = 0$. Then $\sup_{x \in K} \{ |f(x)| \} = 0$ . So $|f(x)| \leq 0$ for all $x \in K$. Hence $f(x) = 0$ for all $x \in K$. Thus $f = 0$. Conversely, suppose that $f = 0$. Then $|f(x)| \leq 0$ for all $x \in K$. Thus $\| f \|_{\sup} = 0$.

- **Showing that $\| \alpha f \|_{\sup} = |\alpha| \| f \|_{\sup}$. Then:

\begin{align} \quad \| \alpha f \|_{\sup} = \sup_{x \in K} \{ |\alpha f(x)| \} = \sup_{x \in K} \{ |\alpha| |f(x)| \} = |\alpha| \sup_{x \in K} \{ |f(x)| \} = |\alpha| \| f \|_{\sup} \end{align}

**Showing that $\| f + g \|_{\sup} \leq \| f \|_{\sup} + \| g \|_{\sup}$:**Let $f, g \in C(K)$. Then:

\begin{align} \quad \| f + g \|_{\sup} = \sup_{x \in K} \{ |f(x) + g(x)| \} \leq \sup_{x \in K} \{ |f(x)| + |g(x)| \} = \sup_{x \in K} \{ |f(x)| \} + \sup_{x \in K} \{ |g(x)| \} = \| f \|_{\sup} + \| g \|_{\sup} \end{align}

- Therefore $\| \cdot \|_{\sup}$ is a norm on $C(K)$. Hence $(C(K), \| \cdot \|_{\sup})$ is a normed linear space. $\blacksquare$