infinity norm for continuous bounded function space

[[concept]]

Proposition

Then $C_{\infty }(X)$ is a vector space and we can define the norm $||u|{|}_{\infty }={\sup }_{x\in X}|u(x)|$on $C_{\infty }(X)$

Proof

NTS that this is indeed a norm by verifying each of the properties. identifiability and homogeneity are satisfied from the definition of the norm. It suffices then to show that the triangle inequality holds.

If $(u,v)\in C_{\infty }(X)$ then for all $x\in X$

\lvert u(x)+v(x) \rvert &\leq \lvert u(x) \rvert + \lvert v(x) \rvert \quad \text{by }\triangle \text{ inequality} \\ &\leq \lvert \lvert u \rvert \rvert_{\infty} + \lvert \lvert v \rvert \rvert_{\infty} \\ \implies \lvert \lvert u + v \rvert \rvert_{\infty } &=\sup_{x \in X} \lvert u(x) + v(x) \rvert \\ &\leq \lvert \lvert u \rvert \rvert_{\infty} + \lvert \lvert v \rvert \rvert_{\infty} \end{align}$$

NOTE

Convergence in this norm means

u_{n} \to u \text{ in }C_{\infty}(X) &\iff \lvert \lvert u_{n} -u \rvert \rvert_{\infty} \to 0 \text{ as } n \to \infty \\ &\iff \forall \epsilon > 0, \, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall x \in X, \lvert u_{n}(x) - u(x) \rvert < \epsilon \\ &\iff u_{n} \to u \text{ uniformly on } X \end{align}$$ Thus convergence in this [[Concept Wiki/distance\|metric]] is uniform convergence when the functions $u$ are bounded and continuous.

Mentions

Mentions

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