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Example 1

Given $$C_{0}(\mathbb{R}^{n})=\{f\in C(\mathbb{R^n} \ | \ \ \exists R \ge 0 \ \text{such that } f(x)=0 \ \text{for} \ ||x||\ge R \}$$ and $$||f(x)||_{\infty} = \max_{x\in R^n}|f(x)| $$

What exactly is : $||x||$ in this context? Is it : $||x||_\infty = \max |x|$ ? If the $||_||\infty$ norm is defined for functions? Is it meant for $f(x)=x$?

Example 2

Given $$C^\alpha ( \mathbb{R}^\alpha)=\{ f\in B(\mathbb{R}^n)\ \ | \ \sup_{x,y \in \mathbb{R}^n , x\ne y} \frac{|f(x)-f(y)|}{||x-y||^\alpha} \}$$

and for $$f\in C^\alpha(\mathbb{R^n}) :||f||_\alpha = \sup_{x\in \mathbb{R}^n}|f(x)| + \sup_{x,y\in \mathbb{R}^n. x\ne y} \frac{|f(x)-f(y)|}{||x-y||^\alpha} < \infty$$

What exactly is $||x-y||^\alpha$ ? Is it : $||x-y||_\alpha ^\alpha = (\sup_{x\in \mathbb{R}^n}|x-y| + \sup_{x,y\in \mathbb{R}^n. x\ne y} \frac{|f(x)-f(y)|}{||x-y||^\alpha} )^\alpha < \infty$ ???

Why are the norms defined for functions and then only used for vectors??

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1 Answer 1

up vote 1 down vote accepted

Your $x \in \mathbb{R}^n$ so $\| x \|$ is a norm on $\mathbb{R}^n$. All norms on finite dimensional spaces are equivalent, so it doesn't matter much which one you take. The most common choice would be the Euclidean norm $\|x\| = (x_1^2 + \cdots x_n^2)^{1/2}$.

For your edited example 2 $\|x-y\|^\alpha$ almost certainly means the Euclidean norm raised to $\alpha$, i.e. $$\|x-y\|^\alpha = \big( (x_1-y_1)^2 + \cdots + (x_n - y_n)^2 \big)^{\alpha/2}$$

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Thank you very much! – bakabakabaka Jan 22 '13 at 12:49

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