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Let $X$ be a non empty set and let $C[a,b]$ denote the set of all real or complex valued continuous functions on $X$ with a metric induced by the supremum norm.

How to find open and closed balls in $C[a,b]$? Can we see them geometrically? For example what is an open ball $B(x_0;1)$ i.e. ball centered at $x_0$ with radius $1$ in $C[a,b]$. I can visualize them in $\mathbb R^n$ but when it comes to functional spaces I have no clue how to identify them?

Thanks for helping me.

Rudy the Reindeer
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Srijan
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    I assume $X$ is the interval $[a,b]$? Given a continuous function $f:[a,b]\to\Bbb R$ I would visualize the open ball of radius one around $f$ to be the set of all functions whose graphs exist in the strip between the graphs of $y=f(x)+1$ and $y=f(x)-1$. – anon May 22 '12 at 19:11

2 Answers2

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Yes, you should think of it just like you think of any other metric space. Every norm $\|\cdot\|$ induces a metric $d(x,y) := \|x-y\|$.

In your example, $$ B(x_0, 1) = \{ f: X \to \mathbb R \Big \vert \|f - x_0\|_\infty < 1 \}$$

In the $\sup$-norm, these $f$ are all functions that are never further away from $x_0$ at any given $x$ in $\mathbb R$. This is what it looks like: enter image description here

Rudy the Reindeer
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So I'm new here and sorry if I have this wrong, but apparently I'm not meant to respond to other answers; yet I also need 50 reputation to add a comment, so I can't do that? Anyway, I believe the above answers are incorrect. Just take the center of an open ball to be the zero function (radius 1) and $f$ to be $2/\pi \arctan(x)$. Then $f(x)$ is always within a distance of 1 from the center (strictly) in R, but $f$ is a distance of 1 from the zero function (exactually 1, so not in the open ball) in the function space, because the supremum of $\{f(x)-0 | x \in R\}$ is indeed 1. In fact f is on the boundry in the function space despite its values never touching 1 or -1 in R.

Neil Du Toit
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    OP's question was specifically about $C[a,b]$ not $C(\mathbb R)$ which indeed behaves a bit differently. – kahen Sep 06 '13 at 19:37
  • $\mathbb{R}$ behaves differently because it isn't bounded; hence it isn't compact. Since $[a,b]$ is compact, any function attains its supremum value somewhere on the interval, thus we won't run into this problem. – Caleb Stanford Sep 06 '13 at 19:48
  • I agree with the comments give above. – Srijan Sep 10 '13 at 10:23