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Let $C([a,b],\mathbb{R})$ denote the space of continuous functions from $[a,b]$ to the real numbers. For a function $f\in C([a,b],\mathbb{R})$ and $d\gt 0$, define

$$p_d(f) :=sup\{\lvert f(x)-f(y)\rvert : \lvert x-y\rvert =d,\ x,y \in [a,b]\}.$$

Does this define a family of seminorms on $C([a,b],\mathbb{R})$ indexed by $d$?

If yes, does it follow that $C([a,b],\mathbb{R})$ is locally convex? Which topology does this family induce on $C([a,b],\mathbb{R})$ (if any of the standard topologies)? Thanks.

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Can you tell us something about why you're interested? Also when you say "does it follow that $C([a,b],R)$ is locally convex?" - with respect to which topology do you mean? –  Anthony Quas Jun 8 '12 at 5:50
    
@Stan I put in some LaTeX, which I hope you do for further questions :) –  David Roberts Jun 8 '12 at 6:22
    
Notice that $d$ is for all intents and purposes bounded above by $b-a$, and I would guess that you could probably consider rational $d$ without loosing any information. –  David Roberts Jun 8 '12 at 6:25
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Yes, this is a family of seminorms, just check the triangle inequality plus multiplication by a scalar. So indeed your space is locally convex. But do you really need $|x−y|=d$, not $\le d$? –  Yulia Kuznetsova Jun 8 '12 at 8:39

1 Answer 1

The $\rho_d$ are indeed seminorms; this can be verified by direct calculations or by noting that

  1. $\sup\lbrace |g(x,y)|:|x-y|=d, x,y\in[a,b]\rbrace$ defines a seminorm on $C([a,b]^2,\mathbb{R})$ (as a supremum of function values on a set), and

  2. the map $F$, defined by $F(f)(x,y)=f(x)-f(y)$, is linear; from $C([a,b],\mathbb{R})$ to $C([a,b]^2,\mathbb{R})$.

Every topology induced by seminorms is locally convex because the seminnorm-balls are convex.

Finally, every constant function gets seminorm zero for all $d$, so this topology is not Hausdorff and therefore not equal to any standard topology on $C([a,b],\mathbb{R})$ that I am aware of.

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