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Let $\mathcal{C}([0, 1])$ be the set of all continuous functions $f : [0, 1] \to \mathbb{R}$. For $f, g \in \mathcal{C}([0, 1])$, define $d(f, g) = \max|f(x) − g(x)|$. Show that $d$ is a metric on $\mathcal{C}([0, 1])$.

Clarinetist
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Step199
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    What have you got so far, where are you stuck? – konewka Oct 16 '14 at 11:54
  • I am trying to understand metric spaces and the axioms involved.. this is all new to me. I can see how d(f,g) is >= 0 (non-negative). I can see how it is symmetric but I dont know how to show triangle inequality.. – Step199 Oct 16 '14 at 11:57
  • For $(f,g,h) \in \mathcal{C}([0,1],\mathbb{R})$ and $x \in [0,1]$, note that :

    $$ \vert f(x) - g(x) \vert \leq \vert f(x) - h(x) \vert + \vert h(x) - g(x) \vert.$$

    which follows from the triangle inequality. It should help you prove that $d$ satisfies to the triangle inequality.

     – pitchounet Oct 16 '14 at 12:03
  • Thanks for that.. how do I show that |f(x) - g(x)|<=|f(x) - h(x)| + |h(x) - g(x)| – Step199 Oct 16 '14 at 12:20
  • If you've proven the triangle inequality in $\mathbb{R}$, note $f(x) - g(x) = f(x)-h(x)+h(x)-g(x)$ and then use the triangle inequality for elements in $\mathbb{R}$. – Clarinetist Oct 16 '14 at 12:23

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If it is zero, functions are same, since they differ nowhere. Always positive otherwise. Triangle inequality comes from the triangle inequality in complex numbers by just putting max in front.

ploosu2
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  • Thank you for the response.. I'm not quite clear on how to proove the triangle inequality still. And why do complex numbers come into it? I am just looking at R2.. thanks – Step199 Oct 16 '14 at 12:03
  • I somehow saw C in there, but anyhow the numbers you have are a part of complex numbers ;) – ploosu2 Oct 16 '14 at 12:17