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To start I want to say, this is for homework. Please don't give me the answer, I'm just looking for a little help.

Let $\mathbb{P}_1[-1,1] = \{f:[-1,1] \to \mathbb{R}: f(t)=a+bt\}$. I need to show that the function defined by:

$d(f,g) = \text{sup} \{|f(t)-g(t)|: -1 \leq t \leq 1\}$

is a metric. I have already done the easy parts of showing it's symmetric, and that is always non-negative, but I'm not sure how to approach the last part.

Thank you for your help.

Matthew Riley
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  • If it helps, this metric is actually given by a norm so you can write a little less – Calvin Khor Oct 09 '18 at 18:58
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    the function is continuous, so there exists a point $ x \in [-1,1] $ such that d(f,g) = |f(x) - g(x)|, now add and substract something and use triangle rule – Presage Oct 09 '18 at 18:59

1 Answers1

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Suppose

$$d(f,g) + d(g,h) < d(f,h)=\sup\{|f(t)-h(t)|:t\in [-1,1]\}$$

then there is a $t\in [-1,1]$ so that

$$|f(t) - h(t)| > d(f,g) + d(g,h) \ge |f(w) - g(w)| + |g(v) - h(v)|$$ for all $w, v \in [-1,1]$. And that includes all possible values of $w,v$ including....

Well, you said you want a hint and not the answer so I'll just stop talking now.

fleablood
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