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Uniform distance: $|f-g|_A= \sup \{ f(x)-g(x), x \in A\}, f(x)-g(x) \geq 0 \forall x \in A$

Find the uniform distance of $f(x)=x, g(x)=1 \forall x \in \mathbb{R}$

My attempt is to take cases for $x$.

$x \geq 1: h(x)=x-1 \geq 0, h'(x)=1$,so $h$ increasing, $\sup \{h(x) | x \geq 1\}= \lim_{x \to +\infty} (x-1)=+\infty$

$x \leq 1:h(x)=1-x \geq 0, h'(x)=-1<0$,so $h$ decreasing, $\sup \{h(x) | x \leq 1 \}=\lim_{x \to -\infty} (1-x)=+\infty$

uniform distance=$+\infty$

Make my thoughts sense?

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    It is standard to define the uniform distance as $$ | f-g |_A = \sup{ |f(x) - g(x)| : x \in A} $$ which doesn't require the inequality $g \leq f$. – Tom Jun 24 '14 at 13:13
  • What do I have to do to find $\sup {|x-1|: x \in \mathbb{R} }$ ? –  Jun 24 '14 at 13:17
  • Perhaps you can notice that for every $n \in \Bbb{N}$ by taking $x = n+1$ you get $|x-1| = n$. From this you know that $\sup{|x-1| : x \in \Bbb{R}} \geq |(n+1)-1| = n$. Since $n$ was arbitrary... – Tom Jun 24 '14 at 13:28

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You overcomplicated your answer, if you ask me. For $x\geq 1$, it is clear that $\displaystyle\lim_{x\to\infty}x-1 = \infty$ and there is no need for calculating $h'$.

As for $x\leq 1$, since for $x\leq 1$ the statement $f(x)-g(x)$ does not hold, there is some confusion. In your definition, actually, you demand that $f(x)-g(x)\geq 0$ for all $x\in A$ which is not true for $f$ and $g$ in your case.

5xum
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  • Can we not find the uniform distance for $x \leq 1$ ? –  Jun 24 '14 at 13:08
  • Is it better to take the absolute value of $f-g$,or does it make no sense ? –  Jun 24 '14 at 13:11
  • I am saying that your definition of the distance is confuzing. You say that it is the supremum of $f(x) - g(x)$ over all $x\in A$ where $f(x)-g(x) \geq 0$ for all $x\in A$. But what about if "$f(x)-g(x) \geq 0$ for all $x\in A$" does not hold? – 5xum Jun 24 '14 at 13:12
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    In the textbook,at the definition it is $|f(x)-g(x)| \geq 0 \forall x \in A$,the professor might accidentally have forgotten the absolute value. –  Jun 24 '14 at 13:15
  • Can I find the uniform distance,with the new definition? –  Jun 24 '14 at 13:16
  • That is probably the case. In any case, it is trivial to see that the value $|x-1|$ can be arbitrarily large for large (or small) values of $x$ so the supremum is $\infty.$ – 5xum Jun 24 '14 at 13:16
  • So,do we take the limit $x \to +\infty$ and $x \to -\infty$ or only $x \to +\infty$? –  Jun 24 '14 at 13:18
  • It is enough to show that one of the limits is $\infty$, since $\sup A \leq \sup B$ for $A\subseteq B$. – 5xum Jun 24 '14 at 13:21
  • I didn't get it.What is $A$ and $B$ in the case now? –  Jun 24 '14 at 13:23
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    $[0,\infty)$ and $\mathbb R$. – 5xum Jun 24 '14 at 13:23
  • In general,do we always take only $\lim_{x \to +\infty} |f-g|$ to find the uniform distance? –  Jun 24 '14 at 13:26
  • Absolutely not. For example, the uniform distance between $\sin x$ and $\cos x$ exists, while the limit of $|\cos x - \sin x|$ does not. – 5xum Jun 24 '14 at 13:28
  • A ok....Thanks! –  Jun 28 '14 at 19:45