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let $X= C^1([0,1], \mathbb{R})$ be the space of real value contnious function defined , differentiable and having a continious derivative on $[0,1]$. define a function fro $X \times X$ into $\mathbb{R}^{+}$ by

$d(f,g) = \sup_{x \in [0,1]} | f'(x) - g'(x)|$ where $f '$ stand for the derivative of $f$ .

Is $d$ a metric on $X$?

My attempt :yes, d is ametric on $X$. I take $f(x) = x^2$ and $g(x) =x$. Then all metric properties are satisfied

Is its true?

Any hints/ guidenace solution will be appreciated

thanks u

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    $\begingroup$ What if $f\equiv 0$ and $g\equiv 1$? $\endgroup$
    – user140541
    Mar 3, 2019 at 8:55

2 Answers 2

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Concentrate on the question whether $d(f,g)=0$ implies $f=g.$

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    $\begingroup$ Yes, but if you take $f(x) = x$ and $g(x)=x+1$ the condition $d(f,g)=0$ is also satisfied, even though the functions are different. $\endgroup$ Mar 3, 2019 at 9:19
  • $\begingroup$ thanks u @Reiner now i understands $\endgroup$
    – jasmine
    Mar 3, 2019 at 9:22
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Another important point to make is that when you need to decide if something is a metric (or generally anything that has some "for all ..." qualifier), you need to actually consider all functions if you want to prove that is indeed true, while a single counterexample to one part is enough to prove it is not true.

That alone should make you understand that "it works for $f(x)=x^2, g(x)=x$" can never be proof for something that requires to work for all functions (with certain restrictions that leave more than those 2 functions in the considered set).

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