# Is $d$ a metric on $X$?

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

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

Concentrate on the question whether $$d(f,g)=0$$ implies $$f=g.$$
• 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. Mar 3, 2019 at 9:19
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).