Does max|f - g| Define a Metric?

[Streltsy]

Technically, this is not a homework question, since I solely seek an answer for self-indulgence.

Homework Statement

Example 1.1.4. Suppose f and g are functions in a space X = {f : [0, 1] → R}. Does
d(f, g) =max|f − g| define a metric?

Homework Equations

(1) d(x, y) ≥ 0 for all x, y ∈ X
(2) d(x, y) = 0 if and only if x = y
(3) d(x, y)=d(y, x)
(4) d(x, z) ≤ d(x, y) + d(y, z)

The Attempt at a Solution

So, from my understanding: for d(f, g) to define a metric on X, it has to satisfy all the given properties of a metric.
Well, my question is not necessarily whether d(f, g) defines a metric (though I wouldn't mind a proof of it); I was wondering if property (2) is satisfied.
Because in my pursuit of an understanding in topology, I stumbled across a compilation of notes, in which the note-taker mentions that the second property is not satisfied.
The reasoning is: that, "by considering two arbitrary functions at any point within the interval [0, 1]. If |f(x) − g(x)| = 0, this does
not imply that f = g because f and g could intersect at one, and only one, point."
However, I was wondering if that could also be said about d(f, g) =max|f − g|, which is the function being originally considered; since if d(f, g) = 0, then max|f − g|= 0, which means that for all points in [0,1], 0 ≤|f − g| ≤ max|f − g| = 0, or |f − g|= 0; which would further imply f = g.

 

[SqueeSpleen]

If f and g are bounded, then that's a metric.
I know that metric by the name "supremum distance" (It has to be a supremum, not maximum, because the maximum doesn't always exists).

 

[Streltsy]

Oh ok.
Would it be safe to say, then, that d(f, g) =max|f − g| does not define a metric on X for this particular case, because X is a set of functions that map [0,1] to R, and R is unbounded?

So one might be able to prove that "if d(f, g) =max|f − g| = 0, then f = g", but not the converse; that is "if f = g, then max|f − g|= 0", since max|f − g| might not even exist.

 

[jbunniii]

So one might be able to prove that "if d(f, g) =max|f − g| = 0, then f = g", but not the converse; that is "if f = g, then max|f − g|= 0", since max|f − g| might not even exist.

Yes, that's correct. Here is an example where the functions are bounded but the max still doesn't exist. Let
$$f(x) = \begin{cases}
x & \text{ if } 0 \leq x < 1 \\
0 & \text{ if } x = 1 \\
\end{cases}$$
and let $g(x) = 0$ for all $x \in [0,1]$. Then $|f - g| = f$ has no maximum value. Since $d(f,g)$ is not even defined for every choice of $f$ and $g$, it certainly can't be a metric.

 

[SqueeSpleen]

Oh ok.
Would it be safe to say, then, that d(f, g) =max|f − g| does not define a metric on X for this particular case, because X is a set of functions that map [0,1] to R, and R is unbounded?

So one might be able to prove that "if d(f, g) =max|f − g| = 0, then f = g", but not the converse; that is "if f = g, then max|f − g|= 0", since max|f − g| might not even exist.

Sorry, I wasn't clear enough the condition 2) always work, if if f = g, then max|f − g|= 0, and it will exists.
The student who toke the notes was wrong.

I was only saying what amends are needed for this to be a metric.

The point that fails, is that this isn't always a function from (X,X) to R, existence fails sometimes.

But once existence is satisfied, the other 4 are always satisfied.

 

[PeroK]

If you restrict the set X to continuous functions, then max will always be defined. And you will have a metric.

 

[Streltsy]

Thank you guys.