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I am relatively new to proofs so I am struggling to understand what really counts as "proved" and what doesn't.

Consider the vector space $\mathbb M $ of real numbers $\mathbb R$. Show that for $z,z_1 \in \mathbb R$, $d:\mathbb M \times \mathbb M \rightarrow\mathbb R^+$ defined as $$d(z,z_1)=f(\lvert z-z_1 \rvert), \space \space \space f:\mathbb R^+\rightarrow \mathbb R^+$$ is a metric provided that for $x,y \in \mathbb R^+$

$\text{(A)} \space \space f(0)=0 \\ \text{(B)} \space \space f(x)>0 \space \text{for} \space x>0 \\ \text{(C)} \space \space f(y) \ge f(x) \space \text{for} \space y \ge x \space \text{i.e. $f$ is a monotonically growing function}\\ \text{(D)} \space \space\frac{f(x)}{x}\ge \frac{f(y)}{y} \space \text{for} \space y \ge x$

This is what I have tried so far:

In order for a space to be a metric space the following axioms have to be true:

$\text{(1)} \space \space d \space \text{is real valued, finite and nonnegative } \\ \text{(2)} \space \space d(x,y)=0 \iff x=y \\ \text{(3)} \space \space d(x,y)=d(y,x) \\ \text{(4)} \space \space d(x,y) \le d(x,z)+d(z,y)$

For (1): $\forall x,y \in \mathbb R^+, \lvert x-y\rvert \ge 0$. From (B) it follows that if $\lvert x-y \rvert > 0 \implies f(\lvert x-y \rvert) \ge 0 \implies d\ge 0 \checkmark$

For (2): $\lvert x-y \rvert=0 \iff x=y \implies f(\lvert x-y \rvert)=f(0)=0 \iff x=y \checkmark$

Does this count as proving? I am not sure how to show 3) and 4). Maybe someone can show me or give me a hint. Thanks.

Mark S.
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bluemoon
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    Yes, the first two look correct. Symmetry(3) follows from the fact that $|x-y| = |y-x|$. For the last one, use monotonicity of $f$ along with the triangle inequality you already know for the abs. value ie. $|x-y| \leq |x-z| + |y-z|$. – Eman Yalpsid Apr 24 '16 at 13:50
  • @Andrew Thanks. So can I just assume $\lvert x-y\rvert=\lvert y-x\rvert$ is true? I always get confused what I can assume to be true and what I have to show. – bluemoon Apr 24 '16 at 13:57
  • Absolute value over the reals can be defined as $|x| = \operatorname{max}{x,-x}$. This function has the nice property that for all $ x\in \mathbb R$ one has $|-x| = \operatorname{max}{-x,-(-x)} = \operatorname{max}{-x,x} = |x|$. So since $y-x = -(x-y)$ we get the result you are asking about. – Eman Yalpsid Apr 24 '16 at 14:02
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    By the way I feel that you could benefit from checking that $\operatorname{d}: \mathbb R \times \mathbb R \to \mathbb R; \operatorname{d} (x,y) : = |x-y|$ is a metric! Also check out the following post: http://math.stackexchange.com/questions/987602/composition-of-a-function-with-a-metric – Eman Yalpsid Apr 24 '16 at 14:04
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    @Andrew Thanks. That happens to be a question on my worksheet! :). For 3) my reasoning looks like this: $$\lvert x-y \rvert=\lvert y-x\rvert \ \implies f(\lvert x-y \rvert)=f(\lvert y-x \rvert) \ \iff d(x,y)=d(y,x) \checkmark$$. I am still working in 4) – bluemoon Apr 24 '16 at 14:09
  • @Andrew I am stuck on trying to show 4). Could you maybe give me an additional hint. I don't understand how monotonicity can help me here. – bluemoon Apr 24 '16 at 14:52

1 Answers1

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We need the following

Lemma. A function $f$ with the properties (A)–(D) is subadditive, i.e., one has $$f(x+y)\leq f(x)+f(y)\qquad(x>0, \>y>0)\ .$$ Proof. From $${f(x+y)\over x+y}\leq {f(x)\over x},\qquad {(x+y)\over x+y}\leq {f(y)\over y}$$ it follows that $${f(x+y)\over x+y}\leq{x\over x+y}\>{f(x)\over x}+{y\over x+y}\>{f(y)\over y}\ .$$ Now multiply with $x+y$.$\qquad\square$

This allows to complete the proof that for any metric $d$ on a set $X$ the function $d':=f\circ d$ is again a metric if $f$ has the properties (A)–(D): Given any three points $x$, $y$, $z\in X$ one has $$d'(x,z)=f\bigl(d(x,z)\bigr)\leq f\bigl(d(x,y)+d(y,z)\bigr)\leq f\bigl(d(x,y)\bigr)+f\bigl(d(y,z)\bigr)=d'(x,y)+d'(y,z)\ .$$

Christian Blatter
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