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A. Let (M, d) be a metric space and $R>0$ be any real number. Show that $(M, R\cdot d)$ is also a metric space.

B. Is $(M,d^2)$ also a metric space if (M, d) is a metric space? If yes, then prove it. If no, then give a counterexample. What if you use the square root instead of squaring?

$\textbf{Part A:}$ Assume that $(M,d)$ is a metric space. To prove that $(M,R \cdot d)$ is a metric space, we must show that the distance function satifies the following properties:

  1. $R \cdot d(x,x)=0 \iff x=0$

Since $(M,d)$ is a metric space and $R>0$ then we have $d(x,x)=0 \iff x=0$ so if we multiply the first equation by R we will of course obtained the following result: $R \cdot d(x,x)=0 \iff x=0$.

  1. $R \cdot d(x,y)=R \cdot d(y,x)$

Since $(M,d)$ is a metric space and $R>0$ then we have $d(x,y)=d(y,x)$ so if we multiply both sides of the equation by R we will of course obtained the following result: $R \cdot d(x,y)=R \cdot d(y,x)$.

  1. $R\cdot d(x,z) \leq R \cdot d(x,y)+ R \cdot d(y,z)$

Since $(M,d)$ is a metric space and $R>0$ then we have $d(x,z) \leq d(x,y)+d(y,z)$ so if we multiply the both sides of the inequality by R we will of course obtained the following result: $R\cdot d(x,z) \leq R \cdot d(x,y)+ R \cdot d(y,z)$.

Hence $(M, R \cdot d)$ is a metric space.

$\textbf{Part B: Is $(M,d^2)$ also a metric space if $(M, d)$ is a metric space?}$

$\textbf{Claim:}$ $(M,d^2)$ is not a metric space, if $(M,d)$ is a metric space.

$\textbf{Proof ( by contradition):}$ Assume that $(M,d^2)$ is a metric space. Also, assume that $(M,d)$ is a metric space. To prove that $(M,R \cdot d)$ is a metric space, we must show that the distance function satifies the following properties:

  1. $d^2(x,x)=0 \iff x=0$

Since $(M,d)$ is a metric space then we have $d(x,x)=0 \iff x=0$ so if we square the first equation we will of course obtained the following result: $d^2(x,x)=0 \iff x=0$.

  1. $d^2(x,y)=d^2(y,x)$

Since $(M,d)$ is a metric space then we have $d(x,y)=d(y,x)$ so if we square the both sides of the equation we will of course obtained the following result: $d^2(x,y)=d^2(y,x)$.

  1. $d^2(x,z) \leq d^2(x,y)+ d^2(y,z)$

Since $(M,d)$ is a metric space then we have $d(x,z) \leq d(x,y)+d(y,z)$. Hence $$d^2(x,z)=(d(x,z))^2 \leq (d(x,y)+d(y,z))^2=d^2(x,y)+2d(x,y)d(y,z)+d^2(y,z)$$ However $d^2(x,y)+2d(x,y)d(y,z)+d^2(y,z)$ will never be less than $d^2(x,y)+d^2(y,z)$. So we need to hope that they are equal. If both quantities are equal then $2d(x,y)d(y,z)=0$ which only happens if $x=y$ or $y=z$. Hence we have a contradiction because this doesn't happen for all $x,y,z \in M$.

Hence $(M, d^2)$ isn't a metric space.

$\textbf{Part B: What if you use the square root instead of squaring?}$

$\textbf{Claim:}$ $(M,d^\frac{1}{2})$ is not a metric space, if $(M,d)$ is a metric space.

$\textbf{Proof ( by contradition):}$ Assume that $(M,d^\frac{1}{2})$ is a metric space. Also, assume that $(M,d)$ is a metric space. To prove that $(M,R \cdot d)$ is a metric space, we must show that the distance function satifies the following properties:

  1. $d^\frac{1}{2}(x,x)=0 \iff x=0$

Since $(M,d)$ is a metric space then we have $d(x,x)=0 \iff x=0$ so if we square-root the first equation we will of course obtained the following result: $d^\frac{1}{2}(x,x)=0 \iff x=0$.

  1. $d^\frac{1}{2}(x,y)=d^\frac{1}{2}(y,x)$

Since $(M,d)$ is a metric space then we have $d(x,y)=d(y,x)$ so if we square-root the both sides of the equation we will of course obtained the following result: $d^\frac{1}{2}(x,y)=d^\frac{1}{2}(y,x)$.

  1. $d^\frac{1}{2}(x,z) \leq d^\frac{1}{2}(x,y)+ d^\frac{1}{2}(y,z)$

Since $(M,d)$ is a metric space then we have $d(x,z) \leq d(x,y)+d(y,z)$. Hence $$d^\frac{1}{2}(x,z)=(d(x,z))^\frac{1}{2} \leq (d(x,y)+d(y,z))^\frac{1}{2}$$

I just don't know how to finish this proof.

Andrés E. Caicedo
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Username Unknown
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  • Try proving that $\sqrt{d(x,y)}$ is a metric. Either you will be successful, or run in to a road block which you can use to prove it is not a metric. – Mike Earnest Oct 12 '14 at 20:00
  • I understand that but I do not know how to continue with the triangle inequality. – Username Unknown Oct 12 '14 at 20:03
  • For $x,y$ positive, how can you compare $\sqrt{x+y}$ with $\sqrt x+\sqrt y$? (which one is bigger?) Is this relevant to your question? – Andrés E. Caicedo Oct 12 '14 at 20:04
  • Of course it is relevant to my question. I was thinking of using Holder's inequality or Minsoski but I am not sure – Username Unknown Oct 12 '14 at 20:05
  • No, no! Square both expressions and compare. Squaring preserves the ordering of positive numbers (right?) – Andrés E. Caicedo Oct 12 '14 at 20:10
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    Now, $(M,d^2)$ is a metric space sometimes. For instance, if $d$ is discrete: $d(x,y)=1$ iff $x\ne y$. It is not a metric space some other times, but your argument does not prove this. You need to exhibit an example of a metric $(M,d)$ where $(M,d^2)$ is not metric. Your work gives you a hint on how to do that, but it is not a proof yet. (It cannot be a proof, since in some cases $(M,d^2)$ is indeed metric.) – Andrés E. Caicedo Oct 12 '14 at 20:18
  • I want to show it is a metric all the time – Username Unknown Oct 12 '14 at 20:21
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    @UsernameUnknown: you've got it backward. $d^2$ is not necessarily a metric. However $d^{1/2}$ is always a metric, assuming that $d$ is. – Cheerful Parsnip Oct 12 '14 at 20:24
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    Not only that, but the "proof" you gave that $d^2$ is not a metric is not a proof, because it is false that $d^2$ is not a metric. The "necessarily" in the previous comment is there for a reason. – Andrés E. Caicedo Oct 12 '14 at 20:28
  • So what is a counterexample for $d^2$? – Username Unknown Oct 12 '14 at 20:39
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    The real line with the usual metric (which raises the question of the examples you checked before asking). – Did Oct 12 '14 at 21:05

1 Answers1

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If $d^{1/2}$ were a metric, it would satisfy $$ d^{1/2}(x,z)\le d^{1/2}(x,y)+d^{1/2}(y,z) $$ Squaring both sides, this becomes $$ d(x,z)\le d(x,y)+d(y,z)+2\sqrt{d(x,y)\cdot d(y,z)} $$ Edit: This last inequality is true, suggesting the triangle inequality does hold. To prove it, $$ d^{1/2}(x,z)\le \sqrt{d(x,y)+d(y,z)} \le \sqrt{d(x,y)+d(y,z)+2\sqrt{d(x,y)d(y,z)}} \\=\sqrt{(d^{1/2}(x,y)+d^{1/2}(y,z))^2}= d^{1/2}(x,y)+d^{1/2}(y,z) $$

Mrcrg
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Mike Earnest
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