Metric on Cartesian Product

Question

I'm currently stuck on an exercise where I need to show that if $(M,d)$ and $(N,\rho)$ are metric spaces, then $d_2((a,x),(b,y)) := \sqrt{d(a,b)^2 + \rho(x,y)^2}$ defines a metric on $M\times N$. I've already shown that the basic criteria of a metric are met, but I still need to show that the triangle ineqaulity holds.

I've tried using the triangle inequality on $d$ and $\rho$ and estimating the square root with $\sqrt{a^2+b^2}\leq a+b$, but regardless of which order I've used them in, I seem to overestimate my target. I thought maybe I somehow need to use Cauchy-Schwarz, but I don't know how.

Answer 1

Let $(a,b,c) \in M$ and $(x,y,z) \in N$. You want to prove that $$\sqrt{d(a,c)^2 + \rho(x,z)^2} \leq \sqrt{d(a,b)^2 + \rho(x,z)^2} + \sqrt{d(b,c)^2 + \rho(y,z)^2}$$

You have, by triangle inequality for $d$ and $\rho$ : $$d(a,c)^2 + \rho(x,z)^2 \leq d(a,b)^2 + d(b,c)^2 + 2 d(a,b)d(b,c) + \rho(x,y)^2 + \rho(y,z)^2 + 2 \rho(x,y)\rho(y,z) \quad (1)$$

But on the other hand, you know that $$(\rho(x,z)d(b,c) - \rho(y,z)d(a,b))^2 \geq 0$$ so $$\rho(x,z)^2 d(b,c)^2 + \rho(y,z)^2d(a,b)^2 \geq 2 \rho(x,z)d(b,c)\rho(y,z)d(a,b)$$ Adding $d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2$ to this relation, you get $$d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2 +\rho(x,z)^2 d(b,c)^2 + \rho(y,z)^2d(a,b)^2 \geq d(a,b)^2d(b,c)^2 + \rho(x,y)^2 \rho(y,z)^2 + 2 \rho(x,z)d(b,c)\rho(y,z)d(a,b)$$ i.e. $$(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2) \geq (d(a,b)d(c,d)+\rho(x,y)\rho(y,z))^2$$ so $$2 \sqrt{(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2)} \geq 2(d(a,b)d(c,d)+\rho(x,y)\rho(y,z)) $$ Injecting this in the equation $(1)$, you get $$d(a,c)^2 + \rho(x,z)^2 \leq d(a,b)^2 + d(b,c)^2 + \rho(x,y)^2 + \rho(y,z)^2 + 2\sqrt{(d(a,b)^2 + \rho(x,z)^2)(d(b,c)^2 + \rho(y,z)^2)}$$ which is equivalent to $$d(a,c)^2 + \rho(x,z)^2 \leq \left(\sqrt{d(a,b)^2+\rho(x,z)^2}+\sqrt{d(b,c)^2 + \rho(y,z)^2}\right)^2$$ i.e. $$\sqrt{d(a,c)^2 + \rho(x,z)^2} \leq \sqrt{d(a,b)^2 + \rho(x,z)^2} + \sqrt{d(b,c)^2 + \rho(y,z)^2}$$

Answer 2

Let $m,m',m''\in M$ and $n,n',n''\in N .$ Let $p=d(m,m'), q=d(m',m''),r=d(m,m'').$ Let $s=\rho(n,n'), t=\rho (n',n''), u=\rho (n,n'').$

Let $A= (p^2+s^2)^{1/2}$ and $B=(q^2+t^2)^{1/2}.$ We wish to prove that $A+B\geq (r^2+u^2)^{1/2}.$

Let $C= ((p+q)^2+(s+t)^2)^{1/2}.$ Since $d$ and $\rho$ are metrics, we have $p+q\geq r\ge 0$ and $s+t\geq u\ge 0,$ so $C\ge (r^2+u^2)^{1/2}.$ So it suffices to prove that $A+B\geq C.$

Since $A,B,C \geq 0$ we have $$ A+B\geq C \iff (A+B)^2\geq C^2 \iff$$ $$\iff p^2+s^2+q^2+t^2+2 ((p^2+s^2)(q^2+t^2))^{1/2}\geq p^2+q^2+s^2+t^2+2pq+2st$$ $$\iff ((p^2+s^2)(q^2+t^2))^{1/2}\geq pq+st \iff$$ $$\iff (p^2+s^2)(q^2+t^2)\geq (pq+st)^2\iff$$ $$\iff (pt-qs)^2\geq 0.$$

In the last two lines above, we have used the identity $(p^2+s^2)(q^2+t^2)=(pq+st)^2+(pt-qs)^2.$

Answer 3

I saw this question on Page 18, Exercise 38 of the following reference

C. C. Pugh, Real Mathematical Analysis, 2nd Edition, Springer, 2015.

There is a hint at the end of this exercise in the book that says, use the Cauchy-Schwarz Inequality. To recall this, in an abstract inner product vector space $V$ over the Real field $\mathbb{R}$, we have

$$\left|\langle x,y\rangle\right|\leq\sqrt{\langle x,x\rangle}\sqrt{\langle y,y\rangle}.$$

If $V=\mathbb{R}^n$, $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$ and $\langle x,y\rangle:=\sum_{i=1}^nx_iy_i$ then this inequality turns out to be

$$\left|\sum_{i=1}^nx_iy_i\right|\leq\sqrt{\sum_{i=1}^nx_i^2}\sqrt{\sum_{i=1}^ny_i^2}.$$

Keep this in mind! Let your cartesian product to be $M=X\times Y$ with $(X,d_X)$ and $(Y,d_Y)$ being metric spaces. Introduce $p_1=(x_1,y_1),\,p_2=(x_2,y_2),\,p_3=(x_3,y_3)\in M$ and by definition we have $$d_M(p_1,p_2)=\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}.$$ We want to show that $d_M$ is indeed a metric. Positive definiteness and symmetry are immidiate and easy to prove. The triangle inequality for $d_M$ reads as follows

\begin{align*} d_M(p_1,p_3)&\leq d_M(p_1,p_2)+d_M(p_2,p_3), \\ \sqrt{d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2}&\leq \sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2} + \sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2}. \end{align*}

To show that this is indeed true, we start with what we know about the metrics $d_X$ and $d_Y$

\begin{align*} d_X(x_1,x_3)\leq d_X(x_1,x_2)+d_X(x_2,x_3), \qquad d_Y(y_1,y_3)\leq d_Y(y_1,y_2)+d_Y(y_2,y_3). \end{align*}

Taking square and then adding the result leaves us with

\begin{align*} &d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq \\ &d_X(x_1,x_2)^2+d_X(x_2,x_3)^2+2d_X(x_1,x_2)d_X(x_2,x_3)+ \\ &d_Y(y_1,y_2)^2+d_Y(y_2,y_3)^2+2d_Y(y_1,y_2)d_Y(y_2,y_3), \\ \\ &d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq \\ &\big(d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2\big)+\big(d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2\big)+ \\ &2\big(d_X(x_1,x_2)d_X(x_2,x_3)+d_Y(y_1,y_2)d_Y(y_2,y_3)\big), \\ \end{align*}

Using the Cauchy-Schwarz Inequality for $n=2$, we observe that

\begin{align*} &d_X(x_1,x_2)d_X(x_2,x_3)+d_Y(y_1,y_2)d_Y(y_2,y_3)\leq \\ &\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}\sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2} \end{align*}

Combining the last two results gives us

\begin{align*} &d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq \\ &\Big(\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}\Big)^2+\Big(\sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2}\Big)^2+ \\ &2\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}\sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2}, \\ \\ &d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq \\ &\Big(\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}+\sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2}\Big)^2, \\ \\ &\sqrt{d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2}\leq \\ &\sqrt{d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}+\sqrt{d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2}. \end{align*}

This completes the proof. You can use the same way to prove for the Cartesian product of $n$ metric spaces $M=X_1\times\dots \times X_n$. For this purpose, you can use mathematical induction or a direct approach by employing Cauchy-Schwarz for an arbitrary $n$.