(Couldn't find it here, if it is duplicated I'm sorry)

Given that $(X_1,d_1),\cdots,(X_n,d_n)$ are metric spaces. Define the function $q: \prod\limits_{i=1}^nX_i \times \prod \limits_{i=1}^nX_i \to[0,\infty)$ by $$q(x,y)=(\sum\limits_{i=1}^nd_i(x_i,y_i)^2)^{\frac{1}{2}}$$

Then prove that $q$ is a metric in $\prod \limits_{i=1}^nX_i $.

I'm stuck on prooving that $q$ satisfy tha triangle inequality.

I know that $q_1(x,y)=\sum\limits_{i=1}^nd_i(x_i,y_i)$ is a metric. Using this, I can prove that for all $x,y$ such that $d_i(x_i,y_i)\leq1 \ \forall i$ we have $q(x,y)\leq \sqrt{q_1(x,y)}$ and we're done because square root is subadditive. But I don't think I could do this for all other $x,y$ in a similar way.

Could you helpe me? Thank's in advance!

  • 3
    $\begingroup$ Apply triangle inequality for the $d_i$'s and then apply triangle inequality in $\mathbb R^{n}$. $\endgroup$ – Kavi Rama Murthy Sep 5 '18 at 6:11

With the definition, $$ q(x,y)=(\sum\limits_{i=1}^nd_i(x_i,y_i)^2)^{\frac{1}{2}}$$

You need to show $$q(x,y)\le q(x,z)+q(z,y)$$

That is $$(\sum\limits_{i=1}^nd_i(x_i,y_i)^2)^{\frac{1}{2}}\le (\sum\limits_{i=1}^nd_i(x_i,z_i)^2)^{\frac{1}{2}} +(\sum\limits_{i=1}^nd_i(z_i,y_i)^2)^{\frac{1}{2}}$$

Upon squaring both sides we get $$\sum\limits_{i=1}^nd_i(x_i,y_i)^2\le \sum\limits_{i=1}^nd_i(x_i,z_i)^2 +\sum\limits_{i=1}^nd_i(z_i,y_i)^2+2(\sum\limits_{i=1}^nd_i(x_i,z_i)^2)^{\frac{1}{2}} (\sum\limits_{i=1}^nd_i(z_i,y_i)^2)^{\frac{1}{2}} $$

Note that we have $$ d_i(x_i, y_i) \le d_i(x_i, z_i)+d_i(z_i, y_i)$$

Squaing both sides gives us $$d_i(x_i, y_i)^2 \le d_i(x_i, z_i)^2+d_i(z_i, y_i)^2+2d_i(x_i, z_i)d_i(z_i, y_i)$$ Upon adding up we get $$\sum _1^n d_i(x_i, y_i)^2 \le \sum _1^nd_i(x_i, z_i)^2+ \sum _1^n d_i(z_i, y_i)^2+2\sum _1^n d_i(x_i, z_i)d_i(z_i, y_i)$$ It remains to show that $$\sum _1^n d_i(x_i, z_i)d_i(z_i, y_i)\le(\sum\limits_{i=1}^nd_i(x_i,z_i)^2)^{\frac{1}{2}} (\sum\limits_{i=1}^nd_i(z_i,y_i)^2)^{\frac{1}{2}} $$

Which is the Cauchy Schwarz inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.