Is this true? euclidean metric <= taxicab metric

[Mosis]

given sequences $$\left\{x_n\right\}, \left\{y_n\right\}$$, is it true that

$$\sqrt{ \Sigma_{n=1}^{\infty} (x_n - y_n)^2} \leq \Sigma_{n=1}^{\infty} |x_n - y_n|$$

this isn't a homework problem. it's just something that came up - I think it's pretty clear that it's true, but I don't know how to show this.

edit: the sequences are square summable, of course.

 

[D H]

Square both sides. The left hand side is obvious. Squaring the right hand side will yield the square of the left hand side plus another series, each of whose terms is non-negative.

 

[Mosis]

right, I'm retarded. i was thinking that squaring something didn't preserve the inequality, but that's because i was confused about something else. thanks! (sorry, it's late, and if i have to prove that one more set is closed or one more set is dense, then I'm going to be very grumpy tomorrow.)

 

[Mosis]

... although it's not clear to me how squaring an infinite sum must make sense

 

[D H]

... although it's not clear to me how squaring an infinite sum must make sense

If two series [itex]A=\sum_{n=1}^{\infty} a_n[/tex] and [itex]B=\sum_{m=1}^{\infty} b_m[/tex] are absolutely convergent, the product of the two series is absolutely convergent and converges to $AB$. See <a href="http://www.mathreference.com/lc-prod,p2s.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.mathreference.com/lc-prod,p2s.html</a><br /> <br /> In this case, each element of the series on the right-hand side is non-negative, so if the series converges it is absolutely convergent, and thus its square is<br /> <br /> $$\left(\sum_{n=1}^{\infty} |x_n-y_n|\right)^2 =<br /> \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} |x_n-y_n| |x_m-y_m|$$There are four cases to consider in determining the validity of $||x-y||_{L2} \le ||x-y||_{L1}$<ol> <li data-xf-list-type="ol">Both sides of the inequality involve convergent series.<br /> The way to show that the inequality holds in this case is to square both sides of the inequality as mentioned above.</li> <li data-xf-list-type="ol">The left-hand side diverges, the right hand side converges.<br /> If this case ever did occur it would falsify the conjecture. It is easy to show that this case cannot occur.</li> <li data-xf-list-type="ol">The left-hand side converges, the right hand side diverges.<br /> Example: y_n=0, x_n=1/n. The inequality is trivially true in this case.</li> <li data-xf-list-type="ol">Both sides of the inequality involve divergent series.<br /> Informally, this is essentially the equivalent of ∞ ≤ ∞. Formally, this is nonsense. One or both series needs to converge here.</li> </ol>[/itex][/itex]

 

[HallsofIvy]

Essentially, you are just saying that a straight line is the shortest distance between two points. The Euclidean metric measures distance along a straight line. The taxi cab metric doesn't.

 

[statdad]

First, if the absolute value infinite series diverges to $$\infty$$, there is nothing to prove - the left side MUST be $$\le$$ the right. If the right side series is finite, then:

If you are concerned about squaring infinite sums you can approach things this way. For any finite $$n$$ ($$\Leftrightarrow$$ refers to ``if and only if'')
I know your problem is written with $$x_i - y_i$$ - think of my work as referring to this difference by $$z_i$$

$$\begin{align*}<br /> \left(\sum_{i=1}^n z_i^2\right)^{1/2} & \le \sum_{i=1}^n |z_i| \\<br /> & \Leftrightarrow \\<br /> \sum_{i=1}^n z_i^2 & \le \left(\sum_{i=1}^n |z_i|\right)^2 \\<br /> & = \sum_{i=1}^n z_i^2 + 2\sum_{i=1}^{n-1} \sum_{j=i+1}^n |z_i z_j| \\<br /> &\le \sum_{i=1}^\infty z_i^2 + 2\sum_{i<j}^\infty |z_i z_j| \\<br /> &=\left(\sum_{i=1}^\infty |z_i|\right)^2 < \infty<br /> \end{align*}$$

Since this holds for every value of $$n$$, you can let $$n \to \infty$$
on the left to find

$$\sum_{i=1}^\infty z_i^2 \le \left(\sum_{i=1}^\infty |z_i|\right)^2$$

the infinite series converge because you are assuming the sums represent metrics on suitably defined sequence spaces.