1
$\begingroup$

Here is a portion of my paper I am currently writing for real analysis part 2. My question is, can I change the summations from $i=1 \rightarrow \infty$ to $i=1 \rightarrow n$ using the definition? This part of the proof is bolded. I am unsure if I am able to do this; my motivation is I want to apply minkowsky's inequality and Cauchy's inequality to prove the triangle inequality here (just like the euclidean metric) but I need the summation to be finite and not infinite.

The set of all square summable sequences, $l_2=\{x=(x_1,x_2,\ldots)|x_n\in \mathbb{R}, \sum_{n=1}^\infty |x_n|^2 < \infty,$ $y=(y_1,y_2,\ldots) | y_n\in \mathbb{R}, \sum_{n=1}^\infty |y_n|^2 < \infty\}$ is a metric space, where the distance $p(x,y)=\sqrt{\sum_{i=1}^\infty(x_i-y_i)^2}$.

Since it has been proven that $p(x,y)=p(y,x)$, it is sufficient to prove the following to satisfy property 3: \begin{align*} &p(x,y) \leq p(x,z) + p(z,y). \end{align*} Using the above definitions, we have to show: \begin{align*} \left(\sum_{i=1}^\infty(x_i-y_i)^2\right)^{\frac{1}{2}} \leq \left(\sum_{i=1}^\infty(x_i-z_i)^2\right)^{\frac{1}{2}}+\left(\sum_{i=1}^\infty(z_i-y_i)^2\right)^{\frac{1}{2}}. \end{align*} However, by assumption we know that: \begin{align*} l_2=\{x=(x_1,x_2,\ldots) | x_n\in \mathbb{R}, \sum_{n=1}^\infty |x_n|^2 < \infty\} \end{align*} Since we know that $\displaystyle{\sum_{n=1}^\infty |x_n|^2 < \infty}$, we can sum from $i=1$ to $n$ instead of from $i=1$ to $\infty$ due to the fact that $\sum_{i=1}^\infty(x_i-y_i)^2$ is still the sum of square summable sequences since both $x_n$ and $y_n$ are squares. The same holds for $z_n$ and $y_n$.Thus, if we show that this inequality holds true for the summation to $n$, for all values of $n$, we can generalize it to an infinite summation. Thus, we want to show that: \begin{align*} \left(\sum_{i=1}^n(x_i-y_i)^2\right)^{\frac{1}{2}} \leq \left(\sum_{i=1}^n(x_i-z_i)^2\right)^{\frac{1}{2}}+\left(\sum_{i=1}^n(z_i-y_i)^2\right)^{\frac{1}{2}}. \end{align*}

$\endgroup$
  • $\begingroup$ Your statement "we can sum from $1$ to $n$" is completely unjustified. At best, it needs clarification (what is the value of $n$? Are you saying that you can just pick some value of $n$?) and at worst, is completely wrong. $\endgroup$ – 5xum Feb 19 '15 at 8:10
  • $\begingroup$ @5xum I am just trying to use the part of the definition which states that the summation is less than infinity. I want to use this to allow us to take a finite sum and not an infinite sum, in order to use Minkowski and Cauchy's inequality to prove this. I know we are supposed to use these two according to our professor. The $n$ from $l_2$ and the summations are not the same $n$. I might be wrong with what I am doing, but I was thinking that I could not use these two inequalities because we have infinite sums and not finite sums, like the definitions of the inequalities are written. $\endgroup$ – H5159 Feb 19 '15 at 8:13
  • $\begingroup$ But why would the fact that the sum is finite have anything to do with the fact that you can take only a finite number of elements? Also, I ask again, are you saying that you will prove the inequality for every value of $n$ or do you just select some value of $n$? Right now, it is very unclear what $n$ is equal to. $\endgroup$ – 5xum Feb 19 '15 at 8:16
  • 1
    $\begingroup$ Thank you for clarifying your position. It is only now possible ti answer your question: Yes. If you prove your inequality for all values of $n$, then you can generalize it to an infinite sumation. $\endgroup$ – 5xum Feb 19 '15 at 8:33
  • 1
    $\begingroup$ Well, it isn't because the sum is less than $\infty$. Even if $\sum_{i=1}^\infty |x|^n$ diverges, your argument would still hold. $\endgroup$ – 5xum Feb 19 '15 at 8:37

Your Answer

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

Browse other questions tagged or ask your own question.