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I want to prove that the set $K = \{c \in R^n\mid c_{1} + c_{1}t +\dotsb+ c_{n}t^{n-1} ≥ 0 \forall t \in [0,1]\}$ is a cone i.e that for $x \in K$ and $\theta \ge 0$, $\theta x\in K$.

Is the follow attempt a correct proof?


Let's consider $x \in K$.

We have $\sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0, \space t ∈ [0,1]$

For $\theta \ge 0$, we then have

$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}}$

since both $\theta$ and the polynomial $\sum\limits_{i=1}^n{x_{i} t^{i-1}}$ are greater than or equal to $0$, we must have

$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0$

Thus $x \in K$, $\space$ $\theta \ge 0$ $\rightarrow$ $\theta x \in K$.

Hence $K$ is a cone. $\square$

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Yes, your proof is correct. It really is that simple!

Added remark: Correctness aside, I recommend that you study Xander's exemplary answer concerning ways to improve presentation.

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  • $\begingroup$ Thanks! Having a non-technical background I wasn't sure if it was a valid proof. $\endgroup$ – Le Noff Jun 5 at 13:11
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In general, if a question on Math SE may be answered with a simple yes-or-no, then the question is almost certainly off-topic. There seems to be a consensus in the community that these kinds of proof verification problems are not off-topic, hence a correct answer to this question cannot be a simple yes-or-no. As such, I must assume that the question is really more about the style of the presentation, not the actual technical details.

I'll note that much of my answer is a matter of opinion; other author's may have differing thoughts on specific matters of style. Thus if there is some pedantic point below with which you disagree, please disregard it.


Starting from the top, I find the statement of the result a little hard to follow, and would rewrite it as

Exercise: Show that the set $$ K = \{ c \in \mathbb{R}^n \mid c_1 + c_2t + \dotsb + c_n t^{n-1} \ge 0 \forall t \in [0,1] \} $$ is a cone. That is, for all $x\in K$ and all $\theta \ge 0$, show that $\theta x \in K$.

The very minor edits here a mostly for readability: the displayed equation is a little easier to follow, and restructuring the final sentence to avoid notation immediately following a comma is (in my opinion) an improvement.


As to your proof (which is correctly argued), here is your presentation, with some notes:

Let's consider $x \in K$.

Personally, I don't like the phrase "Let's consider..." I find it overly casual, and also kind of meaningless. I prefer to write in a more imperative mode and to use more precise verbs. I think it is better to say "Let $x \in K$," or "Fix an arbitrary $x\in K$," or something similar. You could also say "Consider $x\in K$," if you are really attached to that verb.

You might also want to fix a value of $\theta$ here, too. Remember you want to show that the desired result holds for any $x\in K$ and any $\theta \ge 0$. Why not fix both of those values right at the start?

We have $\sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0, \space t ∈ [0,1]$

Again criticizing style, I don't really like the use of the "mathematical 'we'". I know that lots of people use it---I even use it quite a lot when I am not terribly worried about the quality of the presentation, or when I am writing quickly (by way of example, I spent a couple of hours last week removing every instance of "we" from my thesis). I think that it is better to be more direct. Maybe an "if-then" statement, such as "If $t \in [0,1]$, then...". It might even be better to explain where the inequality comes from, i.e. "Since $x \in K$..."

For $\theta \ge 0$, we then have

$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}}$

If you follow my previous suggestion and fix $\theta$ earlier in the argument, then this entire line is unnecessary. Also, it should be "displayed" using $$; this avoids the use of \limits, which is poor style for inline (rather than displayed) mathematics.

Also, what is $t$? (see below)

since both $\theta$ and the polynomial[1] $\sum\limits_{i=1}^n{x_{i} t^{i-1}}$[2] are greater than or equal to $0$, we[3] must have

$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0$[4]

A couple of comments here:

  1. I am a little uncomfortable calling the sum a polynomial, then declaring that the polynomial is nonnegative. Generally, a polynomial is an abstract object, and $t$ is a formal variable. Alternatively, you can see the sum as a polynomial function, which is being evaluated at a particular value of $t$. I think that it would be better to just call it a "sum".

  2. What is $t$? Where have you declared that $t \in [0,1]$? You say at the beginning that if $t\in [0,1]$, then other nice things happen. This doesn't clearly describe every subsequent use of $t$ in the rest of the argument.

  3. Again, drop the mathematical "we". :\

  4. Again, display the math. Also, sentences end with periods.

Thus $x \in K$, $\theta \ge 0$ $\rightarrow$ $\theta x \in K$.

Hence $K$ is a cone. $\square$

A lot of this is redundant. Both $x$ and $\theta$ are defined above, and you have shown that $\theta x \in K$. Done. Everything else is superfluous. I also rather object to the overuse of notation in an inline environment. If you are going to write out all of these details, please write them out: "Thus if $x \in K$ and $\theta \ge 0$, then $\theta x \in K$. Therefore $K$ is a cone.


My presentation of your proof is as follows:

Proof: Let $x = (x_1, \dotsc, x_n) \in K$ and fix $\theta > 0$. By definition of $K$, $$ \sum_{j=1}^{n} x_j t^{j-1} \ge 0 $$ for all $t \in [0,1]$. Then $$ \sum_{j=1}^{n} (\theta x_j) t^{j-1} = \theta \sum_{j=1}^{n} x_j t^{j-1} \ge 0 $$ for any $t\in [0,1]$, as both $\theta$ and the sum $\sum x_jt^{j-1}$ are nonnegative. Therefore $\theta x = (\theta x_1, \dotsc, \theta x_n) \in K$.

I will note that in the second-to-last sentence, I have written $\sum x_jt^{j-1}$ (and have omitted the limits of summation). This is an abuse of notation, but should be clear from context. I have also replaced the indices of summation with $j$s throughout, as I like to reserve $i$ for the imaginary unit. This is a completely pedantic and unnecessary change, but I think it makes my life better.

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  • $\begingroup$ Thank you a lot for your comment. I appreciate your remarks, they'll help me to write better more readable proofs! $\endgroup$ – Le Noff Jun 7 at 23:08

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