# Is my proof that nonnegative polynomials on $[0,1]$ form a convex set correct?

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. $$\square$$

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.

• Thanks! Having a non-technical background I wasn't sure if it was a valid proof. – Le Noff Jun 5 at 13:11

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]

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.

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