Positiveness of partial sums of type $ \psi * D_N $

Question

In his paper about Extremal Functions for the Fourier Transform (see, for example, here? https://projecteuclid.org/download/pdf_1/euclid.bams/1183552525), Jeffrey Vaaler, while trying to build extremal trigonometric polynomials for the function

$$ \psi (x) = \begin{cases} x - \lfloor x \rfloor - 1/2, \; x \ne 0 \\ 0, \; x = 0 \end{cases} $$

states that 'by a classical result of Féjer, the partial sums

$$ S(x,n) = \sum_{l=1}^n \frac{\sin (2\pi l x) }{\pi l } $$

are always positive for $x \in (0,1/2) $.

After some time looking at this assertion, I tried to look for the reference he gives (apparently, a paper by Féjer from the late 1920's on which he shows that), but I could only find that in a Springer website - which means, of course, that it is really expensive to get full access to it, besides it is in German (and my German skills basically inexist).

Another interesting fact is that this assertion holds in the limit. That is, by convergence of Fourier Series - because $\psi$ is fo bounded variation on $(-1/2,1/2)$ - The sequence $S(x,n) \rightarrow \psi (x) $ pointwisely, so, as $\psi(x) > 0$ for $x \in (0,1/2)$, then $ S(x,n) > 0$ for sufficiently large $n = n(x)$.

So, now I'm turning to proving this claim. I've had some ideas:

1 - Trying to differentiate $S(x,n)$ - which gives $ 2 D_n - 2 $, where $D_n$ is the $n-$th Dirichlet kernel - and looking for the behaviour of this derivative in the considered interval. At first, it seemed a little bit difficult to continue this way, but it may be worth to try for some more time.

2 - Writing $S(x,n) = - \psi * D_n $ and studying this defining integral, maybe with some integration by parts or something.

If anyone can ever give me some reference to that and/or give some nice, short proof of this, I would be very thankful. This is not that "essential" for my study, just some kind of interesting fact that I think I should know how to prove.

Answer 1

Here is a one-page proof by Landau: http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002374331 .

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Here is a rough translation:

Fejer's conjecture $$ S_{n}\left(x\right)=\sum_{\nu=1}^{n}\frac{\sin\left(\nu x\right)}{\nu}>0\qquad\text{ for }\qquad0<x<\pi $$ was first proved independently by Mr. Jackson and Mr. Gronwall. The following proof is shorter than any known one.

Let $n>1$ and assume that the claim holds for $n-1$. If we suppose $$ 0\overset{!}{=}2\sin\frac{x}{2}\cdot S_{n}'\left(x\right)=2\sin\frac{x}{2}\sum_{\nu=1}^{n}\cos\left(\nu x\right)\overset{\left(\ast\right)}{=}\sin\left[\left(n+\frac{1}{2}\right)x\right]-\sin\frac{x}{2}, $$ then we get for $0<x<\pi$ that $$ \sin\left(nx\right)=\sin\left[\left(n+\frac{1}{2}\right)x\right]\cdot\cos\left(\frac{x}{2}\right)-\cos\left[\left(n+\frac{1}{2}\right)x\right]\cdot\sin\left(\frac{x}{2}\right)\overset{\left(\dagger\right)}{=}\left\{ 0\text{ or }\sin\left(x\right)\right\} \geq0, $$ and hence $$ S_{n}\left(x\right)\geq S_{n-1}\left(x\right)>0. $$ Thus, $S_{n}$ does not have a minimum $\leq0$ in the range $0<x<\pi$ and hence does not assume any values $\leq0$.

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Some thoughts added: It seems that at $\left(\ast\right)$ the identity $$ 2\sin\left(x\right)\cos\left(y\right)=\sin\left(x-y\right)+\sin\left(x+y\right), $$ together with a telescoping argument is used to get \begin{eqnarray*} 2\sin\frac{x}{2}\cdot\sum_{\nu=1}^{n}\cos\left(\nu x\right) & = & \sum_{\nu=1}^{n}\sin\left(\left(\frac{1}{2}-\nu\right)x\right)+\sum_{\nu=1}^{n}\sin\left(\left(\nu+\frac{1}{2}\right)x\right)\\ & = & \sum_{\nu=1}^{n}\sin\left(\left(\nu+\frac{1}{2}\right)x\right)-\sum_{\nu=1}^{n}\sin\left(\left(\nu-\frac{1}{2}\right)x\right)\\ & \overset{\ell=\nu-1}{=} & \sum_{\nu=1}^{n}\sin\left(\left(\nu+\frac{1}{2}\right)x\right)-\sum_{\ell=0}^{n-1}\sin\left(\left(\ell+\frac{1}{2}\right)x\right)\\ & = & \sin\left(\left(n+\frac{1}{2}\right)x\right)-\sin\frac{x}{2}. \end{eqnarray*}

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At $\left(\dagger\right)$, the following argument seems to be used: Since $\sin\left(\frac{x}{2}\right)=\sin\left[\left(n+\frac{1}{2}\right)x\right]$, the identity $\cos^{2}=1-\sin^{2}$ yields $$ \left(\cos\left[\left(n+\frac{1}{2}\right)x\right]\right)^{2}=\left[\cos\left(\frac{x}{2}\right)\right]^{2}, $$ so that there are two possibilities:

1. We have $\cos\left[\left(n+\frac{1}{2}\right)x\right]=\cos\left(\frac{x}{2}\right)$. In this case, we get \begin{eqnarray*} & & \sin\left[\left(n+\frac{1}{2}\right)x\right]\cdot\cos\left(\frac{x}{2}\right)-\cos\left[\left(n+\frac{1}{2}\right)x\right]\cdot\sin\left(\frac{x}{2}\right)\\ & = & \sin\left(\frac{x}{2}\right)\cdot\left(\cos\left(\frac{x}{2}\right)-\cos\left[\left(n+\frac{1}{2}\right)x\right]\right)\\ & = & 0. \end{eqnarray*}

2. We have $\cos\left[\left(n+\frac{1}{2}\right)x\right]=-\cos\left(\frac{x}{2}\right)$. In this case, we get \begin{eqnarray*} & & \sin\left[\left(n+\frac{1}{2}\right)x\right]\cdot\cos\left(\frac{x}{2}\right)-\cos\left[\left(n+\frac{1}{2}\right)x\right]\cdot\sin\left(\frac{x}{2}\right)\\ & = & \sin\left(\frac{x}{2}\right)\cdot\left(\cos\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)\\ & = & 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\\ & = & \sin\left(x\right). \end{eqnarray*}

This justifies the $\left\{ 0\text{ or }\sin\left(x\right)\right\} $ bracket at $(\dagger)$.

Answer 2

Just to complement PhoemueX's answer:

I decided to look for "Féjer's Conjecture" and "Féjer-Jackson-Gronwall Inequality" on Google, and turned out that these results are quite known - yet Vaaler's reference was slightly difficult to reach.

For example, there is a nice paper about a generalization of some polynomial inequalities like this here, from Stamatis Koumandos:

http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2006/koumandos.pdf

There, he recomends three books/papers for proofs of Féjer-like results. These are his references [10], [12] and [26].

On [10] (Some lecture notes by Richard Askey), some more general stuff is observed: for example, we may note that

$$ \sum_{k=0} ^{n} \sin (2 \pi (k+1/2)x ) \ge 0 \;\; \forall 0 < x < 1/2 $$

(this can be seen by just computing the sums explicitly).

There, he states that the first two proofs of the result are quite hard (given, as PhoemueX said, by Jackson and Gronwall), and this lead mathematicians to look for simpler proofs. A remarkable result by Turán generalizes this fact:

If, for some absolutely convergent series of numbers $\{a_k\}$ we have the positiveness property, that is, if it satisfies

$$ \sum_{k} a_k \sin2\pi(k+1/2)x \ge 0, \; \forall \; 0<x<1/2 $$

Then it also satisfies the property

$$ \sum_{k} a_k \frac{\sin2\pi k x}{\pi k } \ge 0 , \;\forall \; 0<x<1/2 $$

As mentioned in the Askey's survey, Turán gave a complex-analytic solution. There, he also gives a proof due to himself, Fitch and Gasper.

Still, it is quite remarkable that Landau's proof is probably the sortest one to ever appear. Of course, there's much more that can be said about this inequality, but I'll leave it for the interested people to Google the Theorem to check it.