2
$\begingroup$

On pages $116-117$ of Stein and Shakarchi's Fourier Analysis book, there is a claim about the Fejer Kernel

$$F_N(t) = \frac{1}{N} \frac{\sin^2(Nt/2)}{\sin^2(t/2)}$$

and the claim is

$$|F_N'(t)| \leq \frac{A}{t^2} \qquad \forall |t| \leq \pi$$

where $A$ is independent of $N$. The book states that this bound falls out of the fact that

$$F_N'(t) = \frac{\sin(Nt/2)\cos(Nt/2)}{\sin^2(t/2)} - \frac{1}{N}\frac{\cos(t/2) \sin^2(Nt/2)}{\sin^3(t/2)}$$

and that for $|t| < \pi$, we have $|\sin(Nt/2)| \leq CN|t|$ and $|\sin(t/2)| \geq c|t|$.

The bounds of $\sin$ and the calculation of the derivative are correct, but I can't get the desired bound on $|F_N'(t)|$. Based on the phrasing and the fact that $\cos(Nt/2)$ and $\cos(t/2)$ can't be controlled very well throughout all of $t \in [-\pi, \pi]$, I assume that the best move is to triangle inequality. However, this gives

$$\Big|F_N'(t)| \leq \frac{\alpha N}{t} + \frac{\beta N}{t} = \gamma \frac{N}{t}$$

I'm not sure if expanding the Taylor series would be anymore fruitful, because the claim (as stated in the book) is for all of $t \in [-\pi, \pi]$. However the proof only uses this bound when $|t| \geq 1/N$, which may be helpful though I haven't made any progress with this extra condition.

$\endgroup$

1 Answer 1

3
$\begingroup$

We just need to apply $\left|\sin(Nt/2)\right|\leq CN|t|$ and the trivial $\left|\sin(Nt/2)\right|\leq 1$ more flexibly. In fact $$\left|\frac{\sin(Nt/2)\cos(Nt/2)}{\sin^2(t/2)}\right|\leq\frac{1}{\sin^2(t/2)}\leq \frac C{t^2}$$ and \begin{align}\left|\frac{1}{N}\frac{\cos(t/2) \sin^2(Nt/2)}{\sin^3(t/2)}\right| & \leq \left|\frac{\sin(Nt/2)}N\right|\left|\frac{\sin(Nt/2)}{\sin^3(t/2)}\right|\\ & \leq C|t|\left|\frac 1{t^3}\right|=\frac{C}{t^2}\end{align}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .