$L^2$ and uniform norm of $\text{sinc}\, x$ and its derivatives Looking at the graphs of the derivatives of $\mathrm{sinc}\,x$, it appears that they all are bounded by $1/x$, with $[\mathrm{sinc}\,(x)]'$ the sole exception:

A few questions:
1) With the exception of $\mathrm{sinc}'$, is it true that $\left|\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}} \mathrm{sinc}(x)\right| \le \min\left\{1, 1/|x|\right\}$ and how can it be proved?
2) Can it be shown that $\left\|\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}} \mathrm{sinc}\right\|_{\infty} \le \lambda\left\|\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \mathrm{sinc}\right\|_{\infty}$ (or $\le \lambda^{n}$) for some $\lambda \in (0,1)$?
3) Is there a slick way of showing that $\left\| \frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}} \mathrm{sinc}\right\|_{2}^{2} = \frac{\pi}{2n+1}$?
 A: The sinc function is the inverse Fourier transform of the characteristic function $\chi_{[-1,1]}$, at least up to a constant:
$$
\begin{align}
         \chi_{[-1,1]}^{\vee}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{ixs}ds \\
            & = \left.\frac{1}{\sqrt{2\pi}}\frac{e^{isx}}{ix}\right|_{s=-1}^{1} \\
            & = \frac{1}{\sqrt{2\pi}}\frac{e^{ix}-e^{-ix}}{ix} = \sqrt{\frac{2}{\pi}}\frac{\sin(x)}{x}.
\end{align}
$$
So the Fourier transform of the sinc function is
$$
         \widehat{\mbox{sinc}}(s) = \sqrt{\frac{\pi}{2}}\chi_{[-1,1]}(s).
$$
The Fourier transform preserves $L^{2}$ norms. So the $L^{2}$ norm of sinc is
$$
       \|\mbox{sinc}\|_{L^{2}}=\sqrt{\frac{\pi}{2}}\left(\int_{-1}^{1}\chi_{[-1,1]}^{2}\,ds\right)^{1/2}=\sqrt{\pi}.
$$
Differentiation acting on the Fourier transform of a function is the same as the Fourier transform of $(-ix)$ times the original function, at least assuming all functions are in $L^{2}$. Therefore,
$$
\begin{align}
       \left\|\frac{d^{n}}{dx^{n}}\mbox{sinc}\right\|_{L^{2}} & =
     \sqrt{\frac{\pi}{2}}\left( \int_{-1}^{1}|s^{n}|^{2}\,ds\right)^{1/2} \\
    & = \sqrt{\frac{\pi}{2}}\left(\frac{2}{2n+1}\right)^{1/2} 
   = \sqrt{\frac{\pi}{2n+1}}.
\end{align}
$$
Before taking norms, you have an explicit form for sinc and its derivatives:
$$
             \frac{\sin x}{x}= \frac{1}{2}\int_{-1}^{1}e^{isx}\,ds=\int_{0}^{1}\cos(sx)\,ds.
$$
Therefore,
$$
\begin{align}
      \frac{d^{n}}{dx^{n}}\frac{\sin x}{x} & = \int_{0}^{1}s^{n}\cos^{(n)}(sx)\,ds
\\
    & = x^{-n-1}\int_{0}^{1}(sx)^{n}\cos^{(n)}(sx)xds \\
    & = x^{-n-1}\int_{0}^{x}u^{n}\cos^{(n)}(u)\,du.
\end{align} 
$$
A crude estimate would make you think that the integral over $[0,x]$ with respect to $u$ grows like $x^{n+1}$, but that's not true because the integral over a positive cycle of $\cos^{(n)}(u)$ somewhat cancels the integral over the next negative cycle. To make this explicit, first notice that the n-th derivative of $\cos$ is $\pm \sin$ or $\pm \cos$. Then notice that
$$
         (u+\pi)^{n}-u^{n}=[(u+\pi)-u][(u+\pi)^{n-1}+(u+\pi)^{n-2}u+\cdots+(u+\pi)u^{n-2}+u^{n-1}]
$$
is bounded on $u > 0$ by
$$
        \pi[(u+\pi)^{n-1}+(u+\pi)^{n-1}+\cdots+(u+\pi)^{n-1}] = n\pi(u+\pi)^{n-1}.
$$
Then the integral is reduced to an integral only over the positive cycles of $\cos^{(n)}$ times $n\pi(u+\pi)^{n-1}$, which then gives you an estimate of $C/x$ for the n-th derivative of sinc. If you're careful about all of the constants and the intervals of integration, then I think you'll get what you want, at least for large $x$. There might be something strange going on for $x$ within the first cycle, though, or during any odd cycle. I'll leave that to you, but at least it's in the ballpark.
