I have a question on how to get the correct upper bound of a Schwartz function. Unfortunately, I've never understood this even though I've seen my professors do it a thousand times. I figured it's about time I seek some clarification!
Suppose $K > 0$ and let $Q = [-K,K]^n$. Let $y,z \in Q$ and let $w \in Q$ be on the line segment jointing $y$ and $z$. Finally, let $f$ be a in the Schwartz class and let $M > 0$. I want to show there is a constant $C = C(n,M,K)$ so that $$ \left|\nabla \widehat{f}(w)\right| \leq \dfrac{C}{(2\sqrt{n}K + |w|)^M}. $$ Okay, so here's my understanding. Since $f$ is Schwartz, there is a constant $C_0 = C_0(M)$ so that $$ \left|\nabla \widehat{f}(w) \right| \leq \frac{C_0}{(1 + |w|)^{M}}. $$ Technically, I used that we have similar estimates on the first partials of $\widehat{f}$ to get the above inequality, but that's okay.
Now here is the part I've never understood and I know it should just be a simple geometric argument. I realize that the length of $Q$ is $\ell (Q) = 2\sqrt{n}K$ and so what we want to do is replace the $1$ with $\ell(Q)$. As my professor would say, we can do this for free (up to a constant depending on dimension and $K$ of course), and I believe that it is because the point $w$ lies in $Q$. However, I am not sure how to obtain that result analytically and rigorously. Could someone please explain this for me and close this gap in my education?? Thanks a bunch in advance!
I've figured this one out. Let $A > 0$ be constant and let $x \in \mathbb{R}^n$ be arbitrary. Then $$ \frac{1}{1+|x|} \leq \frac{\max(A,1)}{A+|x|}. $$ Note that this will imply my special case above. If $0 < A \leq 1$, the result is immediate since then $\max(A,1) = 1$ and $1 + |x| \geq A + |x|$. If $A > 1$, then $A + |x| \leq A + A|x| = A(1 + |x|)$ and so the result is obtained since $\max(A,1) = A$ in this case. Wow that was easier than I thought! I really tend to over think things...