So I have been reading about Fourier analysis lately. Now I am starting to learn about the Fourier transform. In the book, I am reading, the topic starts with the definition of Schwartz space.
Definition:
A function $f$ is called a Schwartz-function iff for every $n,k \in \mathbb{N}_0$ there exists a constant $C(n,k,f)>0$ such that
$\langle x\rangle ^k |f^{(n)}(x)| \leq C(n,k,f) \forall x \in \mathbb{R}$.
My question is regarding the meaning of $\langle x\rangle^k$ and its meaning. The only brackets notation I know is that of the inner product $\langle \cdot,\cdot\rangle$, which has two arguments.
First I thought that $\langle x \rangle^k$ may be some short notation of $\langle x,x \rangle^k$, but after some googling I found that $\langle x \rangle:=(1+|x|^2)^{\frac{1}{2}}$ (for example here Notation in harmonic analysis).
The book keeps going on with noting that by defining Schwartz-functions that way, it follows that every function in the Schwartz-space decays faster than $\frac{1}{p}$ for any polynomial $p$.
Question:
What does $\langle x \rangle$ mean?
And if $\langle x \rangle $ really mean $(1+|x|^2)^{\frac{1}{2}}$ ,
why do Schwartz functions decay faster than $\frac{1}{p}$?
Best Answer
You are correct that $\langle x \rangle = (1+|x|^2)^{1/2}$ (this is sometimes called the "Japanese bracket" of $x$).
To see the polynomial decay of Schwartz functions, note that if $f$ is Schwartz then for every $k$, $$|f(x)| \le C(0,k,f) \langle x \rangle^{-k}$$ so that it suffices to see that for every polynomial $p$, there exists a $k$ such that $ \langle x \rangle^{-k} \lesssim |p(x)|^{-1}$. Since $\langle x \rangle^{-1} \le 1$ and $p$ is bounded on $B(0,1)$, we can assume without loss of generality that $|x|>1$.
Then it suffices to note that if $p(x) = \sum_{j = 0}^k a_j x^j$ and $|x| > 1$ $$|p(x)| \le \Big ( \sum_{j = 0}^k |a_j| \Big) |x|^k \lesssim \langle x \rangle^k$$ which immediately gives the desired result.