I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:
In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:
$$|x|^l |g(x-y)| \leq A_l(1 + |y|)^l \ \forall \ l \geq 0$$ Remember, $S(\mathbb{R})$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^{(k)}(x)| < \infty \ \forall \ k,l$
The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.
Thanks
Best Answer
Recall the following trivial inequality; $$\lvert x\rvert ^{l} \leq \left(\, \lvert x-y \rvert \, + \lvert y \rvert \, \right)^{l}\leq (2\max(\, \lvert x-y\rvert, \, \lvert y \rvert \, )\, )^{l} \leq\, 2^{l} \, \lvert x-y\rvert^{l} + 2^{l}\, \lvert y \rvert ^{l} \qquad \forall l\geq 0$$
Set $$C_{l} := \sup_{x\in\mathbb{R} } \, \lvert x\lvert ^{l} \,\lvert g(x) \, \rvert <\infty \qquad \forall l\geq 0$$
Now using the above estimate, we get $$\lvert x \rvert^{l} \, \lvert g(x-y) \rvert \leq 2^{l} \, \lvert x-y\rvert^{l} \, \rvert g(x-y)\rvert + 2^{l}\, \lvert y \rvert^{l} \, \lvert g(x-y) \rvert \leq \\ 2^{l}\, C_{l}+ 2^{l} \, C_{0} \, \lvert y \rvert^{l} \leq \underbrace{2^{l}(C_{l}+C_{0})}_{=A_{l}}\, (1+\lvert y\rvert)^{l}$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ \lvert x \rvert^{l} \, \lvert (g\ast f)(x) \rvert \leq \int _{\mathbb{R}} \lvert x \rvert^{l} \, \lvert g(x-y) \rvert \, \lvert f(y) \rvert \, dy \leq A_{l} \int_{\mathbb{R}} (1+\lvert y \rvert )^{l+2} \, \lvert f(y) \rvert \frac{dy}{(1+\lvert y \rvert)^{2}} \leq \\ A_{l} \, C_{l+2} \,\int_{\mathbb{R}} \frac{dy}{(1+\lvert y \rvert)^{2}}< \infty $$ Since the bound is independent of $x\in \mathbb{R}$, the claim now follows.