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Let $g$ be a function in the Schwartz space $\mathscr S (\mathbb R)$. Show that for any $l \ge 0$, we have $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l$ by considering separately the cases $|x|\le 2|y|$ and $|x|\ge 2 |y|$.

The Schwartz space is defined as the set $\mathscr S (\mathbb R)$ of all indefinitely differentiable functions $f:\mathbb R\to \mathbb R$ such that $\sup_{x\in\mathbb R} |x|^k |f^{(l)}(x)|<\infty$ for all $k, l \ge 0$.

(This was used in a proof in Elias Stein's book on Fourier Analysis and is not a homework problem. The book just didn't go through this particular step.)

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    $\begingroup$ $$(1+|x-y|)^{-1} \le \frac {1+|y|}{1+|x|}$$ $\endgroup$
    – Piero D'Ancona
    Jun 1, 2011 at 11:04
  • $\begingroup$ @Piero D'Ancona: I'm mostly having trouble with finding where to use the two cases $|x|\ge 2|y|$ and $|x|\le 2|y|$. Any help? $\endgroup$
    – Feynmaniac
    Jun 2, 2011 at 6:48
  • $\begingroup$ @Feynmaniac: About an approach which doesn't separate the two cases $|x|\ge 2|y|$ and $|x|\le 2|y|$: couldn't it be possible the following proof? 1)use the triangle inequality $|x|\le |y|+|x-y|$, 2)apply the binomial formula to $(|y|+|x-y|)^l$, 3)apply the hypothesis on $g$, and finally $|y|^k\le(1+|y|)^l$, for any $y\in\mathbb{R}$ and $0\le k\le l$. $\endgroup$
    – agt
    Jun 3, 2011 at 7:40

1 Answer 1

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We have $$ \vert x\vert^l\vert g(x-y)\vert\le {(\vert x-y\vert+\vert y\vert)}^l\vert g(x-y)\vert \le (1+\vert y\vert)^l(1+\vert x-y\vert)^l\vert g(x-y)\vert $$ so that $$ \vert x\vert^l\vert g(x-y)\vert\le (1+\vert y\vert)^l\underbrace{\sup_{z\in\mathbb R} (1+\vert z\vert)^l\vert g(z)\vert}_{A_l} $$ which is the sought inequality, where $A_l$ in a semi-norm of $g$ in the Schwartz space. Note that no differentiability property for $g$ is necessary, we have used only fast decay of the function itself.

Bazin.

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