Schwartz space inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:47:06Z http://mathoverflow.net/feeds/question/66644 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66644/schwartz-space-inequality Schwartz space inequality Feynmaniac 2011-06-01T10:25:53Z 2012-04-11T20:22:00Z <p>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 $|x|\ge 2 |y|$.</p> <p>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 $\sup_{x\in\mathbb R} |x|^k |f^{(l)}(x)|&lt;\infty$ for all $k, l \ge 0$.</p> <p>(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.)</p> http://mathoverflow.net/questions/66644/schwartz-space-inequality/92501#92501 Answer by Bazin for Schwartz space inequality Bazin 2012-03-28T20:18:37Z 2012-03-28T20:18:37Z <p>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.</p> <p>Bazin.</p>