# FULL MediaChance Photo Brush 5.30 Serial

## FULL MediaChance Photo Brush 5.30 Serial /*============================================================================= Copyright (c) 2001-2008 Joel de Guzman Copyright (c) 2001-2008 Hartmut Kaiser Distributed under the Boost Software License, Version 1.0. (See accompanying file LICENSE_1_0.txt or copy at =============================================================================*/ #ifndef BOOST_SPIRIT_INCLUDE_CLASSIC_STORED_RULE #define BOOST_SPIRIT_INCLUDE_CLASSIC_STORED_RULE #include #endif Q: Which are the minimal aires units of the space $L^2(\mathbb R^n)$? If $f\in L^2(\mathbb R^n)$, $n>2$, then can we say that $$\int_{\mathbb R^n}\vert f(x)\vert^2\,dx\geq\frac{1}{n^2}\int_{\mathbb R^n}\vert\hat f(\xi)\vert^2\,d\xi.$$ There are a lot of constants, so you better compute it. My thought is that if the rhs is large, the lhs is not less than than the rhs (by the Riemann-Lebesgue lemma), but that’s a guess. A: The answer is yes. Take $g(x)=e^{ -x^2}$, then $g$ is in $L^2(\Bbb R^1)$, and $g otin L^1(\Bbb R^1)$. Also, $\hat g(\xi)=e^{ -\xi^2}$. Now, $||g||_2=||g||_1$. Indeed, $||g||_1\le \int |g|\le||g||_2$, which implies that $||g||_1=||g||_2$ by the RHS is dominated by the LHS. Also, \$\int |\hat g(\xi)|^2=\ f30f4ceada