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Let $J$ be a mollifier, e.g. a function in $J \in C^\infty(\mathbb R^n)$ with the properties $J\geq 0$ and $\int J(x) \mathrm dx=1$ and $J(x)=0$ for all $x$ with $|x|>1.$ Now define $J_\varepsilon (x):=\varepsilon ^{-n}J(\varepsilon ^{-1}x)$ and $$(J_\varepsilon \star u)(x)=\int_{\mathbb R^n}J_\varepsilon (x-y)u(y) \mathrm dy$$ (the symbol $\star$ denotes convolution). How does one prove, that for every function $u \in C^\infty_0(\mathbb R^n)$ (with compact support) the following inequality holds:

$$\left \| u-J_\varepsilon \star u \right \|_1 \leq c \cdot \varepsilon \left \| u \right \|_{1,1}\quad ?$$

Here, $||u||_{1,1}=||u||_{L^1}+\sum_{j=1}^n||\partial_j u||_{L^1}$.

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What is $||u||_{1,1}$? –  Davide Giraudo Jan 16 '12 at 21:39
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In general, this norm is defined by $$ \left \| u \right \|_{m,p}:=\left ( \sum _{\left | \alpha \right |\leq m} \left \| D^\alpha u \right \|^p_p \right )^{\frac{1}{p}}$$ ($\alpha$ is a multi-index) –  David75 Jan 16 '12 at 21:44
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We have \begin{align*} |u(x)-J_{\varepsilon} \star u(x)|&= \left|\int_{B(0,\varepsilon)}J_{\varepsilon}(y)(u(x-y)-u(x))dy\right|\\ &\leq \int_{B(0,1)}J(t)(u(x-\varepsilon t)-u(x))dt\\ &\leq \int_{B(0,1)}J(t)\sum_{j=1}^n\int_0^1\left|\frac{\partial u}{\partial x_j}(x-\varepsilon ts)\right|dsdt, \end{align*} and integrating over $\mathbb R^n$ (in fact a compact subset) \begin{align*} \lVert u -J_{\varepsilon} \star u(x)\rVert_1 &\leq\int_{B(0,1)}J(t)\sum_{j=1}^n\int_0^1\int_{\mathbb R^n}\left|\frac{\partial u}{\partial x_j}(x-\varepsilon ts)\right|dxdsdt\\ &=\varepsilon\int_{B(0,1)}J(t)\int_0^1\sum_{j=1}^n\int_{\mathbb R^n}\left|\frac{\partial u}{\partial x_j}(\varepsilon x'-\varepsilon ts)\right|dx'dsdt\\ &\leq \varepsilon \int_{B(0,1)}J(t)\int_0^1\lVert u\rVert_{1,1}dsdt\\ &= \varepsilon \lVert u\rVert_{1,1}. \end{align*}

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