# prove the following inequality for integral

Mathematics Asked by yi li on December 24, 2020

Let $$eta$$ be the smooth function supported in $$B_1(0)$$.such that $$inteta = 1$$

Let $$u$$ be a smooth function defined on an open set $$Vsubset Bbb{R}^n$$ that is $$uin C_c^infty(V)$$.

Prove the following inequality holds：

$$int_{B_1(0)}eta(y)int_0^1int_V|Du(x-epsilon ty)|dxdtdyle int_V|Du(z)|dz$$

I do it as follows extend the integral domain $$Vto Bbb{R^n}$$ then the inequality $$int_V|Du(x-epsilon ty)dxle int_mathbb{R^n}|Du(x-epsilon ty)|dx = int_V|Du(z)|dz$$.Is my proof correct?If not how to do it?

begin{aligned} int_{B(0,1)} eta(y) int_{0}^{1} int_{V}left|D u_{m}(x-epsilon t y)right| d x d t d y &=int_{B(0,1)} eta(y) int_{0}^{1} int_{V}left|D u_{m}(x)right| d x d t d y \ &=int_{V}left|D u_{m}(x)right| d x int_{B(0,1)} eta(y) d y int_{0}^{1} d t \ &=int_{V}left|D u_{m}(x)right| d x end{aligned}

Correct answer by GAUSS1860 on December 24, 2020

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