prove the following inequality for integral

Question

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

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

Prove the following inequality holds：

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

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

Answer 1

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