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Let $varphi:mathbb R^dtomathbb R_+$ be given as
$$
varphi(x) := start{circumstances}
cexpbig(1/(|x|^2-1)large) & mbox{if } |x|le 1
0 & mbox{in any other case},
finish{circumstances}
$$

the place $c>0$ is chosen such that $int_{mathbb R^d}varphi(x)dx=1$. May we show the next convergence for $varphi_t(x):=varphi(x/t)/t^d$
$$lim_{tto 0+} int_{mathbb R^d} large|dx ~=~0?$$
Right here $f:mathbb R^dtomathbb R$ is a set Lipschitz perform and $varphi_tast f$ denotes the convolution of $varphi_t$ and $f$, i.e.
$$(varphi_tast f)(x):=int_{mathbb R^d}varphi_t(y)f(x-y)dy.$$
Certainly, via a change of variable, it follows that
$$ int_{mathbb R^d} large|dx ~=~int_{mathbb R^d} left(int_{mathbb R^d}varphi(y)large |nabla f(x-ty)-nabla f(x)large |dyright)dx.$$
If we all know $nabla f$ is a.e. steady, then we might conclude utilizing the dominated convergence theorem. However I cannot discover any reference on the continuity of $nabla f$. Any proof, feedback or references are extremely appreciated!

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