Let $K\in L^1(\mathbb R^d)$ be bounded and supported on a set $E$ of finite measure and let $\int_{\mathbb R^d}K=1.$ Define $K_{\delta}(x)=\delta^{-d}K(\delta^{-1}x)$ for every $\delta>0$. Show that $(K_{\delta})_{\delta}$ is an approximation of the identity, that is, it satisfies: $$\text{i) }\int_{\mathbb R^d}K_\delta(x)dx=1\quad\quad\text{ii) }\vert K_\delta(x)\vert\leq A\delta^{-d}\quad\quad\text{iii) }\vert K_\delta(x)\vert\leq A\delta\vert x\vert^{-(d+1)}$$ for all $\delta>0$ and $x\in\mathbb R^d$.
$\textbf{My Attempt & Concerns:}$ To verify (i), I use the dilation property of the Lebesgue integral in $\mathbb R^d,$ which can be found here How to prove Dilation property of Lebesgue integral. Mainly, we have $$\int_{\mathbb R^d}K_\delta(x)dx=\int_{\mathbb R^d}\delta^{-d}K(\delta^{-1}x)dx=\delta^{-d}\delta^{d}\int_{\mathbb R^d}K(x)dx=1.$$
To verify (ii), we use the hypothesis that $K$ is bounded and supported on a set of finite measure, so that if $\delta^{-1}x\notin E$ then we have that $\vert K_{\delta}(x)\vert=0,$ so it is surely bounded by the claimed quatity, with say $A$ being the bound for $K$. If $\delta^{-1}x\in E$ then we use that $\vert K\vert\leq A$ in $\mathbb R^d,$ to obtain that $$\vert K_{\delta}(x)\vert=\vert\delta^{-d}K(\delta^{-1}x)\vert\leq A\delta^{-d}.$$
The third property is the one I do not see a way to prove. If $\delta^{-1}x\notin E$ then it is fine, as $K$ is supported on $E$ so we certainly have the bound. However, if $\delta^{-1}x\in E$, then we have that $$\vert K_\delta(x)\vert=\delta^{-d}\vert K(\delta^{-1}x)\vert=\frac{\delta\vert K(\delta^{-1}x)\vert}{\delta^{d+1}}\leq\frac{\delta A}{\delta^{d+1}},$$ but I do not see a way to get rid of $\delta^{d+1}$ at the bottom and end up with $\vert x\vert^{d+1}.$
Any hints on how to proceed on the third part? Please, only hints. Any feedback on the other parts are also welcomed.
Thank you for time.
Prove it for the case when $K$ is compactly supported, say $supp(K)\subset [-n,n]$ . In general case for any $\epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $\epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $\epsilon$ and big $n$ should give you the result.