Please check my proof for the following result:
I want to prove a result for $D\eta_{\epsilon}$ the gradient of the standard mollifier $\eta$. The function $\eta$ is defined as follows:
Let $\eta_{\epsilon}(x) = \frac{1}{\epsilon^{n}}\eta(\frac{x}{\epsilon})$ where $\eta(x) = Ce^{\frac{1}{|x|^{2}-1}}$ if $|x|<1$ and $0$ if $|x| \geq 1$.
We know that $\eta(x)$ is smooth and bounded with compact support $\text{spt}(\eta_{\epsilon}) \subset B(0,\epsilon) \subset U$ where $U \subset \mathbb{R}^{n}$, obtaining its maximum at 0.
I want to show what $D\eta_{\epsilon}(x)$ is and conclude that $D\eta_{\epsilon} \in L^{\infty}(\mathbb{R}^{n})$. Where $D$ is the gradient vector. To use the chain rule I will let $g_{i}(x_{i}) = \frac{x_{i}}{\epsilon}$ and then by the chain rule we have $\frac{\partial \eta}{\partial x_{i}} = \frac{\partial \eta}{\partial g_{i}}\frac{\partial g_{i}}{\partial x_{i}}$.
$D\eta_{\epsilon}(x) := D(\frac{1}{\epsilon^{n}} \eta(\frac{x}{\epsilon}))$ $= \frac{1}{\epsilon^{n}}D(\eta(\frac{x}{\epsilon}))$ =$\frac{1}{\epsilon^{n}}(\frac{\partial \eta(\frac{x}{\epsilon})}{\partial x_{i}}, ...,\frac{\partial \eta(\frac{x}{\epsilon})}{\partial x_{n}})$ = $\frac{1}{\epsilon^{n}}(\frac{1}{\epsilon}\frac{\partial \eta}{\partial g_{i}}, ...,\frac{1}{\epsilon}\frac{\partial \eta}{\partial g_{n}})$ = $\frac{1}{\epsilon^{n+1}}(\frac{\partial \eta}{\partial g_{i}}, ...,\frac{\partial \eta}{\partial g_{n}})$
Since $\eta$ is bounded with compact support it follows that each $\frac{\partial \eta}{\partial g_{i}}$ will be bounded on compact supports and therefore $|D \eta_{\epsilon}| < \infty$ and $D \eta_{\epsilon} \in L^{\infty}(\mathbb{R}^{n})$.
Is this proof fine? Thanks for assistance.
I don't think you need the computation of $D\eta_\epsilon$. The gradient of any smooth compactly supported function is bounded, being itself a continuous function with compact support. Since $\eta$ is smooth and has compact support, the rescaled function $\eta_\epsilon$ also has those properties. The conclusion follows.