Let $\varphi \in C^{\infty}(\mathbb R^n)$ with compact support. For $\epsilon > 0$ define $\varphi_{\epsilon}(x):=\epsilon^{-n}\varphi(x/\epsilon)$ such that $\varphi_{\epsilon} \in C^{\infty}(\mathbb R^n)$ with compact support.
Determine $\vert \vert \varphi_{\epsilon}\vert \vert_{p}$ with $1 \leq p \leq \infty$ in dependence on $\epsilon$
First question: I'm still confused as to why the condition that $\varphi$ and $ \varphi_{\epsilon}$ are continuous and have compact support is actually necessary?
Otherwise my ideas: $\vert \vert \varphi_{\epsilon}\vert \vert_{\infty}=\vert \vert \epsilon^{-n}\varphi(x/\epsilon)\vert \vert_{\infty}$ and since $\varphi_{\epsilon}$ is continuous then $\vert \vert \epsilon^{-n}\varphi(x/\epsilon)\vert \vert_{\infty}=\epsilon^{-n}\sup_{x \in X}|\varphi(x/\epsilon)|$
what else am I supposed to do here?
For $p<\infty$, it is $\|\varphi_\varepsilon\|_p^p=\displaystyle{\int_{\mathbb{R}^n}\varepsilon^{-n}|\varphi(x/\varepsilon)|^pdx}$. Set $x'=x/\varepsilon$. Then we have that $dx'=\varepsilon^{-n}dx$, since we are on $n$ dimensions. Hence $\|\varphi_\varepsilon\|_p^p=\displaystyle{\int_{\mathbb{R}^n}|\varphi(x')|^pdx'=\|\varphi\|_p^p}$.
For $p=\infty$, it is $\|\varphi_\varepsilon\|_\infty=\displaystyle{\sup_{x\in\mathbb{R}^n}|\varepsilon^{-n}\varphi(x/\varepsilon)|=\varepsilon^{-n}\sup_{x\in\mathbb{R}^n}|\varphi(x/\varepsilon)|=\varepsilon^{-n}\|\varphi\|_\infty}$, since as $x$ ranges all over $\mathbb{R}^n$, so does $x/\varepsilon$.