Let $\varphi \in C^{\infty}(\mathbb R^n)$ and 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 $\epsilon$
I understand the case $p =\infty$, namely $\vert \vert \varphi_{\epsilon}\vert \vert_{\infty}=\sup_{x \in \mathbb R^{n}}|\varphi_{\epsilon}(x)|=\sup_{x \in \mathbb R^{n}}|\epsilon^{-n}\varphi(x/\epsilon)|=\epsilon^{-n}\sup_{ x\in \mathbb R^{n}}|\varphi(x/\epsilon)|=\epsilon^{-n}||\varphi||_{\infty}$
I get confused by the case $p \in [1, \infty[$
Surely, by definition:
$\vert \vert \varphi_{\epsilon}\vert \vert_{p}^{p}=\int_{\mathbb R^{n}}|\epsilon^{-n}\varphi(x/\epsilon)|^{p}dx$ and then by substitution $( x/\epsilon = x^{'}\Rightarrow dx/\epsilon^{n}=dx^{'})(*)$
First question: surely differentiating $n-$times has no effect on $\epsilon$, so surely $(*)$ should be $dx/\epsilon=dx^{'}$ rather than $dx/\epsilon^{n}=dx^{'}$
In any case, assuming $(*)$ holds: $\int_{\mathbb R^{n}}\epsilon^{-np}|\varphi(x^{'})|^{p}\epsilon^{n}dx^{'}=\epsilon^{-n(p-1)}\int_{\mathbb R^{n}}|\varphi(x^{'})|^{p}dx^{'}\Rightarrow \vert \vert \varphi_{\epsilon}\vert \vert_{p}=\epsilon^{\frac{-n(p-1)}{p}}||\varphi||_{p}$
Second Question: I have been told that the answer must be $\vert \vert \varphi_{\epsilon}\vert \vert_{p}=\vert \vert \varphi\vert \vert_{p}$
But I cannot see what I did wrong.