# Confusion on $\vert\vert \varphi\vert\vert_{p}$ where $p \in [1,\infty[$

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.

• Your answer seems fine to me! Maybe I'm missing something too xd. With regards to your first point, you're not differentiating n times but are differentiating n variables each with an $\varepsilon$ scaling. This is what gives you the $\varepsilon^n$! – Drefain Feb 8 at 13:27
• Indeed the $L^p$ norm will in general be different between the two. The situation is obvious for $L^\infty$ (you haven't changed the extrema of $\varphi$), and it is slightly less obvious that it is the same for $L^1$ (essentially you've concentrated the "mass" of $\varphi$ more tightly but also stretched it by exactly the right amount to compensate for the change in "volume"). But for $L^p$ the requisite rescaling is different. – Ian Feb 8 at 13:39
• As for this issue about the change of variable, this is part of why people sometimes use explicit notations like $\vec{x}$ or $\mathbf{x}$ to denote vectors. Here it is true that say $dx_1/\epsilon=dx_1'$ but there is one such $\epsilon$ for every component of $x$. – Ian Feb 8 at 13:40