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.

  • 4
    $\begingroup$ 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 $\varepsilon^n$! $\endgroup$ – Drefain Feb 8 at 13:27
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    $\begingroup$ 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. $\endgroup$ – Ian Feb 8 at 13:39
  • $\begingroup$ 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 $x$. $\endgroup$ – Ian Feb 8 at 13:40

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