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I have one question of one of the steps in the proof of Morrey's inequality in Evans second edition. That is in the changing of variable, with $y = x+tw$, $t = |x-y|$. Evans do the following:

$$ \int_0^s \int_{ \partial B(0,1) } |Du(x+tw)|dS(w) dt = \int_0^s \int_{\partial B(x,t)} \frac{|Du(y)|}{t^{n-1}}dS(y)dt.$$

Here $x,y,w \in \mathbb{R}^n$.I'm wondering how the $t^{n-1}$ appears in the denominator? There should be a jacobian, due to the change of variable, however I am unsure how this is treated in the change from $dS(w)$ to $dS(y)$.

Henrik Finsberg
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  • it is without doubt coming from the Jacobian. If $f(w) = x + tw$, then calculate the Jacobian of $f$ from its definition. – MBM Mar 31 '13 at 10:54
  • yes, but why $ \frac{1}{t^{n-1}} $ ? – Henrik Finsberg Mar 31 '13 at 11:04
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    if these were single variables, you would get $dy = tdw$. $(n - 1)$ because happening on a surface. – MBM Mar 31 '13 at 11:10
  • Like MBM says, that is the way the surface elements changes under scaling. To prove it rigorously you should use polar coordinates. Otherwise just content yourself with the intuitive idea. – Giuseppe Negro Mar 31 '13 at 11:30

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Hausdorff Measure and The Coarea formula are what you need to make it really rigorous. The Coarea formula though is in the appendix and working through Evans PDE book I reckon you can take it as a sort of 'black box' result. As a reference though have a look at sections 2 and 3 of Gariepy and Evans book: Measure Theory and Fine Properties of Functions.

In Evans's notation in the surface integral over an n-dim ball in $\mathbb{R}^n$ the $dS$ corresponds to the Hausdorff $n-1$ dim measure of the surface of the ball. In other words, you get the surface area of the $n$ dim ball. In particular, \begin{equation} \mathcal{H}^{n-1}(\partial B(x, r))=\int_{\partial B(x, r)}\mathrm{d}S=n\alpha(n)r^{n-1}. \end{equation}

So I guess one way to 'see' the change of variables which is just a scaling map in your work above is as follows. We have \begin{equation} \int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\mathrm{d}S\mathrm{d}t= \int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t. \end{equation} Let $y=x+tw$, then \begin{equation} t=\vert x-y\vert\Rightarrow y\in\partial B(x, t) \end{equation} and (symbolically) \begin{equation} dS_y=n\alpha(n)t^{n-1}=t^{n-1}dS \end{equation} because \begin{equation} \mathcal{H}^{n-1}(\partial B(0, 1))=\int_{\partial B(0, 1)}\mathrm{d}S=n\alpha(n). \end{equation} So now putting everything together we have \begin{equation} \int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t=\int_{0}^{s}\int_{\partial B(x, t)}\vert\frac{\vert Du(y)\vert}{t^{n-1}}\mathrm{d}S_y\mathrm{d}t \end{equation}

Nirav
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