Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let

$$\sigma(U)=\iint_U d(x,y)dxdy$$

Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.

I think I have a proof of $\sigma(kU)=k^{2n+1}\sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:

Each of its points of $U$ will have coordinates $x=(x_1,\ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',\ldots,x_n')$ with the relations



Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to $\{x\mid \exists x'\in kU:x'=kx\}=U$:

\begin{align} \sigma(kU) &= \iint_{kU}d(x',y')dx'dy' \\ &= \iint_U d(kx,ky)kdx\ kdy \\ &= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(kx_1-ky_1)^2+\cdots+(kx_n-ky_n)^2}kdx_1\ldots kdx_nkdy_1\ldots kdy_n \\ &= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} k\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}k^{2n}dx_1\ldots dx_ndy_1\ldots dy_n \\ &= k^{2n+1}\int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}dx_1\ldots dx_ndy_1\ldots dy_n \\ &= k^{2n+1}\iint_Ud(x,y)dxdy \\ \sigma(kU)&= k^{2n+1}\sigma(U) \end{align}

Is this correct? Thanks.


One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$\int_{x_1^0}^{x_1^1}...\int_{x_n^0}^{x_n^1}\int_{y_0^0}^{y_0^1}...\int_{y_n^0}^{y_n^1}d(x,y)\,dx\,dy.$$

But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,y\in U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.