I wondered whether it holds, for a function $f:A\subset\mathbb{R}^3\to\mathbb{R}$, $f\in C(A)$, that, for all $(x_0,y_0,z_0)\in A$, $$\lim_{\sqrt{h_x^2+h_y^2+h_z^2}\to 0}\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz=f(x_0,y_0,z_0)$$I think that, if such an identity held, it would allow us to define quantities like the densities used in physics.

Obviously, for all $\varepsilon>0$ there exists a $\delta$ such that, if $\sqrt{h_x^2+h_y^2+h_z^2}<\delta$ (and therefore $|h_x|,|h_y|,|h_z|$ $<\delta$), then $$\Bigg|\frac{1}{h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdy\Bigg|<\varepsilon$$$$\Bigg|\frac{1}{h_y}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdy-\int_{x_0}^{x_0+h_x}f(x,y_0,z_0)dx\Bigg|<\varepsilon$$$$\Bigg|\frac{1}{h_x}\int_{x_0}^{x_0+h_x}f(x,y,z_0)dxdydz-f(x_0,y_0,z_0)\Bigg|<\varepsilon$$but I am not able to use these inequalities to find an arbitrarily small $\tilde{\varepsilon}$ to majorate $\big|\frac{1}{h_xh_yh_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-f(x_0,y_0,z_0)\big|$ with.

Does what I am trying to prove hold and, if it does, how can it be proved? I $\infty$-ly thank anybody for any answer.

  • 2
    $\begingroup$ You might want to look up Lebesgue's differentiation theorem. $\endgroup$ – Mike Sep 8 '15 at 7:45
  • $\begingroup$ to make that integral well-defined you should assume that $(x_0, y_0, z_0)$ is an interior point and exclude $h_xh_yh_z = 0$. $\endgroup$ – user251257 Sep 8 '15 at 13:27

You don't need Lebesgue density / differentiation theorem, as $f$ is continuous.

As I am lazy, I will prove it in 2D and assume without loss of generality $x_0 = y_0 = 0$ and $f(0,0) = 0$. We need to show that $$ \lim_{r\to 0} \sup_{\|h\| < r, h_xh_y\ne 0} \left | \frac1{h_x h_y} \int_0^{h_y} \int_0^{h_x} f(x,y) \; \mathrm dx\mathrm dy \right | = 0.$$

Now, let $\epsilon > 0$. Then, by continuity of $f$ there is some $r > 0$ such that for every $h$ with $\|h\|< r$ it follows $|f(h_x,h_y)| < \epsilon.$ Notice that for every $(x,y)$ with $|x| \le |h_x|$ and $|y|\le |h_y|$ we have $$ \sqrt{x^2 + y^2} \le \| h \| < r. $$ Thus, for $h_xh_y \ne 0$ and $s_x = \operatorname{sign}(h_x), s_y = \operatorname{sign}(h_y)$ we obtain \begin{align} \left | \frac1{h_x h_y} \int_0^{h_y} \int_0^{h_x} f(x,y) \; \mathrm dx\mathrm dy \right | &= \frac1{|h_x| |h_y|} \left | \int_0^{|h_y|} \int_0^{|h_x|} f( s_x x, s_y y) \; \mathrm dx\mathrm dy \right | \\ &\le \frac1{|h_x| |h_y|} \int_0^{|h_y|} \int_0^{|h_x|} \left | f( s_x x, s_y y)\right | \; \mathrm dx\mathrm dy \\ &< \frac1{|h_x| |h_y|} \int_0^{|h_y|} \int_0^{|h_x|} \epsilon \; \mathrm dx\mathrm dy \\ &= \epsilon. \end{align}

  • $\begingroup$ Thank you so much! The Heine-Borel theorem allows us to see that, if $\|h\|$ is small enough for a closed ball $\bar{B}((x_0,y_0,z_0),\|h\|)$ to be contained in $A$, then $f-f(x_0,y_0,z_0)$ is uniformly continuous and therefore for all $\varepsilon>0$ there is a $\delta$ such that, if $\|h\|<\delta$, for all $(x,y,z)\in A$ such that $\|(x,y,z)-(x_0,y_0,z_0)\|\le \|h\|$, $|f(x,y,z)-f(x_0,y_0,z_0)|<\varepsilon$ and (I write the endpoint of the integrals as if $h_x,h_y, h_z>0$ but reversing then when they are negative doesn't change the absolute value) therefore... $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
  • $\begingroup$ ...$\big|\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)dxdydz-f(x_0,y_0,z_0)\big|$ $=\big|\frac{1}{h_x h_y h_z}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}f(x,y,z)-f(x_0,y_0,z_0)dxdydz\big|$ $\le \frac{1}{|h_x h_y h_z|}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}|f(x,y,z)-f(x_0,y_0,z_0)| dxdydz $ $\le \frac{1}{|h_x h_y h_z|}\int_{z_0}^{z_0+h_z}\int_{y_0}^{y_0+h_y}\int_{x_0}^{x_0+h_x}\varepsilon dxdydz$ $=\varepsilon$. $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
  • $\begingroup$ Interestingly, I would say that the same applies to spaces of any dimension, so that we can say that, if $\boldsymbol{x}_0\in\mathring{A}$, then $$\lim_{\|\boldsymbol{h}\| \to 0,\prod_{i=1}^n h_i\ne 0}\frac{1}{\prod_{i=1}^n h_i}\int_{x_{0,n}}^{x_{0,n}+h_n}...\int_{x_{0,n}}^{x_{0,1}+h_1}f(\boldsymbol{x})dx_1...dx_n=f(\boldsymbol{x_0})$$ Thank you so much again!!! $\endgroup$ – Self-teaching worker Sep 8 '15 at 16:24
  • 1
    $\begingroup$ @Self-teachingDavide: You don't need Heine-Borel for that. It is just continuity at $(x_0,y_0,z_0)$. Yes, it applies to any finite dimension. $\endgroup$ – user251257 Sep 8 '15 at 16:26
  • 1
    $\begingroup$ @Self-teachingDavide: sure. $\endgroup$ – user251257 Sep 8 '15 at 16:53

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.