# The Laplacian as the difference between mean value and a point

I am reading a book (Spectral Theory in Riemannian Geometry), which opens with an interpretation of the Laplacian.

To get a better feeling about the Laplacian operator, consider for example a one-dimension $\mathcal{C}^3$ function $u:\mathbb{R} \to \mathbb{R}$. The mean value of $u$ on the compact set $[-h,h]$ is given by $$\bar{u} := \frac{1}{2h}\int_{-h}^{h}u(x)dx.$$ Now, using the Taylor expansion of $u$ around the origin we get: for all $[-h,h]$, $$u(x) = u(0) + u'(0)x + u''(x)\frac{x^2}{2} + u'''(0)\frac{x^3}{6} + o(x^4).$$ Therefore, $$\bar{u} = \frac{1}{2h}\left(\int_{-h}^{h}u(0) + u'(0)x + u''(0)\frac{x^2}{2} + u'''(0)\frac{x^3}{6}\right) + o(h^4)$$ i.e., $\bar{u} = u(0) + \frac{u''(0)}{12}h^2 + o(h^4),$ hence $\bar{u} - u(0) = \frac{u''(0)}{12}h^2 + o(h^4).$ In other words, $$\Delta u(0) = u''(0) = \frac{12}{h^2}(\bar{u} - u(0)) + o(h^2),$$ thus the Laplacian of $u$ measures the difference between the function $u$ at $0$ and the mean value of $u$ on the neighborhood $[-h,h].$

I have a few simple questions mostly pertaining to $o$.
(1) Wikipedia explicitly gives the definition for $f(x)=o(g(x))$ assuming $x\to\infty$, but in our case $x$ is approaching $0$. What is our new definition?
(2) Why is it $o(h^4)$ rather than some other value in the interval?
(3) Why do we add $o(h^2)$?

First, it appears that there’s an error in the quoted material: The expansion of $u(x)$ via Taylor’s formula should have $O(x^4)$ as the error term instead of $o(x^4)$.

(1) $\phi(\mathbf v)\in o(g(\mathbf v))$ if it’s defined in some ball around the origin and for all $\varepsilon>0$ there is a $\delta>0$ such that for all $\mathbf v$, $||\mathbf v||\le\delta\Rightarrow||\phi(\mathbf v)||\le\varepsilon\;||g(\mathbf v)||$. However, what you really need here is the definition of big-Oh: $\psi(\mathbf v)\in O(\mathbf v)$ if there is some constant $k>0$ such that $||\psi(\mathbf v)||\le k\;||\mathbf v||$ for all $\mathbf v$ in a neighborhood of the origin. More generally, for $\psi(\mathbf v)\in O(g(\mathbf v))$ we need $||\psi(\mathbf v)||\le k\;||g(\mathbf v)||$ in the preceding. Informally, $\phi(\mathbf v)\in o(g(\mathbf v))$ means that $\phi(\mathbf v)$ vanishes “faster” than $g(\mathbf v)$, while $\psi(\mathbf v)\in O(g(\mathbf v))$ vanishes “as fast as” $g(\mathbf v)$.

(2) The $O(h^4)$ term comes from integrating the $O(x^4)$ term of the Taylor expansion. The last term in the expression stands for some function $g(x)\in O(x^4)$. Then we have $$\left|\int_{-h}^hg(x)\;dx\right| \le \left|\int_{-h}^h\left|g(x)\right|\;dx\right| \le \left|\int_{-h}^hk\left|x^4\right|\;dx\right| \le k\left|\int_{-h}^h\left|x^4\right|\;dx\right|,$$ which means that the error term after integrating is $O(h^5)$. Dividing this by $\frac1{2h}$ gives $O(h^4)$.

(3) In the last step, when you solve $\bar{u} - u(0) = \frac{u''(0)}{12}h^2 + O(h^4)$ for $u''(0)$, you get the term $O(h^4)/h^2$, which is $O(h^2)$.

• For (2), what about the term $-o((-h)^4=-o(h^4)?$ Does $o(h^4) - o(h^4) = o(h^4)$? – user281519 Oct 19 '15 at 0:11
• And I meant $-o((-h)^4)$. – user281519 Oct 19 '15 at 0:22
• Before going any further, I believe that there’s a systematic error in the derivation you’ve quoted. The little oh in Taylor’s formula should be a big Oh—the error term in the expansion of $u(x)$ doesn’t vanish faster than $x^4$. That makes more sense, since in general you can’t say anything about the integral of a little oh, but you can move an integral “inside” of a big Oh. See the answer for more detail about (2). – amd Oct 19 '15 at 0:44
• Wikipedia says it's little-$o$. Perhaps we are switching to big-$O$ at the point of integration? – user281519 Oct 19 '15 at 0:53
• Sounds like there’s a mistake in Wikipedia, then. It can’t be little-oh since the next term of the Taylor expansion is a multiple of $x^4$ plus an error term, which together is clearly $O(x^4)$, not $o(x^4)$. I looked up Taylor’s formula in other sources to make sure, and it is indeed big-Oh everywhere else. – amd Oct 19 '15 at 1:15