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I am reading Differentiable manifolds from Warner.

In order to prove that the dimension of the tangent space is the same as the dimension of the manifold, they use the following calculus lemma -

If $g$ is of class $C^k$ ($k \geq 2$) on a convex open subset $U$ about $p$ in $\mathbb{R}^d$, then for each $q \in U$, $g(q)\ =\ g(p) + \sum_{i=1}^d \frac{\partial g}{\partial r_i}|_p (r_i(q)-r_i(p))+\sum_{i,j}(r_i(q)-r_i(p))(r_j(q)-r_j(p))\int_0^1(1-t)\frac{\partial^2g}{\partial r_i\partial r_j}|_{(p+t(q-p))} dt.$

This is the Taylor expansion. It further says, if $g\in C^\infty$, then the integral as a function of $q$ is of class $C^\infty$. How is this? Do we have to use fundamental theorem of calculus or something like that?

user52991
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    Read the condition here. You can show by induction that the integral is of the class $C^k$ for any $k\in \Bbb N$ – SBF May 30 '13 at 12:20
  • I am still not clear. Are we to consider the integrand as a function of $q$ and $t$? If so how is the partial derivative of the integrand with respect to q continuous, i.e. how do we say that? – user52991 May 30 '13 at 18:00

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The expression $\int_0^1(1-t)\frac{\partial^2g}{\partial r_i\partial r_j}|_{(p+t(q-p))} dt$ can be differentiated under the integral sign to obtain all of its partial derivatives. Since $g$ is $C^\infty$, so is the integral.

Mikhail Katz
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  • That's not true right? $g(q)-g(p) - \sum_{i=1}^d \frac{\partial g}{\partial r_i}|p (r_i(q)-r_i(p)) = \sum{i,j}(r_i(q)-r_i(p))(r_j(q)-r_j(p))\int_0^1(1-t)\frac{\partial^2g}{\partial r_i\partial r_j}|_{(p+t(q-p))} dt$. So can we still conclude the integral is $C^\infty$? – user52991 May 30 '13 at 17:55
  • Sorry, I was just relying on your formula too quickly, which I misread slightly. Divide by the expression $\sum_{i,j}(r_i(q)-r_i(p))(r_j(q)-r_j(p))$ to get a formula for the integral in terms of $g(q)$ and its first partials at $p$, and the same argument applies. You don't even need differentiability of the partials since you are only using them at a single point $p$. – Mikhail Katz May 31 '13 at 08:24
  • How can I do that? The summation runs over the integral too, there are terms in the integrand which depend on $i, j$. – user52991 May 31 '13 at 14:29
  • Right. I really answered this in too much of a hurry. Then you should be asking about "the integrals as a function of q" (in the plural) rather than "the integral as a function of q"? – Mikhail Katz Jun 02 '13 at 07:24
  • Yes that's right. Integrals as functions of q. – user52991 Jun 02 '13 at 18:22