Why are the remainder terms smooth in Tu's Manifold Lemma 1.4?

Question

I'm trying to start learning about manifold through Loring Tu's book. However, the first result has me somewhat confused. I was for the most part able to follow Lemma 1.4 which proves Taylor's theorem with remainder. However, I am confused as to how Tu was able to easily conclude $g_i(x)$ is in $C^\infty(U)$. Looking at its definition (see picture), it isn't clear to me why this is immediate true. The outer integral "blocking" the smooth function $f$ is what concerns me.

I am convinced from equation 1.1 $$ f(x) = f(p) + \sum (x^i - p^i) g_i(x) $$ that the mixed partials of $g_i(x)$ must exist otherwise $f$ would not be in $C^\infty$. But, why must these mixed partials be continuous over $U$? After all, aren't there examples of partials existing at a point while not being continuous?

Answer 1

I convinced myself that $g_i(x)$ is in fact in $C^\infty$ over $U$. The basic strategy I used involved first thinking about $\frac{\partial g_i}{\partial x^j}(x)$ which by Leibniz integral rule is $\int_0^1 \frac{\partial f}{\partial x^j \partial x^i}(p + t(x-p))dt$. Because $f$ is in $C^\infty$ over $U$, the integrand is also in $C^\infty$ over $U$. By repeatedly taking partial derivatives of $g_i$ and using Leibniz rule, we get a function of the form $G(x) = \int_0^1 F(p + t(x-p))dt$ where $F$ is in $C^\infty$ over $U$. By showing that $G(x)$ is continuous over $U$, it would effectively be proven that $g_i$ and all its mixed partials exist and are continuous over $U$.

Consider forming a closed box $B = [a_1, a_1'] \times \ldots \times [a_n, a_n']$ around $x \in U$, which exists as $U$ is open. Then form a "cone" $K = \{p + \lambda(b - p) \vert b \in B, \lambda \in [0,1]\}$ which lies entirely in $U$ as $U$ is star-shaped. The set $K$ is closed and bounded and thus compact. Because $F$ is continuous over $U$, $F$ is continuous over the compact set $K \subseteq U$ and is thus uniformly continuous over $K$.

Uniform continuity over $K$ implies that $$ \begin{align*} | G(x + \Delta x) - G(x) | &= \left \vert \int_0^1 F(p + t(x + \Delta x - p)) - F(p + t(x - p)) dt \right \vert \\ &\leq \int_0^1 | F(p + t(x + \Delta x - p)) - F(p + t(x - p))| dt \end{align*} $$ can be made arbitrarily small so long as $\Delta x$ has sufficiently small length proving continuity at $x$.