How to obtain integral remainder in Taylor's formula for functions between Banach spaces?

Question

In Amann's textbook Analysis II, the authors present below version of Taylor's theorem.

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Let $E,F$ be Banach spaces. Suppose $X$ is open in $E, q \in \mathbb{N}^{\times}$, and $f$ belongs to $C^{q}(X, F)$. We use the following notation: $$ \partial^{k} f(x)[h]^{k}:=\left\{\begin{array}{lr} \partial^{k} f(x)[\underbrace{h, \ldots, h}_{k \text {-times }}], & 1 \leq k \leq q, \\ f(x), & k=0 \end{array}\right. $$ for $x \in X, h \in E$, and $f \in C^{q}(X, F)$. Then $$ f(x+h)=\sum_{k=0}^{q} \frac{1}{k !} \partial^{k} f(x)[h]^{k}+R_{q}(f, x ; h) $$ for $x \in X$ and $h \in E$ such that $[\![ x, x+h ]\!] \subset X$. Here $$ R_{q}(f, x ; h):=\int_{0}^{1} \frac{(1-t)^{q-1}}{(q-1) !}\left[\partial^{q} f(x+t h) \color{blue} {-\partial^{q} f(x)} \right][h]^{q} \mathrm d t \in F. $$

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I have come across a version (at page 64 of this note) with below integral remainder, i.e., $$ R_{q}(f, x ; h):=\int_{0}^{1} \frac{(1-t)^{q-1}}{(q-1) !} \partial^{q} f(x+t h) [h]^{q} \mathrm d t. $$

Is it possible to deduce this remainder from the version in my textbook?

Answer 1

It's a simple integration by parts: \begin{align} R_{q,\text{book}}(f,x,h)&:=\int_0^1\frac{(1-t)^{q-1}}{(q-1)!}[\partial^qf(x+th)-\partial^qf(x)][h^q]\,dt\\ &=\int_0^1-\left(\frac{d}{dt}\frac{(1-t)^q}{q!}\right)[\partial^qf(x+th)-\partial^qf(x)][h^q]\,dt\\ &=\bigg[-\frac{(1-t)^q}{q!}[\partial^qf(x+th)-\partial^qf(x)][h^q]\bigg]_0^1\\ &- \int_0^1\left(-\frac{(1-t)^q}{q!}\right)\frac{d}{dt}[\partial^qf(x+th)-\partial^qf(x)][h^q]\,dt\tag{IBP}\\\\ &= 0 +\int_0^1\frac{(1-t)^q}{q!}\partial^{q+1}f(x+th)[h^{q+1}]\,dt\\ &=R_{q,\text{notes}}(f,x,h). \end{align}