Integrating a linear-map valued function My textbook for an Analysis course I am taking presents the Mean Value Equality theorem as
Suppose $\mathbb{X}, \mathbb{Y}$ are Banach Spaces.
Let $U\subseteq \mathbb{X}$ be an open set, and let $f\colon U\to \mathbb{Y}$ be Fréchet differentiable everywhere on $U$. Let $x,v\in\mathbb{X}$ be such that the line segment from $x$ to $x+v$ lies completely in $U$, then
$$f(x+v)-f(x)=\left[\int_0^1 df(x+tv)\,dt \right](v).$$
My question is: How do I interpret this integral? Integrating over a series of functions and then evaluating afterwards confuses me, as this doesn't fit in with what I've learnt about integration (I'm only used to integrating over $\mathbb{R}$-valued functions).
note: i should say $df(x+tv)\colon\mathbb{X}\to\mathbb{Y}$ is the Fréchet derivative at $x+tv$.
 A: It is standard to assume that $df(x)$ is a bounded linear operator on $X$. You should probably also assume that $x\mapsto df(x)$ is continuous as a function from $X$ into $\mathcal{L}(X)$, the space of bounded linear operators on $X$. The chain rule applies to the function $t \mapsto f(x+tv)$ to give a derivative
$$
            \frac{d}{dt} f(x+tv) = df(x+tv)\frac{d}{dt}(x+tv)=df(x+tv)v.
$$
Therefore
\begin{align}
              f(x+v)-f(x) & =\int_{0}^{1}\frac{d}{dt}f(x+tv)dt \\
                & = \int_{0}^{1}df(x+tv)v\;dt \\
                & = \left[\int_{0}^{1}df(x+tv)dt\right]v
\end{align}
The first two integrals on the right can be viewed as Riemann integrals on $[0,1]$ of continuous functions which take values in $X$. The last integral can be viewed as a Riemann integral of a continuous function from $[0,1]$ into the vector space of bounded linear operators $\mathcal{L}(X)$; the result is another bounded linear operator, which is then applied to the vector $v$. The last identity is easily proved using Riemann sums, assuming that $t\mapsto df(x+tv)$ is continuous from $[0,1]$ into $\mathcal{L}(X)$. Keep in mind that $\mathcal{L}(X)$ is just another Banach space.
