$\int_a^b x'(t)dt=x(b)-x(a)$ in Banach space $X$

Question

Let $X$ be a Banach space. Then $\int_a^b x'(t)dt=x(b)-x(a)$ if $x:[a,b]\rightarrow X $ is continuously differentiable.

I have a few problems understanding what I have to show.

First of all, what is $x'(t)$? From the deffinition of differentiability of functions $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$, it should be something like:

There exists a linear map $J:[a,b]\rightarrow X$ such that $\lim_{h\rightarrow0}\frac{\lVert x(t_0+h)-x(t_0)-J(h)\rVert_X}{\lvert h\rvert}=0$. What does the integral mean? And how do I aproach this? Can I somehow use the Newton-Leibniz formula for real valued functions by taking images inder linear functionals?

Answer 1

In fact, the case of a function $x: [a,b] \to X$ is in some ways somewhat simpler than that of a function $f: \mathbb{R}^n \to \mathbb{R}^m$.

Since the domain of your function is a subset of $\mathbb{R}$, $x$ is differentiable at $t \in [a,b]$ if and only if $$x'(t) := \lim_{h \to 0} \frac{x(t + h) - x(t)}{h}$$ exists in $X$.

The easiest way to transfer the usual results of calculus to this setting is to use continuous linear functionals. If $\phi \in X^*$ then we have $\frac{d}{dt} \phi(x(t)) = \phi(x'(t))$. So we have

$$\phi \bigg(\int_a^b x'(t) dt \bigg) = \int_a^b \phi(x'(t)) dt = \int_a^b \frac{d}{dt} \phi(x(t)) dt$$ Now $\phi \circ x$ is real-valued so by the usual fundamental theorem of calculus we have $$\phi \bigg(\int_a^b x'(t) dt \bigg) = \phi(x(b)) - \phi(x(a)) = \phi(x(b)-x(a)).$$ Finally, since $\phi \in X^*$ was arbitrary, by one of the standard corollaries to Hahn-Banach, we must have that $$\int_a^b x'(t) dt = x(b)-x(a)$$ as desired.

Answer 2

One typical approach is to reduce the problem to real valued functions first and then 'stitch' the result together.

One convenient way of defining the integral $\int f$ is the unique $I \in X$ such that $\int \phi (f) = \phi(I)$ for all $\phi \in X^*$.

It is straightforward to show that the Frechet derivative of $x$ is equivalent to $x'(t) = \lim_{h \to 0} {x(t+h)-x(t) \over h}$. Note that $(\phi(x))'(t) = \phi(x'(t))$ for $\phi \in X^*$.

Now apply the usual theorem of calculus to $\int_a^t \phi(x'(\tau)) d \tau$.