> [!theorem] > > Let $\cx, \cy$ be [[Banach Space|Banach spaces]], then the map > $ > L(\cx, \cy) \times \cx \quad (T, x) \mapsto Tx > $ > is [[Bounded Linear Map|bounded]] and [[Multilinear Map|bilinear]]. Let $J = [a, b]$ be a closed interval and $\alpha: J \to L(\cx, \cy)$ be a [[Continuity|continuous]] map, then we can integrate along the curve > $ > \int_a^b \alpha(t)dt \in L(\cx, \cy) > $ > If $\alpha$ is [[Derivative|differentiable]], then $\frac{d\alpha}{dt}(t) \in L(\cx, \cy)$ as well. > [!theorem] > > Let $\alpha: J \to L(\cx, \cy)$ be a continuous map and $x \in \cx$, then > $ > \int_a^b \alpha(t)(x)dt = \int_a^b \alpha(t)dt \cdot x > $ > *Proof*. The map $\lambda \mapsto \lambda x$ is linear and bounded. Therefore it can be taken out of the integral. > [!theorem] Mean Value Theorem > > Let $\cx, \cy$ be [[Banach Space|Banach spaces]], $U \subset \cx$ be [[Open Set|open]] and $x \in U$. Let $y \in \cx$, and $f: U \to \cy$ be a $C^1$ map. If the line $\bracs{x + ty: t \in [0, 1]} \subset U$, then > $ > f(x + y) - f(x) = \int_0^1Df(x + ty)(y) dt = \int_0^1Df(x + ty)dt \cdot y > $ > *Proof*. Let $g(t) = f(x + ty)$, then $Dg(t) = Df(x + ty) \circ y$ by the chain rule. Since $g$ is continuous on $[0, 1]$, by the Fundamental Theorem of Calculus, > $ > \begin{align*} > g(1) - g(0) &= \int_0^1 Dg(t)dt \\ > f(x + y) - f(x) &= \int_0^1 Df(x + ty)(t)dt \\ > &= \int_0^1Df(x + ty)dt \cdot y > \end{align*} > $ > [!theorem] > > Let $U \subset \cx$ be open and $x, z \in U$ such that the line segment > $ > L = \bracs{tx + (1 - t)z: t \in [0, 1]} > $ > lies in $U$. If $f \in C^1$, then > $ > \norm{f(z) - f(x)} \le \norm{z - x} \cdot \sup_{v \in L}\norm{Df(v)} > $ > *Proof*. By the mean value theorem with $y = z - x$, > $ > \begin{align*} > f(z) - f(x) &= \int_0^1 Df(x + t(z - x))dt \cdot (z - x) \\ > \norm{f(z) - f(x)}&= \norm{\int_0^1 Df(x + t(z - x))\cdot (z - x)dt} \\ > &\le \int_0^1\norm{Df(x + t(z - x)) \cdot (z - x)}dt \\ > &\le \int_0^1 \sup_{t \in [0, 1]}\norm{Df(x + t(z - x))} \cdot \norm{z - x}dt \\ > &= \sup_{z \in L}\norm{Df(z)} \cdot \norm{z - x} > \end{align*} > $ > where the supremum exists since $f \in C^1$ and $Df \in C^0$. > [!theorem] > > Let $U \subset \cx$ be open and $x, z, x_0 \in U$ such that the line segment > $ > L = \bracs{tx + (1 - t)z: t \in [0, 1]} > $ > lies in $U$. Then > $ > \norm{f(z) - f(x) - Df(x_0)(z - x)} \le \norm{z - x} \sup_{v \in L}\norm{Df(v) - Df(x_0)} > $ > *Proof*. > $ > \begin{align*} > &f(z) - f(x) \\ > &= \int_0^1 Df(x + t(z - x))(z - x)dt \\ > &= \int_0^1 \braks{Df(x + t(z - x)) - Df(x_0) + Df(x_0)}(z - x)dt \\ > &= Df(x_0)(z - x) + \int_0^1 \braks{Df(x + t(z - x)) - Df(x_0)}(z - x)dt \\ > &\le Df(x_0) + \sup_{z \in L}\norm{Df(v) - Df(x_0)} \cdot \norm{z - x} > \end{align*} > $