> [!definition] > > Let $\cx = \prod_{k = 1}^{n}\cx_k$ be a product of [[Banach Space|Banach spaces]], $U_i \in \cx_i$ be [[Open Set|open]] for each $i$, $\cy$ be a Banach space and let > $ > f: \prod_{k = 1}^{n}U_k \to \cy > $ > be a [[Function|function]]. Let $(x_1, \cdots, x_n) \in \cx$ and define a *partial map* > $ > x_i \mapsto f(x_1, \cdots, x_i, \cdots, x_n) > $ > with $x_j$ fixed for all $j \ne i$. If this map is [[Derivative|differentiable]], then its derivative is the **partial derivative** of $f$, and denote it by $D_if(x)$ at the point $x$. If it exists, then > $ > D_if(x) = \lambda: \cx_i \to \cy > $ > is the unique [[Bounded Linear Map|bounded linear map]] $\lambda \in L(\cx_i, \cy)$ such that > $ > f(x_1, \cdots, x_i + h, \cdots, x_n) - f(x_1, \cdots, x_n) = \lambda h + o(h) > $ > [!theorem] > > Let $\cx_1, \cx_2, \cy$ be Banach spaces, $\cx = \cx_1 \times \cx_2$, $U_1 \subset \cx_1$ and $U_2 \subset \cx_2$ be open sets, and $f: U_1 \times U_2 \to \cy$. Then $f \in C^p$ if and only if $D_1f, D_2f \in C^{p - 1}$, in which case, > $ > Df(x)(v_1, v_2) = D_1f(x)v_1 + D_2f(x)v_2 > $ > *Proof*. Let $(x, y) \in U_1 \times U_2$, and $h = (h_1, h_2) \in \cx$. Suppose that the partial derivatives exist and are continuous, then by the [[Mean Value Theorem|mean value theorem]] > $ > \begin{align*} > &f(x + h_1, y + h_2) - f(x, y) \\ > &= f(x + h_1, y + h_2) - f(x + h_1, y) \\ > &+ f(x + h_1, y) - f(x, y) \\ > &= \int_0^1D_2f(x + h_1, y + th_2)h_2dt \\ > &+ \int_0^1D_1 f(x + th_1, y)h_1 dt > \end{align*} > $ > where we can create an error term > $ > \psi(h_1, th_2) = D_2f(x + h, y + th_2) - D_2f(x, y) > $ > Since $D_2f$ is continuous, $\lim_{h \to 0}\psi(h_1, th_2) = 0$. Rewrite the first term as > $ > \begin{align*} > &\int_0^1D_2f(x + h_2, y + th_2)h_2dt \\ > &= \int_0^1 D_2f(x, y)h_2dt + \int_0^1\psi(h_1, th_2)h_2dt > \end{align*} > $ > where the error term > $ > \begin{align*} > \abs{\int_0^1\psi(h_1, th_2)h_2dt} &\le \underbrace{\sup_{t \in [0, 1]}\abs{\psi(h_1, th_2)}}_{\to 0} \cdot \abs{h_2} = o(h) > \end{align*} > $ > is little-oh of $h$. Using the same technique on the second part and combining the results yields > $ > f(x + h_1, y + h_2) - f(x, y) = D_1f(x, y)h_1 + D_2f(x, y)h_2 + o(h) > $ > where > $ > Df(x, y)(h_1, h_2) = D_1f(x, y)h_1 + D_2f(x, y)h_2 > $ > If each partial is of class [$C^p$](Space%20of%20Continuously%20Differentiable%20Functions), then $Df$ is also of class $C^p$, making $f$ of class $C^{p + 1}$. > [!theorem] Partial Derivatives Commute > > Let $U \subset \cx_1 \times \cx_2$ be open, $f: U \to \cy$ be a map such that $D_1f$, $D_2f$, $D_1D_2f$, $D_2D_1f$ exist and are all continuous. Then $D_1D_2f = D_2D_1f$. > > *Proof*. Let $h = (h_1, h_2) \in \cx_1 \times \cx_2$, and consider the difference > $ > \Delta = f(x + h_1, y + h_2) - f(x + h, y) - f(x, y + h) + f(x, y) > $ > where if > $ > g(x) = f(x, y + h_2) - f(x, y) > $ > then > $ > \Delta = g(x + h_1) - g(x) > $ > By the mean value theorem, > $ > g(x + h_1) - g(x) = \int_0^1 Dg(x + th_1)h_1dt > $ > where > $ > \begin{align*} > Dg(x + th_1) \cdot h_1 &= Df(x + th_1, y + h_2) \cdot h_1 - Df(x + th_1, y) \cdot h_1 \\ > &= D_1f(x + th_1, y + h_2) \cdot h_1 - D_1f(x + th_1, y) \cdot h_1 \\ > &= \braks{D_1f(x + th_1, y + h_2) - D_1f(x + th_1, y)} \cdot h_1 > \end{align*} > $ > Using the fact that $D_1 f \in C^1$, we can apply the mean value theorem again > $ > \begin{align*} > &D_1f(x + th_1, y + h_2) - D_1f(x + th_1, y) \\ > &= \int_0^1 D \cdot D_1f(x + th_1, y + th_2) \cdot h_2 dt \\ > &= \int_0^1 D_2 D_1f(x + th_1, y + th_2) \cdot h_2dt > \end{align*} > $ > In other words, > $ > \begin{align*} > g(x + h_1) - g(x) &= \int_0^1 Dg(x + th_1)h_1dt \\ > &= \int_0^1\braks{\int_0^1D_2D_1f(x + th_1, y + sh_2)h_2ds}h_1dt \\ > &= \int_0^1\int_0^1 D_2D_1f(x + th_1, y + sh_2)(h_2, h_1)dsdt > \end{align*} > $ > Since $D_2D_1f$ is continuous, we can again bound the error term > $ > \psi(th_1, sh_2) = D_2D_1f(x + th_1, y + sh_2) - D_2D_1f(x, y) > $ > which has the property $\lim_{h \to 0}\psi(h_1, h_2) = 0$. Integrating the error yields > $ > \begin{align*} > \abs{\int_0^1 \psi(th_1, sh_2)h_2ds} &\le \abs{h_2} \sup_{s \in [0, 1]}\abs{\psi(th_1, sh_2)} \\ > \abs{\int_0^1\int_0^1 \psi(th_1, sh_2)(h_2, h_1)dsdt} &\le \abs{h_1} \cdot \abs{h_2} \sup_{s, t}\abs{\psi(th_1, sh_2)} > \end{align*} > $ > Finally, this gives > $ > \Delta = D_2D_1(x, y)(h_2, h_1) + \int_0^1\int_0^1 \psi(th_1, sh_2)(h_2, h_1)ds > $ > Repeating all of the above also yields > $ > \Delta = D_1D_2(x, y)(h_1, h_2) + \int_0^1\int_0^1 \psi'(th_1, sh_2)(h_1, h_2)ds > $ > where the difference between the two derivatives is the zero map due to the error integral being "little-oh". # Difference Quotient > [!theorem] > > Let $U \subset \real^d$ be an open subset, $\cy$ be a Banach space, and $f: U \to \cy$ be a map of class $C^1$, then for any $x \in U$, > $ > \frac{f(x + he_j) - f(x)}{h} \to \partial^jf(x) > $ > [[Uniform Convergence on Compact Sets|uniformly on compact sets]]. > > *Proof*. Let $K$ be compact, then $K' = K + \bracs{te_j: t \in (-1, 1)}$ is compact as well. Since $f \in C^1$, $\partial^j f|_{K'}$ is uniformly continuous. Let $\eps > 0$, then there exists $\delta > 0$ such that $\norm{\tau_x \partial^j f - \partial^j f}_u < \eps$ for all $x$ with $\norm{x} < \delta$. In which case, by the mean value theorem > $ > \limsup_{h \to 0}\frac{f(x + he_j) - f(x)}{h} \le \partial^jf(x) + \eps > $ > and > $ > \liminf_{h \to 0}\frac{f(x + he_j) - f(x)}{h} \ge \partial^jf(x) - \eps > $ > as the above holds for all $\eps > 0$, and the choice of $\delta$ does not depend on $x$, the convergence is uniform.