Mathematics_for_Physicists/chapters/sections/partial_total_diff.tex

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\begin{document}
\section{Partial and Total Differentiability}

\begin{defi}
    Let $U \subset \realn^n$ be open, $x \in (x_1, \cdots, x_n) \in U$ and define the function $f: U \rightarrow \realn^m$.
    The mapping $f$ is said to be partially differentiable in $x$ in terms of $x_i$ if 
    \[
        t \longmapsto f(x_1, \cdots, x_{i-1}, t, x_{i+1}, \cdots, x_n)
    \]
    is differentiable in $x_i$, i.e.
    \[
        \partial_i f(x) = \limes{h}{0} \frac{f(x_1, \cdots, x_{i-1}, x_i + h, x_{i+1}, \cdots, x_n) - f(x_1, \cdots, x_n)}{h}
    \]
    exists. $\partial_i f(x)$ is said to be the partial derivative of $f$ in $x$ in terms of $x_i$. Another notation is 
    \[
        \pdv{f}{x_i}
    \]
    This mapping is said to be partially differentiable in $x$ if it is partially differentiable in terms of $x_i ~~\forall i \in \set{1, \cdots, n}$.
\end{defi}

\begin{eg}
    Consider 
    \begin{align*}
        f: \realn^2 &\longrightarrow \realn \\
        (x, y) &\longmapsto \begin{cases}
            1, & x = 0 \vee y = 0 \\
            0, & \text{else}
        \end{cases}
    \end{align*}
    $f$ is partially differentiable in $(0, 0)$, but not continuous.
\end{eg}

\begin{thm}
    Let $U \subset \realn$ be open, $x \in U$ and $f: U \rightarrow \field$.
    \begin{gather*}
        f \text{ is differentiable in } x \\
        \iff \\
        \exists a \in \field, \phi: U \rightarrow \field: ~~f(y) = f(x) + a(y - x) + \phi(y) ~~\forall y \in U
    \end{gather*}
    and 
    \[
        \limes{y}{x} \frac{\phi(x)}{\abs{y - x}} = 0
    \]
\end{thm}
\begin{proof}
    We will first prove the "$\impliedby$" direction. So let $a, \phi$ be as demanded in the theorem. Then 
    \begin{equation}
        \frac{f(y) - f(x)}{y - x} = a + \frac{\phi(y)}{\abs{y - x}} \cdot \frac{\abs{y - x}}{y - x} \conv{y \rightarrow x} a
    \end{equation}
    which means $f$ is differentiable in $x$ and $f'(x) = a$. Now let $f$ be differentiable, and set
    \begin{equation}
        \phi(y) = f(y) - f(x) - f'(x)(y - x)
    \end{equation}
    Which is equivalent to the equation in the theorem, with $a = f'(x)$. Then
    \begin{equation}
        \limes{y}{x} \frac{\phi(x)}{\abs{y - x}} = \left( \frac{f(y) - f(x)}{y - x} - f'(x) \right) \cdot \frac{y - x}{\abs{y - x}} = 0
    \end{equation}
\end{proof}

\begin{defi}
    Let $U \subset \realn^n$, $x \in U$ and $f: U \rightarrow \realn^m$.
    $f$ is said to be (totally) differentiable in $x$ if a matrix $A \in \realn^{m \times n}$ and a mapping $\phi: U \rightarrow \realn^m$ exist, such that
    \[
        f(y) = f(x) + A(y - x) + \phi(x) ~~\forall y \in U
    \]
    and 
    \[
        \limes{y}{x} \frac{\phi(y)}{\norm{y - x}} = 0
    \]
    $f$ is said to be (totally) differentiable if it is (totally) differentiable in every point $x \in U$.
\end{defi}

\begin{thm}
    Let $U \subset \realn^n$ be open, $x \in U$ and $f: U \rightarrow \realn^m$ with 
    \[
        f = (f_1, \cdots, f_m), ~~f_1, \cdots, f_m: U \longrightarrow \realn
    \]
    If $f$ is totally differentiable in $x$, then it is partially differentiable as well, and the matrix $A$ is given by
    \[
        a_{ji} = \partial_i f_j(x)
    \]
\end{thm}
\begin{proof}
    Let $A, \phi$ be as demanded above. Let $e_1, \cdots, e_n$ be the canonical basis for $\realn^n$. 
    We insert $y = x + he_i$ and receive 
    \begin{equation}
        f(x + he_i) = f(x) + h \cdot Ae_i + \phi(x + he_i)
    \end{equation}
    By rearranging this yields
    \begin{equation}
        \frac{f(x + he_i) - f(x)}{h} = Ae_i + \frac{\phi(x + he_i)}{|h|} \cdot \frac{|h|}{h} \conv{h \rightarrow 0} Ae_i
    \end{equation}
    Thus, $f$ is partially differentiable in $x$ in terms of $x_i$ with $\partial_i f(x) = Ae_i$.
\end{proof}

\begin{defi}
    The matrix $(\partial_i f_j(x))_{ij}$ is called the Jacobian matrix of $f$ in $x$. 
    We write $Df(x)$. If $f$ is totally differentiable, then $Df(x)$ is said to be the (total) derivative of $f$ in $x$.

    For $m = 1$ (so $f: \realn^n \rightarrow \realn$), the Jacobian matrix has one column, and we call it gradient
    \[
        Df(x) =: \grad f(x)
    \]
    Note: I will adhere to the physical notation of the gradient, using the Nabla operator $\nabla$.
\end{defi}

\begin{eg}
    Let $A \in \realn^{m \times n}$ and define 
    \begin{align*}
        f_A: \realn^n &\longrightarrow \realn^m \\
        x &\longmapsto Ax
    \end{align*}
    Then we have 
    \[
        f_A(y) = Ay = Ax + A(y - x) = f_A(x) - f_A(y - x)
    \]
    Thus, $f_A$ is differentiable $(\phi = 0)$ and the derivative is 
    \[
        Df_A(x) = A ~~\forall x \in \realn^n
    \]
    For another example, let 
    \begin{align*}
        f: (0, \infty) \times (0, 2\pi) &\longrightarrow \realn^2 \\
        (r, \phi) &\longmapsto (r \cos \phi, r \sin \phi)
    \end{align*}
    Then $f$ is partially differentiable.
    \[
        Df(r, \phi) = \begin{pmatrix}
            \cos\phi & -r\sin\phi \\
            \sin\phi & r\cos\phi
        \end{pmatrix}
    \]
    So $f$ is also totally differentiable (We'll get back to this later).
\end{eg}

\begin{rem}
    \begin{enumerate}[(i)]
        \item Let $U \subset \realn^n$ be open and $f: U \rightarrow \realn^m$ differentiable, 
        then the derivative $Df$ is a function $U \rightarrow \realn^{m \times n}$

        \item Total differentiability is also called local linear approximation. Linearity is the property
        \[
            A(x + \lambda y) = Ax + \lambda Ay ~~\forall x, y \in \realn^n ~\lambda \in \realn
        \]

        \item For arbitrary vector spaces $V, W$, a mapping $V \rightarrow W$ is said to be linear if 
        \[
            A(x + \lambda y) = Ax + \lambda Ay ~~\forall x, y \in \realn^n ~\lambda \in \realn
        \]
        So we can analogously define differentiability for mappings $f: V \rightarrow W$ between arbitrary normed vector spaces.

        \item $f$ is totally differentiable in $x$ if and only if the Jacobian matrix  exists and 
        \[
            \limes{x}{y} \frac{f(y) - f(x) - Df(x)(y-x)}{\norm{y-x}} = 0
        \]

        \item Let $f = (f_1, \cdots, f_m)$ with $f_1, \cdots, f_m: U \rightarrow \realn$.
        \[
            f \text{ totally differentiable} \iff f_i \text{ totally differentiable } ~\forall i \in \set{1, \cdots, n}
        \]
        The Jacobian matrix $Df_i(x)$ is the $i$-th row of $Df(x)$.

        \item Total differentiability implies continuity.
        \item Partial and total differentiability are local properties.
        \item The mapping $h \mapsto Df(x) \cdot h$ is linear.
        \item The derivative $x \mapsto Df(x)$ is not linear in general.
    \end{enumerate}
\end{rem}

\begin{thm}[Chain rule]
    Let $U \subset \realn^n$ be open, $V \subset \realn^m$ open, $x \in U$, $g: U \rightarrow V$ differentiable in $x$, and $f: V \rightarrow \realn^k$ differentiable in $g(x)$.
    Then $f \circ g$ is differentiable and 
    \[
        D(f \circ g) = Df(g(x)) \cdot Dg(x)
    \]
\end{thm}
\begin{proof}
    Differentiability of $g$ in $x$ means
    \begin{equation}
        \exists \phi_g: U \longrightarrow \realn^m: ~~g(y) - g(x) = D_g(x)(y-x) + \phi_g(y)
    \end{equation}
    Differentiability of $f$ in $g(x)$ means 
    \begin{equation}
        \exists \phi_f: V \rightarrow \realn^k:: ~\limes{z}{g(x)} \phi_f(z) \inv{\norm{z - g(x)}} = 0
    \end{equation}
    and 
    \begin{equation}
        f(z) = f(g(x)) + D_f(g(x))(z - g(x)) + \phi_f(z)
    \end{equation}
    Now set $z = g(y)$, then 
    \begin{equation}
        \begin{split}
            \underbrace{f(g(y))}_{(f \circ g)(y)} = \underbrace{f(g(x))}_{(f \circ g)(x)} &+ D_f(g(x)) \cdot D_g(x)(y-x) \\
            &+ (D_f(g(x)) \phi_g(y) + \phi_f(g(y)))
        \end{split}
    \end{equation}
    And we finally need to show 
    \begin{equation}
        \frac{D_f(g(x)) \phi_g(y) + \phi_f(g(y))}{\norm{y - x}} \conv{y \rightarrow x} 0
    \end{equation}
    We know that 
    \begin{equation}
        Df(g(x)) \frac{\phi_g(y)}{\norm{y - x}} \conv{} 0
    \end{equation}
    because
    \begin{equation}
        z \longmapsto Df(g(x)) z \text{ linear and thus continuous}
    \end{equation}
    We define a new mapping 
    \begin{equation}
    \begin{split}
        \psi: U &\longrightarrow \realn \\
        z &\longmapsto \begin{cases}
            \phi_f(z) - \inv{\norm{z - g(x)}}, & z \ne g(x) \\
            0, & z = g(x)
        \end{cases}
    \end{split}
    \end{equation}
    $\psi$ is continuous in $g(x)$. Then $\forall y \in U$ we have 
    \begin{equation}
        \frac{\phi_f(g(y))}{\norm{y - x}} = \underbrace{\psi(g(y))}_{\conv{y \rightarrow x} 0} \cdot \frac{\norm{g(y) - g(x)}}{\norm{y - x}}
    \end{equation}
    and
    \begin{equation}
    \begin{split}
        \frac{\norm{g(y) - g(x)}}{\norm{y - x}} &= \norm{Dg(x) \frac{y - x}{\norm{y - x}} + \frac{\phi_g(y)}{\norm{y - x}}} \\
        &\le \underbrace{\norm{Dg(x) \frac{y - x}{\norm{y - x}}}}_{\le \norm{Dg(x)}} + \underbrace{\norm{\frac{\phi_g(y)}{\norm{y - x}}}}_{\conv{y \rightarrow x} 0}
    \end{split}
    \end{equation}
    thus $\psi$ is bounded.
    \begin{equation}
        \implies \psi(g(y)) \cdot \frac{\norm{g(y) - g(x)}}{\norm{y - x}} \conv{} 0
    \end{equation}
\end{proof}

\begin{thm}
    Let $U \subset \realn^n$ and $f: U \longrightarrow \realn^m$. If $\forall x \in U$ the partial derivatives $\partial_i f(x)$ exist and are continuous $\forall i \in \set{1, \cdots, n}$.
    then $f$ is totally differentiable.
\end{thm}
\begin{proof}
    Without proof.
\end{proof}

\begin{defi}
    Let $U \subset \realn^n$ be open. $f: U \rightarrow \realn^m$ is said to be continuously differentiable if all partial derivatives exist and are continuous.
    The vector space of all such functions is denoted as $C^1(U, \realn^m)$, or in the special case $m=1$ as $C^1(U)$.
\end{defi}

\begin{eg}
    \begin{enumerate}
        \item Coming back to a previous example, we consider 
        \[
            Df(r, \phi) = \begin{pmatrix}
                \cos \phi & -r \sin \phi \\
                \sin \phi & \cos \phi
            \end{pmatrix}
        \]
        Thus, $f$ is continuously differentiable, and therefore totally differentiable.

        \item Let $N \in \natn$ and $c_{\eta} \in \field$ for every multiindex $\eta \in \natn_0^n$ with $\abs{\eta} \le N$.
        Then the polynomial
        \begin{align*}
            P: \realn^n &\longrightarrow \field \\
            x &\longmapsto \sum_{\substack{\eta \\ \abs{\eta} \le N}} c_{\eta} x^{\eta}
        \end{align*}
        is continuously differentiable, and therefore totally differentiable.
        \begin{align*}
            \partial_i x^{\eta} &= \partial_i \left( x_1^{\eta_1}, x_2^{\eta_2}, \cdots, x_n^{\eta_n} \right) \\
            &= \eta_i x_1^{\eta_1} \cdots x_{i-1}^{\eta_{i-1}} x_i^{\eta_{i-1}} x_{i+1}^{\eta_{i+1}} \cdots x_n^{\eta_n} 
        \end{align*}
        This is another polynomial, and therefore continuous.
    \end{enumerate}
\end{eg}

We introduce the following new notation, for $x, y \in \realn^n$:
\begin{align*}
    \oline &:= \set[t \in (0, 1)]{x + t(y - x)} \\
    \cline &:= \set[{t \in [0, 1]}]{x + t(y - x)}
\end{align*}
They denote the connecting line between $x$ and $y$.

\begin{center}
    \begin{tikzpicture}
        \draw (0, 0) circle [radius=2pt] node[above left] {$x$};
        \draw (3, 1.5) circle [radius=2pt] node[above right] {$y$};

        \draw (0, 0) -- node[below right] {$\oline$} (3, 1.5);
    \end{tikzpicture}
\end{center}

\begin{thm}[Intermediate value theorem for $\realn$-valued functions]
    Let $U \subset \realn^n$ be open, $x, y \in U$ and $\cline \subset U$.
    Now let $f: U \rightarrow \realn$ differentiable on $\oline$ and continuous in $x, y$. Then 
    \[
        \exists \xi \in \cline: ~~f(y) - f(x) = Df(\xi) (y-x)
    \]
\end{thm}
\begin{proof}
    Consider 
    \begin{equation}
    \begin{split}
        g: [0, 1] &\longrightarrow \realn \\
        t &\longmapsto f(x + t(y - x))
    \end{split}
    \end{equation}
    Apply the one dimensional intermediate value theorem. Due to the chain rule, $g$ fulfils the prerequisites.
    $\exists \theta \in (0, 1)$ such that
    \begin{equation}
        f(y) - f(x) = g(1) - g(0) = g(\theta) = Df(x + \theta(y - x))(y - x)
    \end{equation}
    For $\xi = x + \theta(y - x)$ follows the initial statement.
\end{proof}

\begin{thm}[Intermediate value theorem]
    Let $U \subset \realn^n$ be open, $\cline \subset U$ and $f: U \rightarrow \realn^m$ differentiable on $\oline$ and continuous in $x, y$. Then 
    \[
        \exists \xi \in \oline: ~~\norm{f(y) - f(x)} \le \norm{Df(\xi)(y - x)}
    \]
\end{thm}
\begin{proof}
    For $a \in \realn^m$, consider the (real) helper function 
    \begin{equation}
        a^Tf(x) = \innerproduct{a}{f(x)}
    \end{equation}
    According to the previous theorem
    \begin{equation}
        \exists \xi \in \oball: ~~a^T f(y) - a^T f(x) = a^T D f(\xi)(y - x)
    \end{equation}
    In this implication the chain rule has been applied. We can rewrite this using the scalar product
    \begin{equation}
        \begin{split}
            \norm{f(y) - f(x)}^2 &= \abs{\innerproduct{f(y) - f(x)}{Df(\xi)(y-x)}} \\
            &\le \norm{f(y) - f(x)}\norm{Df(\xi)(y - x)}
        \end{split}
    \end{equation}
\end{proof}

\begin{cor}
    Let $U \subset \realn^n$ be open and $f: U \rightarrow \realn^m$ a differentiable function.
    \[
        Df = 0 \text{ on } U \implies \exists V \subset U: f \text{ constant on } V
    \]
\end{cor}
\begin{proof}
    Let $x \in U$, choose $\epsilon > 0$ such that $\oball(x) \subset U$. Then 
    \begin{equation}
        \forall y \in \oball(x) ~\exists \xi \in \oline: ~~\norm{f(y) - f(x)} \le \norm{Df(\xi)(y - x)} = 0
    \end{equation}
    This implies 
    \begin{equation}
        \norm{f(y) - f(x)} = 0 \implies f(y) = f(x) ~~\forall y \in \oball(x)
    \end{equation}
\end{proof}

\begin{rem}
    Functions with vanishing derivatives must be constant. Consider 
    \begin{align*}
        f: (-2, -1) \cup (1, 2) &\longrightarrow \\
        x &\longmapsto \begin{cases}
            -1, & x < 0 \\
            1, & x > 0
        \end{cases}
    \end{align*}
    Local constancy implies constancy on connected sets.
\end{rem}
\end{document}