第七周。
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连续映射不一定把开集映射为开集。
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\end{remark}
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\begin{remark}
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上面的定理也等价于\boldf 在$D$中处处连续的充分必要条件是对任意的$\realnum^m$中的闭集$G$,原像集$\boldf^{-1}(G)$是$\ndreal$中的闭集。
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\end{remark}
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\begin{theorem}[复合函数的极限]
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设$\bvec{g}: D \to \realnum^m$,$\boldf: \Omega \to \realnum^k$,$\bvec{G}(D) \subset \Omega$。又设$\bvec{a} \in \deriv{D}$,$\tolim{\bvec{x}}{\bvec{a}} \bvec{g}(\bvec{x}) = \bvec{p}, \tolim{\bvec{y}}{\bvec{p}} \boldf(\bvec{y}) = \boldf(\bvec{p})$,那么
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\[\tolim{\bvec{x}}{\bvec{a}} \boldf \circ \bvec{g}(\bvec{x}) = \tolim{\bvec{x}}{\bvec{a}} \boldf(\bvec{g}(\bvec{x})) = \boldf(\bvec{p})\eqper\]
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100
14多变量函数的微分学.tex
100
14多变量函数的微分学.tex
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\[\Delta x = x - x_0, \Delta y = y - y_0, \Delta z = z - z_0\]
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\begin{definition}[函数的微分]
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设开集$f: D \to \realnum$。取定一点$\bvec{x}_0 \in D\interior$。如果存在$n$维向量$\bvec{A} = \lambda_1, \lambda_2, \dots, \lambda_n$,满足
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设$D \subset \ndreal$,$f: D \to \realnum$。取定一点$\bvec{x}_0 \in D\interior$。如果存在$n$维向量$\bvec{A} = (\lambda_1, \lambda_2, \dots, \lambda_n)$,满足
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\[f(\bvec{x}_0 + \Delta \bvec{x}) - f(\bvec{x}_0) = \brak{\bvec{A}, \Delta \bvec{x}} + o(\norm{\Delta \bvec{x}})\]
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那么称函数$f$在点$\bvec{x}_0$处可谓,并称$\brak{\bvec{A}, \Delta \bvec{x}}$为$f$在$\bvec{x}_0$处的微分,记作
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那么称函数$f$在点$\bvec{x}_0$处可微,并称$\brak{\bvec{A}, \Delta \bvec{x}}$为$f$在$\bvec{x}_0$处的微分,记作
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\[\dif f(\bvec{x}_0) = \brak{\bvec{A}, \Delta \bvec{x}}\]
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其中$\bvec{A}$称为微分系数。
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\end{definition}
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\end{definition}
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设$f$在$\bvec{a}$点可微,$\dif f(\bvec{a}) = \brak{A, \Delta \bvec{x}}$,$A = (\lambda_1, \lambda_2, \dots, \lambda_3)$。因此
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\[\tolim{\norm{\Delta \bvec{x}}}{0} \frac{\abs{f(\bvec{a} + \Delta \bvec{x}) - f(\bvec{a}) - \brak{A, \Delta \bvec{x}}}}{\norm{\Delta \bvec{x}}} = 0\]
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记
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\begin{align*}
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\bvec{u}_1 & = (1, 0, 0, \dots, 0)\\
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\bvec{u}_2 & = (0, 1, 0, \dots, 0)\\
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& \quad \dots\\
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\bvec{u}_n & = (0, 0, \dots, 0, 1)
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\end{align*}
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为确定微分系数,取$\Delta \bvec{x} = t \bvec{u}_i$,那么
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\[\norm{\Delta \bvec{x}} = \abs{t}, \brak{A, \Delta \bvec{x}} = \brak{A, t\bvec{u}_i} = t\lambda_i\]
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带入上式
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\[\tolim{t}{0} \abs{\frac{f(\bvec{a} + t \bvec{u}_i) - f(\bvec{a}) - t\lambda_i}{t}} = 0\]
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因此
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\[\lambda_i = \tolim{t}{0} \frac{f(\bvec{a} + t\bvec{u}_i) - f(\bvec{a})}{t} = \frac{\partial f}{\partial x_i} (\bvec{a}), i = 1, 2, \dots, n\eqper\]
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\begin{corollary}
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设$f:D \to \realnum$,$\bvec{a} \in D\interior$。如果$f$在$\bvec{a}$点可微,则在$\bvec{a}$点的偏导数存在,且
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\[\dif f(\bvec{a}) = \frac{\partial f}{\partial x_1}(\bvec{a}) \Delta x_1 + \frac{\partial f}{\partial x_2}(\bvec{a}) \Delta x_2 + \dots + \frac{\partial f}{\partial x_n}(\bvec{a}) \Delta x_n\eqper\]
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\end{corollary}
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\begin{corollary}
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设$f:D \to \realnum$,$\bvec{a} \in D\interior$。如果$f$在$\bvec{a}$点可微,则$f$在$\bvec{a}$点连续。
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\end{corollary}
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\begin{proof}
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\begin{align*}
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\abs{f(\bvec{a} + \Delta \bvec{x}) - f(\bvec{a})} & = \abs{\brak{A, \Delta \bvec{x}} + o(\norm{\Delta \bvec{x}})}\\
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& \leq \norm{A} \cdot \norm{\Delta \bvec{x}} + \abs{o(\norm{\Delta \bvec{x}})} \to 0\qedhere
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\end{align*}
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\end{proof}
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\begin{definition}
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令
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\[Jf(\bvec{x}) = (D_1 f(\bvec{x}), D_2f(\bvec{x}), \dots, D_n f(\bvec{x}))\]
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并称它为函数$f$在点$\bvec{x}$处的Jacobian。函数的Jacobian也常记为$\gra f$或$\nabla f$,即
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\[\gra f(\bvec{x}) = J f(\bvec{x})\]
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称之为数量函数$f$的梯度。
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\end{definition}
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\begin{proposition}
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如果$f$在$\bvec{a}$点可微,则对于任意方向$\bvec{u} \in \ndreal$,$\norm{\bvec{u}} = 1$,那么
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\[D_{\bvec{u}} f(\bvec{a}) = \frac{\partial f}{\partial \bvec{u}} (\bvec{a}) = \brak{\gra f(\bvec{a}), \bvec{u}}\eqper\]
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\end{proposition}
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\begin{corollary}
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对于任意方向$\bvec{u}$,$\abs{D_{\bvec{u}} f(\bvec{a})} \leq \norm{\gra f(\bvec{a})}$。
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若$\gra f(\bvec{a}) \neq 0$,$\bvec{u} = \frac{\gra f(\bvec{a})}{\norm{\gra f(\bvec{a})}}$,则$D_{\bvec{u}} f(\bvec{a}) = \norm{\gra f(\bvec{a})}$。
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这说明,$f$在$\bvec{a}$的梯度向量的方向是$f$值增加最快的方向,大小是$f$在该点所有方向导数的最大值。
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\end{corollary}
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下面的命题给出了一个函数可微的必要条件。
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\begin{proposition}
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若函数$f$在$\bvec{a}$点可微,则存在
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\[\gra f(\bvec{a}) = (D_1 f(\bvec{a}), \dots, D_n f(\bvec{a}))\]
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从而在该点的所有方向导数都存在。
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\end{proposition}
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下面的命题则给出了一个函数可微的充分条件。
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\begin{proposition}
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如果$f$的每个偏导数$D_i f(\bvec{x}), i = 1, 2, \dots, n$在$\bvec{x} = \bvec{a}$点都存在且连续,则$f$在$\bvec{a}$点可微。
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\end{proposition}
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\begin{proof}
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以$n = 2$为例。在$P = (a, b)$点附近考虑函数$f(x, y)$:
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\[f(a + \Delta x, b + \Delta y) - f(a, b) = f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) + f(a + \Delta x, b) - f(a, b)\]
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应用一元函数中值定理,存在$\eta, \theta \in (0, 1)$满足
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\begin{align*}
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f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) & = D_y f(a + \Delta x, b + \eta \Delta y) \Delta y\\
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f(a + \Delta x, b) - f(a, b) & = D_x f(a + \theta \Delta x, b)\Delta x
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\end{align*}
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可以将上式凑配为
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\begin{align*}
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f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) & = D_y f(a, b) \Delta y + [D_y f(a + \Delta x, b + \eta \Delta y) - D_y f(a, b)] \Delta y\\
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f(a + \Delta x, b) - f(a, b) & = D_x f(a, b) \Delta x + [D_x f(a + \theta \Delta x, b) - D_x f(a, b)] \Delta x
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\end{align*}
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记
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\begin{align*}
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[\alpha] & = D_y f(a + \Delta x, b + \eta \Delta y) - D_y f(a, b)\\
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[\beta] & = D_x f(a + \theta \Delta x, b) - D_x f(a, b)
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\end{align*}
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那么
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\[f(a + \Delta x, b + \Delta y) - f(a, b) = D_x f(a, b) \Delta x + D_y f(a, b) \Delta y + [\alpha] \Delta x + [\beta] \Delta y\]
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于是我们只需证明$[\alpha] \Delta x + [\beta] \Delta y = o\left(\sqrt{x^2 + y^2}\right)$。我们已知$D_x f(x, y), D_y f(x, y)$在$P = (a, b)$点连续,因此
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\[\frac{\abs{[\alpha]\Delta x + [\beta] \Delta y}}{\sqrt{\Delta x^2 + \Delta y^2}} \leq \abs{[\alpha]} + \abs{[\beta]} \to 0\]
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综上,
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\[f(a + \Delta x, b + \Delta y) - f(a, b) = D_x f(a, b) \Delta x + D_y f(a, b) \Delta y + o\left[\sqrt{\Delta x^2 + \Delta y^2}\right] \eqper \qedhere\]
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\end{proof}
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总结起来,偏导数在$\bvec{a}$点都连续可以推出函数在$\bvec{a}$点可微,进而可以推出函数在$\bvec{a}$点连续,也可以推出函数在$\bvec{a}$点所有方向导数都存在。
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