MATH118C Real Analysis [118c]

  • Mar 30, 2026 - Present

Lecture 1. Sequences and series of functions [l1]

  • Apr 2, 2026

Theorem 1.1. uniform convergence allows switching limits [br711]

    Suppose $f_{n} \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$ \lim_{ t \to x } f_{n}(t) = A_{n} \quad (n = 1, 2, 3, \dots). $$ Then $\{ A_{n} \}$ converges, and $$ \lim_{ t \to x } f(t) = \lim_{ n \to \infty } A_{n}. $$ In other words, the conclusion is that $$ \lim_{ t \to x } \lim_{ n \to \infty } f_{n}(t) = \lim_{ n \to \infty } \lim_{ t \to x } f_{n}(t). $$

    Proof. sketch [br711A]

      Omitted.

      An immediate corollary follows:

      Theorem 1.2. uniform convergence of continuous fns implies continuity of the limit fn [br712]

        If $\{ f_{n} \}$ is a sequence of continuous functions on $E$, and if $f_{n} \to f$ uniformly on $E$, then $f$ is continuous on $E$.

        Definition 1.3. equicontinuous [br722]

          A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \lvert f(x) - f(y) \rvert < \epsilon $$ whenever $d(x, y) < \delta, x \in E, y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.

          It is clear that every member of an equicontinuous family is uniformly continuous.

          Theorem 1.4. pointwise boundedness of $\mathscr{C}$ in a countable set [br723]

            If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.

            Proof. sketch [br723A]

              Use diagonal magic.

              Let $E = \{ x_{i} \}$. Since $\{ f_{n}(x_{1}) \}$ is a bounded sequence on $\mathbb{C}$, we can find a subsequence that converges to a point of $\mathbb{C}$, denote it by $S_{1} = \{f_{1, k}(x_{1})\}$.

              Now, given $\{ f_{1, k}(x_{2}) \}$, a bounded sequence on $\mathbb{C}$, we can again construct a subsequence that converges, denote it by $S_{2} = \{ f_{2, k}(x_{2}) \}$.

              Similarly, we do the same for every $x_{i}$, and thus obtain sequences $S_{1}, S_{2}, S_{3}, \dots$, which we represent by the array $$ \begin{align*} &S_{1}: \ f_{1,1} \; f_{1, 2} \; f_{1, 3} \; f_{1, 4} \; \dots \\ &S_{2}: \ f_{2,1} \; f_{2, 2} \; f_{2, 3} \; f_{2, 4} \; \dots \\ &S_{3}: \ f_{3,1} \; f_{3, 2} \; f_{3, 3} \; f_{3, 4} \; \dots \\ &\dots \dots \dots \dots \dots \dots \dots \dots \end{align*} $$ Then, $\{ f_{n,n} \}$ is the desired subsequence of $\{ f_{n} \}$.

              Theorem 1.5. compactness and uniform convergence implies equicontinuity [br724]

                If $K$ is a compact metric space, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ converges uniformly on $K$, then $\{ f_{n} \}$ is equicontinuous on $K$.

                Proof. sketch [br724A]

                  搬砖题

                  $\{ f_{n} \}$ converges uniformly on $K$ $\implies$ for all $\epsilon > 0$, there exists an integer $N > 0$ such that for any $n, m > N$ we have $\lVert f_{n} - f_{m} \rVert < \epsilon$.

                  Pick any $k > N$. $f_{k} \in \mathscr{C}(K) \implies$ $f_{k}$ continuous on $K$. Now $K$ is compact $\implies$ $f_{k}$ is uniformly continuous on $K$. So there exists $\delta > 0$ such that for any $x, y \in K$ with $d(x, y) < \delta$, we have $\lvert f_{k}(x) - f_{k}(y) \rvert < \epsilon$. Then for any $n > N$, $$ \lvert f_{n}(x) - f_{n}(y) \rvert = \lvert (f_{n}(x) - f_{k}(x)) + (f_{k}(x) - f_{k}(y)) + (f_{k}(y) - f_{n}(y)) \rvert < 3\epsilon. $$ For $1 \leq n \leq N$, $f_{n}$ is uniformly continuous, just take $\delta' = \min_{1 \leq i \leq N}\delta_{i} > 0$, and our final choice of $\delta^*$ is $\min \{ \delta, \delta' \}$, which is positive.

                  Theorem 1.6. [br725]

                    If $K$ is compact, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ is pointwise bounded and equicontinuous on $K$, then

                    1. $\{ f_{n} \}$ is uniformly bounded on $K$,
                    2. $\{ f_{n} \}$ contains a uniformly convergent subsequence.

                    Proof. sketch [br725A]

                      1. $\{ f_{n} \}$ is equicontinuous on $K$ $\implies$ for all $\epsilon > 0$, there exists $\delta > 0$ such that any $x, y \in K$ with $d(x, y) < \delta$ implies $\lvert f_{n}(x) - f_{n}(y) \rvert < \epsilon$ for all $n$.

                        For all $x \in K$, let $B_{x}$ denote the neighborhood of $x$ with radius $\delta$. Then $\{ B_{x} \}$ is an open cover of $K$. Since $K$ is compact, there exists a finite set of points $\{ p_{1}, \dots, p_{m} \}$ such that $\{ B_{p_{i}} \}$ is a finite subcover of $K$.

                        Now consider $B_{p_{i}}$. Let $M_{i}$ be an upper bound of $\{ \lvert f_{n}(p_{i}) \rvert \}$. Then for all $x \in B_{p_{i}}$, $|f_{n}(x)| < |f_{n}(p_{i})| + \epsilon \leq M_{i} + \epsilon$ for all $n$. Hence $M_{i} + \epsilon$ is a global upper bound of $\{ f_{n} \}$ on $B_{p_{i}}$.

                        Take $M = \max_{i} M_{i} + \epsilon < \infty$, then for all $x \in K$, $\lvert f_{n}(x) \rvert < M$ for all $n$. So $\{ f_{n} \}$ is uniformly bounded on $K$.

                      2. Choose any countable dense set $E$ on $K$. Then there exists a subsequence $\{ f_{n_{k}} \}$ of $\{ f_{n} \}$ such that $\{ f_{n_{k}} \}$ converges at every point of $E$. Let $\{ g_{k} \}$ denote $\{ f_{n_{k}} \}$.

                        Use the same $\{ B_{p_{i}} \}$ as in the first part. Pick any $x \in E$, then $x \in B_{p_{i}}$ for some $i$. Then $\{ g_{k}(x) \}$ converges, i.e., there exists a positive integer $N_{i}$ such that any $n, m > N_{i}$ means $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon / 3$. Now for any $y \in B_{p_{i}} \cap E$, we have that for any $n, m > N_{i}$, $$\lvert g_{n}(y) - g_{m}(y) \rvert = \lvert \left( g_{n}(y) - g_{n}(x) \right) + \left( g_{n}(x) - g_{m}(x) \right) + \left( g_{m}(x) - g_{m}(y) \right) \rvert < \epsilon$$ Now choose $N = \max_{i} N_{i} < \infty$. For any $x \in E$, $n, m > N$ implies $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon$. Hence $\{ g_{k} \}$ is uniformly convergent. Therefore, $\{ f_{n} \}$ contains a uniformly convergent subsequence.

                      Lecture 2. Linear transformations [l2]

                      • Apr 7, 2026

                      Definition 2.1. linear transformation [br94]

                        A mapping $A$ of a vector space $X$ into a vector space $Y$ is said to be a linear transformation if $$ A (\mathbf{x}_{1} + \mathbf{x}_{2}) = A \mathbf{x}_{1} + A \mathbf{x}_{2}, \quad A(c \mathbf{x}) = cA \mathbf{x} $$ for all $\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x} \in X$ and all scalars $c$.

                        Notation 2.1.1. some notations [br94A]

                          We often writes $A \mathbf{x}$ insteand of $A(\mathbf{x})$ if $A$ is linear.

                          Linear transformations of $X$ into $X$ are often called linear operators on $X$. If $A$ is a linear operator on $X$ which is one-to-one and onto, we say that $A$ is invertible.

                          Definition 2.2. $L(X, Y)$, products and norms [br96]

                            1. Let $L(X, Y)$ be the set of all linear transformations of the vector space $X$ into the vector space $Y$. Instead of $L(X, X)$, we shall simply write $L(X)$.

                            2. If $X, Y, Z$ are vector spaces, and if $A \in L(X, Y)$ and $B \in L(Y,Z)$, we define their product $BA$ to be the composition of $A$ and $B$: $$(BA)\mathbf{x} = B(A\mathbf{x}) \qquad (\mathbf{x} \in X).$$ Then $BA \in L(X, Z)$.

                            3. For $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, define the norm $\lVert A \rVert$ of $A$ to be the sup of all numbers $\lvert A\mathbf{x} \rvert$, where $\mathbf{x}$ ranges over all vectors in $\mathbb{R}^{n}$ with $\lvert \mathbf{x} \rvert \leq 1$.

                              Observe that the inequality $$\lvert A\mathbf{x} \rvert \leq \lVert A \rVert \lvert \mathbf{x} \rvert $$ holds for all $\mathbf{x} \in \mathbb{R}^{n}$. Also, if $\lambda$ is such that $\lvert A\mathbf{x} \rvert \leq \lambda \lvert \mathbf{x} \rvert$ for all $\mathbf{x} \in \mathbb{R}^{n}$, then $\lVert A \rVert \leq\lambda$.

                            Theorem 2.3. some results about norms [br97]

                              1. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, then $\lVert A \rVert < \infty$ and $A$ is a uniformly continuous mapping of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$.
                              2. If $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $c$ is a scalar, then$$\lVert A + B \rVert \leq \lVert A \rVert + \lVert B \rVert , \qquad \lVert cA \rVert = \lvert c \rvert \lVert A \rVert.$$With the distance between $A$ and $B$ defined as $\lVert A - B \rVert$, $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ is a metric space.
                              3. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $B \in L(\mathbb{R}^{m}, \mathbb{R}^{n})$, then$$\lVert BA \rVert \leq \lVert B \rVert \lVert A \rVert.$$

                                Proof. sketch [br97A]

                                  1. Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ be the standard basis in $\mathbb{R}^{n}$ and suppose $\mathbf{x} = \sum c_{i}\mathbf{e}_{i}$. Now $\lvert \mathbf{x} \rvert \leq 1 \implies \lvert c_{i} \rvert \leq 1$ for all $i$. Then $$\lvert A \mathbf{x} \rvert = \left\lvert \sum c_{i} A \mathbf{e}_{i} \right\rvert \leq \sum \lvert c_{i} \rvert \lvert A \mathbf{e}_{i} \rvert \leq \sum \lvert A \mathbf{e}_{i} \rvert$$ so that $$\lVert A \rVert \leq \sum_{i=1}^{n} \lvert A \mathbf{e}_{i} \rvert < \infty. $$

                                    Each $A \mathbf{e}_{i}$ is a specific vector in $\mathbb{R}^{m}$, and hence $\lvert A \mathbf{e}_{i} \rvert < \infty$.

                                    Since $\lvert A \mathbf{x} - A \mathbf{y} \rvert \leq \lVert A \rVert \lvert \mathbf{x} - \mathbf{y} \rvert$ if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, we see that $A$ is uniformly continuous.

                                  2. Omitted.

                                  3. Omitted.

                                Theorem 2.4. the set of invertible linear operators [br98]

                                  Let $\Omega$ be the set of all invertible linear operators on $\mathbb{R}^{n}$.

                                  1. If $A \in \Omega, B \in L(\mathbb{R}^{n})$, and$$\lVert B - A \rVert \cdot \lVert A^{-1} \rVert < 1,$$then $B \in \Omega$.
                                  2. $\Omega$ is an open subset of $L(\mathbb{R}^{n})$, and the mapping $A \to A^{-1}$ is continuous on $\Omega$.

                                  Proof. sketch [br98A]

                                    1. Suppose $B$ is not invertible $\iff$ $B$ is not injective $\iff$ there exists $\mathbf{x} \in \mathbb{R}^{n}$ such that $\mathbf{x} \neq \mathbf{0}$ and $B\mathbf{x} = \mathbf{0}$. Since $A \in \Omega$, there exists $\mathbf{y} \in \mathbb{R}^{n}$ such that $\mathbf{y} \neq 0$ and $A^{-1}\mathbf{y} = \mathbf{x}$. Then,$$\begin{align*}\lvert (B - A)A^{-1} \mathbf{y} \rvert &= \lvert (B - A)\mathbf{x} \rvert = \lvert -A\mathbf{x} \rvert = \lvert \mathbf{y} \rvert \\ &\leq \lVert (B - A)A^{-1} \rVert \lvert \mathbf{y} \rvert < \lvert \mathbf{y} \rvert. \end{align*}$$This is impossible. So our assumption cannot be true. Therefore, $B \in \Omega$.
                                    2. Openness is trivial. Continuity can be immediately induced from the observation that$$B^{-1} - A^{-1} = B^{-1}(A - B)A^{-1}.$$

                                    Definition 2.5. matrices [br99]

                                      Suppose $\{ \mathbf{x}_{1}, \dots, \mathbf{x}_{n} \}$ and $\{ \mathbf{y}_{1}, \dots, \mathbf{y}_{m} \}$ are bases of vector spaces $X$ and $Y$, respectively. Then every $A \in L(X, Y)$ determines a set of numbers $a_{ij}$ such that $$ A \mathbf{x}_{j} = \sum_{i=1}^{m} a_{ij} \mathbf{y}_{i} \qquad (1 \leq j \leq n). $$ We represent $A$ by an $m$ by $n$ matrix: $$ [A] = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{mn} \end{bmatrix} $$

                                      $[A]$ depends not only on $A$ but also on the choice of bases in $X$ and $Y$.

                                      If $\mathbf{x} = \sum_{j} \mathbf{x}_{j}$, then the Schwarz inequality shows that $$ \lvert A \mathbf{x} \rvert^{2} = \sum_{i} \left( \sum_{j} a_{ij} c_{j} \right)^{2} \leq \sum_{i} \left( \sum_{j} a_{ij}^{2} \cdot \sum_{j} c_{j}^{2} \right) = \sum_{i, j} a_{ij}^{2} \lvert \mathbf{x} \rvert ^{2}. $$ Thus $$ \lVert A \rVert \leq \sqrt{ \sum_{i, j} a_{ij}^{2} }. $$ Moreover, if we replace $A$ by $B - A$, where $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, and view each $a_{ij}$ as continuous functions of a single parameter, then we have the following:

                                      If $S$ is a metric space, if $a_{11}, \dots, a_{mn}$ are real continuous functions on $S$, and if, for each $p \in S$, $A_{p}$ is the linear transformation of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ whose matrix has entries $a_{ij}(p)$, then the mapping $p \mapsto A_{p}$ is a continuous mapping of $S$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.

                                      Lecture 3. Differentiation [l3]

                                      • Apr 9, 2026

                                      The theory on continuity of single-variable scalar valued functions can be directly generalized to multi-variable vector-valued functions except left and right limits.

                                      We now come to the theory on differentiation.

                                      Example 3.1. preliminaries [910]

                                        If $f: (a, b) \to \mathbb{R}^{1}$ and if $f'(x_{0})$ exists for $x_{0} \in (a, b)$, i.e., the limit $$ \lim_{ x \to x_{0} } \frac{f(x) - f(x_{0})}{x - x_{0}} $$ exists, then let $h = x - x_{0}$, we have the following $$ \lim_{ h \to 0 } \frac{f(x_{0} + h) - f(x_{0}) - hf'(x_{0})}{h} = 0. $$ For a scalar-valued single variable function, the existence of the derivative is equivalent to say that $$ \frac{f(x_{0} + h) - \left( f(x_{0}) + hf'(x_{0}) \right)}{h} \to 0 $$ with respect to $h$.

                                        To generalize to muti-variable vector-valued functions:

                                        1. Replace single variable to multi-variable: $(x, y) = (x_{0}, y_{0}) + \vec{h}$. We have replaced $h$ by $\lVert \vec{h} \rVert$.$$\vec{v} = (v_{1}, v_{2}), \qquad \lVert \vec{v} \rVert = \sqrt{ v_{1}^{2} + v_{2}^{2} }$$
                                        2. Then the needed limit is$$\frac{f(x, y) - \left[ f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0}, y_{0})(x - x_{0}) + \frac{\partial f}{\partial y}(x_{0}, y_{0})(y - y_{0}) \right]}{\lVert \vec{h} \rVert }$$where$$f(x, y) - \left[ f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0}, y_{0})(x - x_{0}) + \frac{\partial f}{\partial y}(x_{0}, y_{0})(y - y_{0}) \right]$$is the error between linear part and $f(x, y)$.
                                          1. Ensure the existence of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.
                                          2. Ensure the limit $\to 0$.

                                        Definition 3.2. differentiation [br911]

                                          Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, and $\mathbf{x} \in E$. If there exists a linear transformation $A$ of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ such that $$ \lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0, $$ then we say $\mathbf{f}$ is differentiable at $\mathbf{x}$, and we write $$ \mathbf{f}'(\mathbf{x}) = A. $$ If $\mathbf{f}$ is differentiable at every $\mathbf{x} \in E$, we say that $\mathbf{f}$ is differentiable in $E$.

                                          Theorem 3.3. uniqueness of $\mathbf{f}'(\mathbf{x})$ [br912]

                                            Suppose $E$ and $\mathbf{f}$ are as in Definition 911, $\mathbf{x} \in E$, and $$ \lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0 $$ holds with $A = A_{1}$ and with $A = A_{2}$. Then $A_{1} = A_{2}$.

                                            Remark 3.4. differentiability $\implies$ continuity [br913]

                                              When $\vec{h} \to 0$, $$ \begin{align*} \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) \rVert &= \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) - A \vec{h} + A \vec{h} \rVert \\ &\leq \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) - A \vec{h} \rVert + \lVert A \vec{h} \rVert \\ &\leq \lVert \vec{h} \rVert + \lVert A \rVert \cdot \lVert \vec{h} \rVert \\ &= (\lVert A \rVert + 1)\lVert \vec{h} \rVert. \end{align*} $$ Hence $\vec{F}$ is continuous at $\vec{x}$, provided that $\vec{F}$ is differentiable at $\vec{x}$.

                                              Theorem 3.5. chain rule [br915]

                                                Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable at $\mathbf{x}_{0} \in E$, $\mathbf{g}$ maps an open set containing $\mathbf{f}(E)$ into $\mathbb{R}^{k}$, and $\mathbf{g}$ is differentiable at $\mathbf{f}(\mathbf{x}_{0})$. Then the mapping $\mathbf{F}$ of $E$ into $\mathbb{R}^{k}$ defined by $$ \mathbf{F}(\mathbf{x}) = \mathbf{g}\left( \mathbf{f}(\mathbf{x}) \right) $$ is differentiable at $\mathbf{x}_{0}$, and $$ \mathbf{F}'(\mathbf{x}_{0}) = \mathbf{g}'\left( \mathbf{f}(\mathbf{x}_{0}) \right) \mathbf{f}'(\mathbf{x}_{0}). $$

                                                Definition 3.6. partial derivatives [br916]

                                                  Consider $\mathbf{f}: E \to \mathbb{R}^{m}$ where $E \subset \mathbb{R}^{n}$, Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ and $\{ \mathbf{u}_{1}, \dots, \mathbf{u}_{m} \}$ be the standard bases of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$. The components of $\mathbf{f}$ are the real functions $f_{1}, \dots, f_{m}$ defined by $$ \mathbf{f}(\mathbf{x}) = \sum_{i=1}^{m} f_{i}(\mathbf{x}) \mathbf{u}_{i} \qquad (\mathbf{x} \in E), $$ or equivalently, by $f_{i}(\mathbf{x}) = \mathbf{f}(\mathbf{x}) \cdot \mathbf{u}_{i}, 1 \leq i \leq m$.

                                                  For, $\mathbf{x} \in E$, $1 \leq i \leq m, 1 \leq j \leq n$, we define $$ (D_{j}f_{i})(\mathbf{x}) = \lim_{ t \to 0 } \frac{f_{i}(\mathbf{x} + t \mathbf{e}_{j}) - f_{i}(\mathbf{x}) }{t}, $$ provided the limit exists. Writing $f_{i}(x_{1}, \dots, x_{n})$ in place of $f_{i}(\mathbf{x})$, we see that $D_{j}f_{i}$ is the derivative of $f_{i}$ with respect to $x_{j}$, keeping the other variables fixed. The notation $$ \frac{\partial f_{i}}{\partial x_{j}} $$ is therefore used in place of $D_{j}f_{i}$, and $D_{j}f_{i}$ is called a partial derivative.

                                                  Theorem 3.7. existence of partial derivatives [br917]

                                                    Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, and $\mathbf{f}$ is differentiable at a point $\mathbf{x} \in E$. Then the partial derivative $(D_{j}f_{i})(\mathbf{x})$ exist, and $$ \mathbf{f}'(\mathbf{x})\mathbf{e}_{j} = \sum_{i=1}^{m}(D_{j}f_{i})(\mathbf{x})\mathbf{u}_{i} \qquad (1 \leq j \leq n). $$

                                                    If $\mathbf{F}(\mathbf{x})$ is a scalar-valued function, then $$D \mathbf{F}(\mathbf{x}) = \begin{bmatrix}D_{1}\mathbf{F} & D_{2}\mathbf{F} & \cdots & D_{n}\mathbf{F}\end{bmatrix}$$ It’s also called the gradient of $\mathbf{F}$.

                                                    Theorem 3.8. Lipschitz condition [br919]

                                                      Suppose $\mathbf{f}$ maps a convex open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable in $E$, and there is a real number $M$ such that $$ \lVert \mathbf{f}'(\mathbf{x}) \rVert \leq M $$ for every $\mathbf{x} \in E$. Then $$ \lvert \mathbf{f}(\mathbf{b}) - \mathbf{f}(\mathbf{a}) \rvert \leq M \lvert \mathbf{b} - \mathbf{a} \rvert $$ for all $\mathbf{a} \in E, \mathbf{b} \in E$.

                                                      Definition 3.9. $\mathscr{C}'$-mapping [br920]

                                                        A differentiable mapping $\mathbf{f}$ of an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ is said to be continuously differentiable in $E$ if $\mathbf{f}'$ is a continuous mapping of $E$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.

                                                        More explicitly, it is required that to every $\mathbf{x} \in E$ and to every $\epsilon > 0$ corresponds a $\delta > 0$ such that $$ \lVert \mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x}) \rVert < \epsilon $$ if $\mathbf{y} \in E$ and $\lvert \mathbf{x} - \mathbf{y} \rvert < \delta$.

                                                        If this is so, we also say that $\mathbf{f}$ is a $\mathscr{C}'$-mapping, or that $\mathbf{f} \in \mathscr{C}'(E)$.

                                                        Theorem 3.10. necessary and sufficient conditions for $\mathbb{f} \in \mathscr{C}'(E)$ [br921]

                                                          Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$. Then $\mathbf{f} \in \mathscr{C}'(E)$ if and only if the partial derivatives $D_{j}f_{i}$ exist and are continuous on $E$ for $1 \leq i \leq m, 1 \leq j \leq n$.