Let \(L(X,Y)\) be the set of all linear transformations of the vector space \(X\) into the vector space \(Y\). For \(A\in L(R^n,R^m)\), 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 \(R^n\) with \(\lvert x\rvert\leq1\) The inequality \[\lvert A\mathbf x\rvert\leq\lVert A\rVert\lvert \mathbf x\rvert\] holds for all \(\mathbf x\in R^n\). If \(\lambda\) is such that \[\lvert A\mathbf x\rvert\leq\lambda\lvert \mathbf x\rvert\] for all \(\mathbf x\in R^n\) then \(\lVert A\rVert\leq\lambda\).
Let \(\{\mathbf e_1,\cdots,\mathbf e_n\}\) be the standard basis in \(R^n\) and suppose \(\mathbf x=\sum c_i\mathbf e_i\), \(|\mathbf x|\leq 1\), so that \(|c_i|\leq1\) for \(i=1,\cdots,n\) Then \[|A\mathbf x|=|\sum c_iA\mathbf e_i|\leq\sum|c_i||A\mathbf e_i|\leq \sum|A\mathbf e_i|\] and \[\lVert A\rVert\leq\sum|A\mathbf e_i|<\infty\] Since \[\lvert A\mathbf x-A\mathbf y\rvert\leq\lVert A\rVert\lvert \mathbf x-\mathbf y\rvert\] if \(\mathbf x,\mathbf y\in R^n\), so that \(A\) is uniformly continuous.
Because \[\lvert (A+B)\mathbf x\rvert=\lvert A\mathbf x+B\mathbf x\rvert\leq\lvert A\mathbf x\rvert+\lvert B\mathbf x\rvert\leq(\lVert A\rVert+\lVert B\rVert)\lvert \mathbf x\rvert\] And if \(\lambda\) is such that \[\lvert A\mathbf x\rvert\leq\lambda\lvert \mathbf x\rvert\] for all \(\mathbf x\in R^n\) then \(\lVert A\rVert\leq\lambda\). So that \[\lVert A+B\rVert\leq\lVert A\rVert+\lVert B\rVert\] and similarly \[\lVert A-C\rVert=\lVert (A-B)+(B-C)\rVert\leq\lVert A-B\rVert+\lVert B-C\rVert\] \[\lVert BA\rVert\leq\lVert B\rVert\lVert A\rVert\]
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 maps \(X\) onto \(X\), we say that \(A\) in invertible.
A function \(\mathbf f\) that maps \((a,b)\subset R^1\) into \(R^m\). In that case \(\mathbf f'(x)\) was defined to be that vector \(\mathbf y\in R^m\) for which \[\lim_{h\to0}\Biggl[\frac{\mathbf f(x+h)-\mathbf f(x)}{h}-\mathbf y\Biggr]=0\] Or \[\mathbf f(x+h)-\mathbf f(x)=h\mathbf y+\mathbf r(h)\] where \(\mathbf r(h)/h\to0\) as \(h\to0\). \(\mathbf y\) or \(\mathbf f'(x)\) is a member of \(L(R^1,R^m)\). If \(\mathbf f\) is a differentiable mapping of \((a,b)\subset R^1\) into \(R^m\) and if \(x\in(a,b)\), then \(\mathbf f'(x)\) is the linear transformation of \(R^1\) into \(R^m\) that satisfies \[\lim_{h\to0}\Biggl[\frac{\mathbf f(x+h)-\mathbf f(x)-\mathbf f'(x)h}{h}\Biggr]=\mathbf 0\] or \[\lim_{h\to0}\frac{|\mathbf f(x+h)-\mathbf f(x)-\mathbf f'(x)h|}{|h|}=0\] Suppose \(E\) is an open set in \(R^n\), \(\mathbf f\) maps \(E\) into \(R^m\) and \(\mathbf x\in E\). If there exists a linear transformation \(A\) of \(R^n\) into \(R^m\) such that \[\lim_{\mathbf h\to0}\frac{|\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)-A\mathbf h|}{|\mathbf h|}=0\] then we say that \(\mathbf f\) 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 \(E\). The relation can be rewritten in the form \[\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)=\mathbf f'(\mathbf x)\mathbf h+\mathbf r\mathbf h\] where the remainder \(\mathbf r\mathbf h\) satisfies \[\lim_{\mathbf h\to\mathbf 0}\frac{|\mathbf r\mathbf h|}{|\mathbf h|}=0\] Then \(\mathbf f'\) maps \(E\) into \(L(R^n,R^m)\).
A function \(\mathbf f\) maps an open set \(E\subset R^n\) into \(R^m\), let \(\{\mathbf e_1,\cdots,\mathbf e_n\}\) and \(\{\mathbf u_1,\cdots,\mathbf u_n\}\) be the standard bases of \(R^n\) and \(R^m\). The components of \(\mathbf f\) are the real functions \(f_1,\cdots,f_m\) defined by \[\mathbf f(\mathbf x)=\sum_{i=1}^{m}f_i(\mathbf x)\mathbf u_i\quad(\mathbf x\in E)\] or equivalently \[f_i(\mathbf x)=\mathbf f(\mathbf x)\cdot\mathbf u_i\] For \(\mathbf x\in E, 1\leq i\leq m, 1\leq j\leq n\) we define \[\frac{\partial f_i}{\partial x_j}=(D_jf_i)(\mathbf x)=\lim_{t\to0}\frac{f_i(\mathbf x+t\mathbf e_j)-f_i(\mathbf x)}{t}\] \(\frac{\partial f_i}{\partial x_j}\) or \(D_jf_i\) is the partial derivative of \(\mathbf f\) with respect to \(x_j\), keeping the other variables fixed.
A function \(\mathbf f\) maps an open set \(E\subset R^n\) into \(R^m\), and \(\mathbf f\) is differentiable at a point \(\mathbf x\in E\). Let \(\{\mathbf e_1,\cdots,\mathbf e_n\}\) and \(\{\mathbf u_1,\cdots,\mathbf u_n\}\) be the standard bases of \(R^n\) and \(R^m\). Then \[\mathbf f(\mathbf x+t\mathbf e_j)-\mathbf f(\mathbf x)=\mathbf f'(\mathbf x)(t\mathbf e_j)+\mathbf r(t\mathbf e_j)\] where \(\frac{|\mathbf r(t\mathbf e_j)|}{t}\to0\) as \(t\to0\) and \[\lim_{t\to0}\frac{\mathbf f(\mathbf x+t\mathbf e_j)-\mathbf f(\mathbf x)}{t}=\mathbf f'(\mathbf x)\mathbf e_j=\lim_{t\to0}\sum_{i=1}^{m}\frac{f_i(\mathbf x+t\mathbf e_j)-f_i(\mathbf x)}{t}\mathbf u_i\quad(1\leq i\leq m, 1\leq j\leq n)\] It follows that each quotient in this sum has a limit, as \(t\to0\), so that each \((D_jf_i)(\mathbf x)\) exists and then \[\mathbf f'(\mathbf x)\mathbf e_j=\sum_{i=1}^{m}(D_jf_i)(\mathbf x)\mathbf u_i\quad(1\leq i\leq m, 1\leq j\leq n)\] The \(\mathbf f'(\mathbf x)\mathbf e_j\) is the \(j^{th}\) column vector of \(\mathbf f'(\mathbf x)\) and \((D_jf_i)(\mathbf x)\) is the \(i^{th}\) and \(j^{th}\) column element of \(\mathbf f'(\mathbf x)\) and \[\underset{m\times n}{\mathbf f'(\mathbf x)}=\begin{bmatrix} (D_1f_1)(\mathbf x)&\cdots&(D_nf_1)(\mathbf x)\\ (D_1f_2)(\mathbf x)&\cdots&(D_nf_2)(\mathbf x)\\ \vdots&\ddots&\vdots\\ (D_1f_m)(\mathbf x)&\cdots&(D_nf_m)(\mathbf x)\\ \end{bmatrix}\]
For \(f\in L(R^n,R^1)\), the \(f'(\mathbf x)\) is a \((1\times n)\) row matrix, which has \(D_jf(\mathbf x)\) element in the \(j^{th}\) column. We can define the gradient of \(f\) at \(\mathbf x\) by \[\underset{(n\times 1)}{(\nabla f)(\mathbf x)}=\sum_{j=1}^{n}(D_jf)(\mathbf x)\mathbf e_j\] and if \(\mathbf x,\mathbf u\in R^n, |\mathbf u|=1, f\in L(R^n, R^1)\) the limit \[\lim_{t\to0}\frac{f(\mathbf x+t\mathbf u)-f(\mathbf x)}{t}=(\nabla f)(\mathbf x)\cdot\mathbf u\] is called the directional derivative of \(f\) at \(\mathbf x\), in the direction of the unit vector \(\mathbf u\) and can also be denoted as \(D_{\mathbf u}f(\mathbf x)\).
Suppose \(\mathbf f\) maps a convex open set \(E\subset R^n\) into \(R^m\), \(\mathbf f\) 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 \(\mathbf a,\mathbf b\in E\). Fix \(\mathbf a,\mathbf b\in E\), define \[\gamma(t)=(1-t)\mathbf a+t\mathbf b\] since \(E\) is convex, \(\gamma(t)\in E\) if \(t\in[0,1]\). Let \(\mathbf g(t)=\mathbf f(\gamma(t))\) Then \[\mathbf g'(t)=\mathbf f'(\gamma(t))\gamma'(t)=\mathbf f'(\gamma(t))(\mathbf b-\mathbf a)\] so that \[\lvert\mathbf g'(t)\rvert\leq \lVert\mathbf f'(\gamma(t))\rVert\lvert\mathbf b-\mathbf a\rvert\leq M\lvert\mathbf b-\mathbf a\rvert\] for all \(t\in[0,1]\). Because there are some \(\lvert\mathbf g'(t)\rvert\) will large than the slope \(\frac{\lvert\mathbf g(1)-\mathbf g(0)\rvert}{1-0}\), so \[\lvert\mathbf g(1)-\mathbf g(0)\rvert=\lvert\mathbf f(\mathbf b)-\mathbf f(\mathbf a)\rvert\leq M\lvert\mathbf b-\mathbf a\rvert\]
A differentiable mapping \(\mathbf f\) of an open set \(E\subset R^n\) into \(R^m\) is said to be continuously differentiable in \(E\) if \(\mathbf f'\) is a continuous mapping of \(E\) into \(L(R^n, R^m)\) \[\forall\mathbf x,\mathbf y\in E, \forall\varepsilon>0, \exists\delta>0, \lVert\mathbf f'(\mathbf y)-\mathbf f'(\mathbf x)\rVert<\varepsilon\] when \(|\mathbf x-\mathbf y|<\delta\) we also say that \(\mathbf f\) is a \(\mathscr C'\)-mapping or that \(\mathbf f\in\mathscr C'(E)\)
Suppose \(\mathbf f\) maps an open set \(E\subset R^n\) into \(R^m\). Then \(\mathbf f\in\mathscr C'(E)\) if and only if the partial derivatives \(D_jf_i, 1\leq i\leq m, 1\leq j\leq n\) exist and are continuous on \(E\). Assume that \(\mathbf f\in\mathscr C'(E)\), \[(D_jf_i)(\mathbf x)=(\mathbf f'(\mathbf x)\mathbf e_j)\cdot\mathbf u_i\] for all \(i,j\). Hence, \[(D_jf_i)(\mathbf y)-(D_jf_i)(\mathbf x)=[(\mathbf f'(\mathbf y)-\mathbf f'(\mathbf x))\mathbf e_j]\cdot\mathbf u_i\] \[|(D_jf_i)(\mathbf y)-(D_jf_i)(\mathbf x)|\leq|(\mathbf f'(\mathbf y)-\mathbf f'(\mathbf x))\mathbf e_j|\leq\lVert\mathbf f'(\mathbf y)-\mathbf f'(\mathbf x)\rVert\] Hence \(D_jf_i\) is continuous. If \(D_jf_i\) is continuous, there is a open ball \(S\subset E\) with center at \(\mathbf x\) and radius \(r\), \[|(D_jf_i)(\mathbf y)-(D_jf_i)(\mathbf x)|<\frac{\varepsilon}{n}\quad(\mathbf y\in S, 1\leq j\leq n)\] Suppose \(\mathbf h=\sum h_j\mathbf e_j, |\mathbf h|<r\), put \(\mathbf v_0=\mathbf 0\) and \(\mathbf v_k=h_1\mathbf e_1+\cdots+h_k\mathbf e_k,\quad(1\leq k\leq n)\) Then \[f(\mathbf x+\mathbf h)-f(\mathbf x)=\sum_{j=1}^{n}[f(\mathbf x+\mathbf v_j)-f(\mathbf x+\mathbf v_{j-1})]\] Since \(|\mathbf v_k|<r\) for \((1\leq k\leq n)\) and since \(S\) is convex, the segments with end points \(\mathbf x+\mathbf v_{j-1}\) and \(\mathbf x+\mathbf v_{j}\) lie in \(S\). Since \(\mathbf v_{j}=\mathbf v_{j-1}+h_j\mathbf e_{j}\), For some \(\theta_j\in (0,1)\) the mean value theorem shows that \[f(\mathbf x+\mathbf v_j)-f(\mathbf x+\mathbf v_{j-1})=h_j(D_jf)(\mathbf x+\mathbf v_{j-1}+\theta_jh_j\mathbf e_j)\] It follows that \[\Biggl|f(\mathbf x+\mathbf h)-f(\mathbf x)-\sum_{j=1}^{n}h_j(D_jf)(\mathbf x)\Biggr|\leq\sum_{j=1}^{n}|h_j|\frac{\varepsilon}{n}\leq|\mathbf h|\varepsilon\] for all \(\mathbf h\) such that \(|\mathbf h|<r\). This says that \(f\) is differentiable at \(\mathbf x\) and that \(f'(\mathbf x)\) is the linear function which assigns the number \(\sum h_j(D_jf)(\mathbf x)\) to the vector \(\mathbf h=\sum h_j\mathbf e_j\). The matrix \([f'(\mathbf x)]\) consists of the row \((D_1f)(\mathbf x),\cdots,(D_nf)(\mathbf x)\) and since \(D_1f,\cdots,D_nf\) are continuous functions on \(E\), then \(\mathbf f\in\mathscr C'(E)\).
Let \(X\) be a metric space, if \(\varphi\) maps \(X\) into \(X\) and if there is a number \(c<1\) such that \[d(\varphi(x),\varphi(y))\leq cd(x,y)\] for all \(x,y\in X\), then \(\varphi\) is said to be a contraction of \(X\) into \(X\).
Suppose \(\mathbf f\) is a \(\mathscr C'\)-mapping of an open set \(E\subset R^n\) into \(R^n\), \(\mathbf f'(\mathbf a)\) is invertible for some \(\mathbf a\in E\) and \(\mathbf b=\mathbf f(\mathbf a)\), then there exist open sets \(U\) and \(V\) in \(R^n\) such that \(\mathbf a\in U, \mathbf b\in V\) \(\mathbf f\) is one-to-one (invertible) on \(U\) and \(\mathbf f(U)=V\). Put \(\mathbf y=\mathbf f(\mathbf x)\), \(\mathbf f'(\mathbf a)=A\). Since \(\mathbf f'\) is continuous at \(\mathbf a\), there is an open ball \(U\in E\), with center at \(\mathbf a\), such that \[\lVert \mathbf f'(\mathbf x)-A\rVert<\frac{1}{2\lVert A^{-1}\rVert}\quad(\mathbf x \in U).\] We associate to each \(\mathbf y\in R^n\) a function \(\varphi\), defined by \[\varphi(\mathbf x)=\mathbf x+A^{-1}(\mathbf y-\mathbf f(\mathbf x))\quad(\mathbf x\in E)\] \[\varphi'(\mathbf x)=\mathbf I-A^{-1}\mathbf f'(\mathbf x)=A^{-1}(A-\mathbf f'(\mathbf x))\quad(\mathbf x\in E)\] Then \[\lVert\varphi'(\mathbf x)\rVert=\lVert A^{-1}\rVert\lVert(A-\mathbf f'(\mathbf x))\rVert<\lVert A^{-1}\rVert\frac{1}{2\lVert A^{-1}\rVert}=\frac{1}{2}\quad(\mathbf x\in U).\] Hence \[|\varphi(\mathbf x_1)-\varphi(\mathbf x_2)|\leq\frac{1}{2}|\mathbf x_1-\mathbf x_2|\quad (\mathbf x_1,\mathbf x_2\in U)\] It follows that \(\varphi\) has at most one fixed point in \(U\), so that \(\mathbf f(\mathbf x)=\mathbf y\) for at most one \(\mathbf x\in U\). Thus \(\mathbf f\) is \(1-1\) in \(U\).
Suppose \(\mathbf f\) is a \(\mathscr C'\)-mapping of an open set \(E\subset R^n\) into \(R^n\), \(\mathbf f'(\mathbf a)\) is invertible for some \(\mathbf a\in E\) and \(\mathbf b=\mathbf f(\mathbf a)\), if \(\mathbf g\) is the inverse of \(\mathbf f\), defined in \(V\) by \[\mathbf g(\mathbf f(\mathbf x))=\mathbf x\] then \(\mathbf g\in \mathscr C'(V)\). Pick \(\mathbf y\in V,\mathbf y+\mathbf k\in V\). Then there exist \(\mathbf x\in U, \mathbf x+\mathbf h\in U\), so that \(\mathbf y=\mathbf f(\mathbf x), \mathbf y+\mathbf k=\mathbf f(\mathbf x+\mathbf h)\) We associate to each \(\mathbf y\in R^n\) a function \(\varphi\), defined by \[\varphi(\mathbf x)=\mathbf x+A^{-1}(\mathbf y-\mathbf f(\mathbf x))\quad(\mathbf x\in E)\] \[\varphi(\mathbf x+\mathbf h)-\varphi(\mathbf x)=\mathbf h+A^{-1}[\mathbf f(\mathbf x)-\mathbf f(\mathbf x+\mathbf h)]=\mathbf h-A^{-1}\mathbf k\] and \[|\varphi(\mathbf x+\mathbf h)-\varphi(\mathbf x)|=|\mathbf h-A^{-1}\mathbf k|\leq \frac{1}{2}|\mathbf h|\] Thus \[|\mathbf h|-|A^{-1}\mathbf k|\leq|\mathbf h-A^{-1}\mathbf k|\leq\frac{1}{2}|\mathbf h|\] \[|A^{-1}\mathbf k|\ge\frac{1}{2}|\mathbf h|\] \[|\mathbf h|\leq2|A^{-1}\mathbf k|\leq2\lVert A^{-1}\rVert|\mathbf k|\] Then \[\lVert \mathbf f'(\mathbf x)-A\rVert<\frac{1}{2\lVert A^{-1}\rVert}\leq \frac{|\mathbf k|}{|\mathbf h|}\] Then \(\mathbf f'(\mathbf x)\) has an inverse, say \(T\). Since \[\mathbf g(\mathbf y+\mathbf k)-\mathbf g(\mathbf y)-T\mathbf k=\mathbf h-T\mathbf k=-T[\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)-\mathbf f'(\mathbf x)\mathbf h]\] implies \[\frac{|\mathbf g(\mathbf y+\mathbf k)-\mathbf g(\mathbf y)-T\mathbf k|}{|\mathbf k|}=\frac{|-T[\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)-\mathbf f'(\mathbf x)\mathbf h]|}{|\mathbf k|}\leq \frac{\lVert-T\rVert|[\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)-\mathbf f'(\mathbf x)\mathbf h]|}{|\mathbf k|}\\ \leq\frac{\lVert T\rVert\lVert2A^{-1}\rVert|[\mathbf f(\mathbf x+\mathbf h)-\mathbf f(\mathbf x)-\mathbf f'(\mathbf x)\mathbf h]|}{|\mathbf h|}\] As \(\mathbf k\to0\), \(\mathbf h\to0\). The right side of the last inequality tends to 0, hence the same is true of the left. Then \(\mathbf g'(\mathbf y)=T\), because \(T\) is the reverse of \(\mathbf f'(\mathbf x)=\mathbf f'(\mathbf g(\mathbf y))\) Thus \[\mathbf g'(\mathbf y)=[\mathbf f'(\mathbf g(\mathbf y))]^{-1}\quad(\mathbf y\in V)\] Finally, note that \(\mathbf g\) is a continuous mapping of \(V\) onto \(U\) (since \(\mathbf g\) is differentiable), that \(\mathbf f'\) is a continuous mapping of \(U\) into the set \(\Omega\) of all invertible elements of \(L(R^n)\) and that inversion is a continuous mapping of \(\Omega\) onto \(\Omega\). Then \[\mathbf g\in\mathscr C'(V)\].
A point in \(R^{n+m}\) can be denoted as \((\mathbf x,\mathbf y)=(x_1,\cdots,x_n,y_1,\cdots,y_m)\in R^{n+m}\). Every \(A\in L(R^{n+m},R^n)\) can be split into two linear transformations \(A_{\mathbf x}\) and \(A_{\mathbf y}\), for any \(\mathbf h\in R^n, \mathbf k\in R^m\) \[\mathbf A_{\mathbf x}\mathbf h=A(\mathbf h,\mathbf 0),\quad A_{\mathbf y}\mathbf k=A(\mathbf 0,\mathbf k)\] \[\underset{n\times(n+m)}{A}\begin{bmatrix} \underset{n\times1}{\mathbf h}\\ \underset{m\times1}{\mathbf k}\\ \end{bmatrix}=\begin{bmatrix} \underset{n\times n}{A_{\mathbf x}}&\underset{n\times m}{A_{\mathbf y}}\\ \end{bmatrix}\begin{bmatrix} \underset{n\times1}{\mathbf h}\\ \underset{m\times1}{\mathbf 0}\\ \end{bmatrix}+\begin{bmatrix} \underset{n\times n}{A_{\mathbf x}}&\underset{n\times m}{A_{\mathbf y}}\\ \end{bmatrix}\begin{bmatrix} \underset{n\times1}{\mathbf 0}\\ \underset{m\times1}{\mathbf k}\\ \end{bmatrix}\] Or \[A(\mathbf h, \mathbf k)=A_{\mathbf x}\mathbf h+A_{\mathbf y}\mathbf k\]
Suppose \(X\) and \(Y\) are vector spaces and \(A\in L(X,Y)\). The null space of \(A\), \(\mathscr N(A)\) is the set of all \(A\mathbf x=\mathbf 0, \mathbf x\in X\), which is a vector space in \(X\). The range of \(A\), \(\mathscr R(A)\), is a vector space in \(Y\). The rank of \(A\) is the dimension of \(\mathscr R(A)\).
Let \(X\) be a vector space. An operator \(P\in L(X)\) is said to be a projection in \(X\) if \(P^2=P\) or \[P(P\mathbf x)=P\mathbf x\] for all \(\mathbf x\in X\).
Suppose \(f\) is a real function defined in an open set \(E\subset R^n\), with partial derivatives \(D_1f,\cdots,D_nf\), if the functions \(D_jf\) are themselves differentiable, then the second-order partial derivatives of \(f\) are \[D_{ij}f=D_iD_jf\quad(i,j=1,\cdots,n).\] If all these functions \(D_{ij}f\) are continuous in \(E\), we say that \(f\) is of class \(\mathscr C''\) in \(E\) or \(f\in \mathscr C''(E)\). Suppose \(E\subset R^2\), \(D_1f,D_2f, D_{21}f\) exist at every point of \(E\) and \(D_{21}f\) is continuous at some point \((a,b)\in E\), then \(D_{12}f\) exists at point \((a,b)\in E\) and \[D_{21}f(a,b)=D_{12}f(a,b)\]. For the rectangle \(Q\) with \((a,b)\) and \((a+h,b+k)\) as opposite vertices. Put \[\Delta(f,Q)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)\] Then based on the mean value theorem, there exists point \((x,y)\) in \(Q\) that \[\Delta(f,Q)=hk(D_{21}f)(x,y)\] Because \(D_{21}f\) is continuous at point \((a,b)\), we can move \((x,y)\) near enough to \((a,b)\) to make \[|(D_{21}f)(x,y)-(D_{21}f)(a,b)|=\Biggl|\frac{\Delta(f,Q)}{hk}-(D_{21}f)(a,b)\Biggr|<\varepsilon\] Fix \(h\), let \(k\to0\), since \(D_2f\) exists in \(E\), then \[\Biggl|\frac{(D_{2}f)(a+h,b)-(D_{2}f)(a,b)}{h}-(D_{21}f)(a,b)\Biggr|\leq\varepsilon\] It follows that \[(D_{12}f)(a,b)=(D_{21}f)(a,b)\]