Suppose X and Y Are Two Bounded Subsets of R and Write X+y, Prove Sup(X+y)

My solutions to some of Spivak's exercises (I skip the few I found non-interesting). I guarantee no correctness. Hopefully any self-learners out there can benefit from this. Please do send comments, corrections and discussion to me by e-mail.

Spivak's Notation Spivak denotes points in a Euclidean space as row vectors in text, however used in matrix equations -- for example, top of Page 4 -- they can magically turn into column vectors. This can be confusing and while reading the book one must always remember to think of vectors as points in an abstract space, not as of 'rows' or 'columns' of some sort. In addition, linear transformation matrices should always be considered \textit{matrices}, even if they are \(n \times 1\) or \(1 \times n\) matrices. To make this clear, even though Spivak will denote a point in \(\mathbf{R}^3\) as \((a,b,c)\), a linear transormation \(f\colon\mathbf{R}^3\rightarrow \mathbf{R}\) is still a \(1\times 3\) matrix. This is particulary confusing in cases like Problem 2-13, where even the author feels compelled to provide an explanatory note.

My Notation I will denote matrices and linear transformations with capital letters; the i-th row of matrix \(A\) by \(A_{i,:}\), and its j-th column by \(A_{:,j}\).

1- Functions on Euclidean Space

1-1 \((\sum_i |x^i|)^2 = \sum_i |x^i|^2 + r\), where \(r\) is a non-negative number. This is trivially greater or equal to \(|x|^2 = \sum_i |x^i|^2\).

1-10 This result comes up many times in the book. Let \(T\) denote the \(n \times m\) matrix of the linear transformation, and set \(k=Th\), an n-dimensional vector with \(T_{i,:} \cdot h\) as its i-th entry. By Thm 1-1(4), \(\|k_i\|\leq \|T_{i,:}\|\cdot \|h\|\) which is smaller or equal to \(\leq m|\max{(T_{i,j})}|\cdot \|h\|\). Since this is independent of i, we can set \(M = nm|max{(T_{i,j})}|\) by Problem 1-1.

1-12 Let \(u,v\in \mathbf{R}^n\) be two different vectors, for example \(u^i \neq v^i\). Then \(\phi_u(e_i) = u^i\) and \(\phi_v(e_i)=v^i\) so \(T\) is 1-1. It is linear by bilinearity of the dot product. The last part follows from the fact that every injective linear transofrmation is also surjective.

1-13 By Thm 1-2(5), \(|x+y|^2 = \sum_i (x^i+y^i)^2 = \sum_i (x^i)^2+(y^i)^2+x^iy^i = |x|^2+|y|^2+\langle x,y \rangle = |x|^2+|y|^2\).

1-14 By Spivak's definition, a set is open iff it is any sum of open rectangles. Thus, any sum of open sets is trivially also open. Since taking the intersection of two open rectangles also gives an open rectangle, finite intersections are open. For the infinite counter-example, take \(\bigcup_{i=1}^\infty (-1/i,1/i) = \{0\}\), a singleton closed set.

1-19 Let \(B = A\cap[0,1]\). Then \(B\) is an intersection of two closed sets and so is itself closed, and is in addition bounded. By Corollary 1-7, \(B\) is compact. Now assume there is some \(x \in [0,1], x \notin B\); then \(\bigcup_{i=1}^\infty (-1,x-1/i) \cup \bigcup_{j=1}^\infty (x+1/j,2)\) is an open cover of \(B\). However, it has no finite subcover, which contradicts the compactness of \(B\).

1-20 Assume \(A\in \mathbf{R}^n\) is not bounded. Then \(\bigcup_{k=1^\infty} (-k,k)^n=(-k,k)\times(-k,k)\times\cdots(-k,k)\) is an open cover of \(A\) that has no finite subcover, a contradiction. Now assume \(A\) is not closed -- that is, there is a point \(x\notin A\) on \(A\)s boundary. Let \(U_k(x)\) be an open square of side length \(k\) centered on \(x\). Then \(\bigcup_{k\in\mathbf{R}}[\mathbf{R}^n-U_k(x)]\), that is the sum of complements of all such open squares, is an open cover of \(A\) (the only point it doesn't contain is \(x\)). However, for any finite subcover there is a neighborhood of \(x\) the subcover doesn't contain, and this neighborhood has a non-empty intersection with \(A\) since \(x\) is on the boundary.

1-21 (a) Assume the premise is not true, and for any \(d\) take an open square centered on \(x\), with side of length \(2d\). Since each such square contains some point \(y\in A\),\(x\) is on the boundary of \(A\). However, \(A\) is closed and thus contains its own boundary, which contradicts the assumption that \(x\notin A\).
(b) Note that by (a), there is already for each \(y\in B\) a number \(d_y\) such that \(|y-x| \geq d \) for all \(x\in A\); we only need to prove that there is a smallest such \(d_y>0\). Now, since \(A\) is closed, its compliment \(A'\) is open. Since \(B \subset A'\), for any \(y\in B\) we can find an open rectangle \(U_y\) disjoint with \(A\). Since all such open rectangles form a cover of the compact set \(B\), there is their finite subset \(\{U_{y_1}, U_{y_2},...,U_{y_N}\}\) also covers. Now choose \(d\) to be the half of the largest side length of these rectangles.
(c) Let \(A\) be the set \(\{(x,y) | y \leq log(x)\}\) and let \(B\) be its reflection along the y-axis. The two sets are closed but not bounded and the distance between them takes on arbitrarily low values.

1-22 Complement of \(U\) is closed and disjoint from \(C\), so we can find a nonzero distance \(d\) between the sets by Problem 1-21. Now cover each point in \(C\) with an open ball of radius \(d/2\) and take a finite subcover \(\hat{D}\). By construction, every point in this subcover is at least \(d/2\) distance away from the complement of \(U\). Its closure \(D\) is thus also contained in \(U\), and is a finite union of closed sets. It is also bounded and contains \(C\) by construction.

1-26 (a) Take any line \(ax-by=0, b\in\{0,1\}\). If \(a \leq 0\) or \(b=0\) then the whole line is in \(\mathbf{R}^2-A\). Now consider the case where \(a>0,b=1\). Then the line intersects the parable \(y=x^2\) at \(x=0 \text{ and } x=a\). Thus there is an interval on the line, corresponding to \(x\in (-1,a)\), outside of the set \(A\).
(b) \(f(0,0)=0\), but the sequence \((f(x_i)), x_i = (1/i, \frac{(1/i)^2}{2})\) is of constant value 1 while \(x_i\rightarrow (0,0)\) -- hence \(f\) is not continuous at \(0\) (a simpler way ). For any sequence \((t_i)\rightarrow 0\), \((g_h(t_i))\) consists of the values of \(f\) along a certain line. However, we've shown in (a) above that every such sequence converges to zero, thus \(g_h\) is continuous at zero.

1-29 This is an important fact. Assume \(f\) has no minimum or maximum. Then \(f(A)\) is unbounded. However, \(f\) is continuous so \(f(A)\) must be compact and thus closed and bounded, a contradiction.

2 - Differentiation

2-1 \( 0 = \lim_{h\rightarrow 0} \frac{|f(a+h)-f(a)-Df(a)(h)|}{|h|} \geq \lim_{h\rightarrow 0}\frac{|f(a+h)-f(a)|}{|h|}-\lim_{h\rightarrow 0}\frac{|Df(a)(h)|}{|h|} \geq \lim_{h\rightarrow 0} \frac{|f(a+h)-f(a)|}{|h|} - M_a\), where \(M_a\) is some nonnegative scalar by Problem 1-10. From this it follows that \(0=\lim_{h\rightarrow 0}|h|M_a \geq \lim_{h\rightarrow 0}|f(a+h)-f(a)|\) and by the "squeeze theorem", \(f(a+h)\) converges to \(f(a)\).

2-4 (a) Note that \(h(t) = |tx|g(x/|x|)=t|x|g(x/|x|)\) if \(t>0\), \(h(0)=0\), \(h(t) = -|tx|g(x/|x|)=t|x|g(x/|x|)\) if \(t<0\). Thus \(h(t) = tf(x)\) and \(Dh = f(x)\).
(b) If \(g\) is identically \(0\), then so is \(f\) and its derivative. Otherwise, assume \(Df(0,0)\) exists, which implies that \[lim_{(h,0)\rightarrow (0,0)}\frac{|f(h,0)-f(0) - Df(0,0)(h,0)|}{|h|}=0\] and \[lim_{(0,k)\rightarrow (0,0)}\frac{|f(0,k)-f(0)-Df(0,0)(0,k)|}{|k|}=0.\] However, since \(f(0)=f(h,0) = f(0,k) = 0\) for all \(h,k\), and by linearity of \(Df(0,0)\), these two imply that \(Df(0,0)(1,0)=0\) and \(Df(0,0)(0,1)=0\). Again by linearity, this gives a zero derivative at \((0,0)\). However, we can find a (nonzero) \(x\) such that \(g(x) \neq 0\). Taking this as the \(x\) in (a), we have \(Dh(0)=0\), which contradicts \(Dh(0) = f(x) \neq 0\). Another way to reach a contradiction is to plug the zero derivative into the definition: \(\lim_{h\rightarrow 0}\frac{|f(h)|}{|h|} = \lim_{h\rightarrow 0}\frac{|h||g(h\|h|)|}{|h|} = \lim_{h\rightarrow 0}|g(h/|h|)|\), but this limit doesn't exist if \(g\) is not a constant (that is zero by description of \(g\)).

2-5 Let \(a\in \mathbf{R}^2\) and define \(g(a/|a|) = \frac{a^1}{|a|}\frac{|a^2|}{|a|}\). This fulfills all the requirements from Problem 2-4, and \(f(x,y) = |(x,y)|g\left(\frac{(x,y)}{|(x,y)|}\right)\) as desired.

2-6 We can show that if the derivative exists it must be \(Df(0,0)=0\) just like in Problem 2-4. However, take for example \(\lim_{t\rightarrow 0}\frac{\sqrt{|t^2|}}{|(t,t)|}=\frac{\sqrt{2}}{2}\), a contradiction.

2-7 We'll show that the derivative at \(0\) is zero, that is \(\lim_{h\rightarrow 0} \frac{|f(h)-f(0)|}{|h|} = 0\). This follows as \(f(0) = 0\) and then \(\lim_{h\rightarrow 0}\frac{|f(h)|}{|h|} \leq \lim_{h\rightarrow 0}\frac{|h|^2}{|h|} = 0\), so by the "squeeze theorem" the limit equals zero.

2-8 The proof is not very illuminating and follows from later sections anyway.

2-9 (a) Let \(f\) be differentiable at \(a\), that is \[\lim_{h\rightarrow 0} \frac{f(a+h)-f(a)-f'(a)h}{h}=0\] for some \(f'(a)\). We need to define \(g\) such that \(g(a+h) = f(a)+f'(a)h\). Using \[g(x) = f(a)+f'(a)(h-a)\] works well enough. For the converse, assume \[\lim_{h\rightarrow 0}\frac{f(a+h)-(a_0+a_1(a+h-a))}{h} = 0,\] for some \(a_0,a_1\). Then \(\lim_{h\rightarrow 0}\frac{f(a+h)-a_0-a_1 h}{h} = 0,\) which implies that \[\lim_{h\rightarrow 0}\frac{f(a+h)-a_0}{h}=a_1.\] Now if \(a_0 \neq f(a)\), the LHS above goes to infinity, so the above equation is satisfied if and only if \(a_0 = f(a), f'(a) = a_1.\)

2-10 In each case, write the function carefully in term of projections and function compositions, for example:
(a) \(f = \exp \circ (\ln \circ \pi^1 \cdot \pi^2)\) so by theorems in this section, \(f'(a,b,c) = \exp'(\ln{a}\cdot b)[\ln{a}(0, 1, 0)+\frac{b}{a}(1, 0, 0)] = a^b(\frac{b}{a}, \ln{a}, 0)=(ba^{b-1}, \ln{a}\cdot a^b,0)\).
(b) By Thm 2-3(3) the first row of \(f'\) is the same as in (a), so \[f'(a,b,c) = \begin{pmatrix} ba^{b-1} & \ln{a}\cdot a^b & 0\\ 0 & 0& 1 \end{pmatrix}. \]
(c) \(f = \sin \circ (\pi^1 \cdot \sin \circ \pi^2)\), hence \begin{align*} f'(a,b) &= \sin'(a \sin{b})[a\sin'(b)(0, 1, 0)+sin{b}(1, 0, 0)]\\ &= \sin'(a\sin{b})[\sin{b}, a\sin'(b),0]\\ &= (\sin{b}\cos{(a\sin{b})}, a\cos{b}\cos{(a\sin{b})}). \end{align*}
(d) Let \(g(y,z) = \sin{(y \sin{z})}\). Then \(f = \sin \circ [\pi^1 \cdot g \circ (\pi^2, \pi^3)]\), and we already know \(g'\) from problem 2-10(c) above. Thus \begin{align*} f'(a,b,c) &= \cos{(a\sin{(b\sin{c})})}[(1, 0, 0)g(b,c)+ a\cdot g'(b,c)\cdot \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}]\\ &=\cos{(a\sin{(b\sin{c})})}(g(b,c), g'(b,c)^1, g'(b,c)^2). \end{align*}
(e) Similarily to the above, let \(g(y,z) = y^z\) as in Problem 2-10(a), and \(f = \exp \circ \ln \circ \pi^1 \cdot g \cdot (\pi^2, \pi^3)\). The rest of the calculations is automatic.
(f) Use \(g\) as above and note that \(f = g \circ (\pi^1, \pi^2+\pi^3)\).
(g) Use \(g\) as above and \(f = g \circ (\pi^1+\pi^2, \pi^3)\).
(h) \(f = \sin \circ (\pi^1 \cdot \pi^2)\), and \(f'(a,b) = \cos{(ab)}[(0,a)+(b,0)] = (b\cos{(ab)},a\cos{(ab)})\).
(i) My version of the book is quite unreadable here, but looks like \(f = g \circ (\sin \circ \pi^1 \cdot \pi^2, \cos \circ [??])\), where \(g\) is defined like in Problem 2-10(e).
(j) This follows directly from Thm 2-3(3) and the problems above. We get \[f'(a,b,c) = \begin{pmatrix}b\cos{(ab)} & a\cos{(ab)}\\ \sin{b}\cos{(a\sin{b})} & a\cos{b}\cos{(a\sin{b})})\\ ba^{b-1} & \ln{a}\cdot a^b\end{pmatrix}. \]

2-11 Define \(Int(x) = \int_a^x g\), and from the Fundamental Theorem of Calculus (e.g. Theorem 5.1 in Apostol), \(DInt(x) = g(x)\). Then we have
(a) \(f = Int \circ (\pi^1+\pi^2)\), so \(f'(x,y) = Int'(x,y)\cdot (1,1) = (g(x+y), g(x+y))\).
(b) \(f = Int \circ (\pi^1\cdot \pi^2)\), so \(f'(x,y) = g(xy)(y,x)=(yg(xy),xg(xy))\).
(c) Let \(h(x,y,z) = \sin(x\sin(y\sin(z)))\). Note that we already know \(h'\) from Problem 2-10(d). Then \(f = Int \circ h - Int \circ (\pi^2\cdot \pi^1)\), by basic properties of the integral (see e.g. various theorems 1.* in Apostol). Thus \(f'(x,y,z) = g(h(x,y,z))h'(x,y,z)-(yg(xy), xg(xy), 0)\).

2-12 (a) Note that for any vector \(a=(a^1,...,a^N)\) we have \(|a|\leq|(a^{i_1}, a^{i_2},...,a^{i_k})|\) for any choice of \(i\). We can thus bound the sequence in question by a sequence with limit zero as follows: \begin{align*} \frac{|f(h,k)|}{|(h,k)|} &\leq \frac{1}{|k|}|f(h,k)| = \frac{1}{|k|}\bigl|\sum_{j=1}^m k^j f(h,e_j)\bigr|\\ &\leq \frac{1}{|k|}\bigl|\sum_{j=1}^m |k| f(h,e_j)\bigr|\\ &\leq \frac{|k|}{|k|}\sum_{j=1}^m |f(h,e_j)|, \end{align*} So \(\lim_{(h,k)\rightarrow 0}\frac{|f(h,k)|}{|(h,k)|} \leq \lim_{(h,k)\rightarrow 0}\sum_{j=1}^m|f(h,e_j)|=\sum_{j=1}^m|f(0,e_j)| = 0\).
(b) We can simply check that the equation defining the derivative holds: \begin{align*} \lim_{(h,k)\rightarrow 0}\frac{|f(x+h, y+k)-f(x,y)-[f(a,y)+f(x,b)]|}{|(h,k)|} &= \lim_{(h,k)\rightarrow 0}\frac{|f(x, y)+f(x,k)+f(h,y)+f(h,k)-f(x,y)-f(a,y)-f(x,b)]|}{|(h,k)|} \\ &= \lim_{(h,k)\rightarrow 0}\frac{|f(h,k)|}{|(h,k)|}\\ &= 0. \end{align*} (c) It's easy to check that \(p\) is bilinear. Then, from (b) above applied to \(f=p\) we have \(Dp(a,b)(x,y) = p(a,y)+p(x,b) = ay+bx\), as in Thm 2-3.

2-13 It's easy to verify that the inner product IP is bilinear, so we can apply Problem 2-12.
(a)\(D(IP)(a,b)(x,y) = IP(a,y)+IP(x,b) = \langle a,y \rangle + \langle b, x \rangle\) (as the inner product is symmetric). From this and the definition of matrix-vector multiplication we get \((IP)'(a,b) = (b, a)\).
(b) \(h = IP \circ (f, g)\), and we can apply the Chain Rule to get \begin{align*} h'(a) &= IP'(f(a),g(a))\cdot \begin{pmatrix}f'(a)\\g'(a)\end{pmatrix}\\ &= (g(a), f(a))\cdot \begin{pmatrix}f'(a)\\g'(a)\end{pmatrix}\\ &= \langle g(a), f'(a)^T \rangle + \langle f(a), g'(a)^T \rangle, \end{align*} where the last line follows from the definition of the inner product.
(c) Since \(|x|^2 = \langle x,x \rangle\) for any \(x\), the constant-norm assumption gives us \[IP'(f(t), f(t)) = 0,\] which gives by (b) above \begin{align*} 0 &= \langle f'(t)^T, f(t) \rangle + \langle f(t), f'(t)^T \rangle\\ &= 2\langle f'(t)^T, f(t) \rangle. \end{align*}
(d) We can simply use \(f(t)=t\) (as \(|t|\) is not differentiable at zero). This doesn't really fit with the rest of the problem though...

2-16 We have \(Df\circ f^{-1} = D(id)\), and from the Chain Rule \(f'(f^{-1}(a))(f^{-1})'(a)=\mathbf{1}\), where \(\mathbf{1}\) is the identity matrix. This directly implies the equality in question.

2-17,2-18 These are just a series of one-dimensional differentiations using the definition of a partial derivative. The most important insight is how much easier it is to perform these calculations than to directly apply the chain rule like in Problem 2-10.

2-19 \(f(1,y) = 1\) which is constant in \(y\), so \(D_2f(1,y)=0\).

2-20 (a) \(D_1f(x,y) = h(y)Dg(x)\), \(D_2f(x,y) = g(x)Dh(y)\).
(b) Using the Chain Rule, \(D_1f(x,y) = h(y)g(x)^{h(y)-1}Dg(x)\), \(D_2f(x,y) = g(x)^{h(y)}\ln{(g(x))}Dh(y)\).
(e) Again using the Chain Rule, \(D_1f(x,y) = D_2f(x,y) = Dg(x+y)\).

2-21 (a) The first term of the sum is independent of \(y\), and the derivative (with respect to y) of the second term equals \(g_2(x,y)\) by the Fundamental Theorem of Calculus.
(b) This question is not quite clear. The function \[ f(x,y) = \int_0^x g_1(t,y)\, dt + \int_0^y g_2(0,t)\!dt\] is of similar for and gives the desired result, but so does for example \[f(x,y) = \int_0^x g_1(t,y)dt + h(y)\] for any function \(h\).
(c) Let \[f(x,y) = \int_0^x t\, dt+\int_0^y t\, dt\] for the first condition, and let \[f(x,y) = xy\] for the second.

2-22 Let \(D_2f = 0\). Then, by one-variable calculus (in particular the Mean Value Theorem, see e.g. Apostol) \(f(x,y_1) = f(x,y_2)\) for all \(y_1,y_2\). That is, \(f\) is independent of the second variable. If in addition \(D_1f=0\), then \(f\) is constant in both variables by simmilar reasoning.

2-23 (a) Let \((x_1,y_1)\) and \((x_2,y_2)\) be two points in \(A\). Because \(D_1f=0\), the value of \(f\) is constant along the lines \(l_1 = \{(x,y_1)|x\in \mathbf{R}\}\) and \(l_2 = \{(x,y_2)|x\in \mathbf{R}\}\). In particular, we have \(f(-1,y_1) = f(x_1,y_1), f(-1,y_2) = f(x_2,y_2)\). Because \(D_2f=0\) we must also have \(f(-1,y_1)=f(-1,y_2)\), hence \(f\) is constant on \(A\).
(b) The line \((x\geq 0, y=0)\) is not in the domain, so the Mean Value Theorem doesn't necessarily hold for intervals of form \([x,y_1], [x,y_2], x\geq 0, y_1<0, y_2>0\). For example, the function \[f(x,y) = \begin{cases} 0 & \text{if } x<0\\ x & \text{if } x\geq 0,y>0\\ -x & \text{if } x\geq 0, y<0 \end{cases}\] is independent of the second variable excapt on such intervals.

2-25 Let first \(x\neq 0\). Then by the Chain Rule\(f^{(1)}(x) = 2x^{-3}e^{-x^{-2}}\). It is easy by induction that any \(f^{(i)}(x)\) will be a sum of product of power functions and the exponential \(e^{-x^{-2}}\), all of which are differentiable and thus by the Product Rule and Chain Rule \(f\) has derivatives of all orders at \(x\neq 0\). At \(x=0\), let \begin{align*} f'(0) = \lim_{h\rightarrow 0}\frac{f(h)}{h} &= \lim_{h\rightarrow 0}\frac{1/h}{e^{h^{-2}}}\\ &= \lim_{h\rightarrow 0}\frac{-h^{-2}}{-2h^{-3}e^{h^{-2}}}\\ &= 0.5 \lim_{h\rightarrow 0} h e^{-1/h^2} = 0, \end{align*} where we used L'Hospital's rule to obtain the second line. Calculating \(f''(0)\) requires using L'Hostpital's rule twice: \begin{align*} f''(0) = \lim_{h\rightarrow 0}\frac{f'(h)}{h} &= \frac{2h^{-3}e^{-h^{-2}}}{h}\\ &=4\lim_{h\rightarrow 0}e^{-h^{-2}} = 0. \end{align*} Since we've proven that \(f^{(i)}\) is differentiable, it must also be continuous by Problem 2-1, so \(f\in C^\infty\) as desired.

2-26 (a) Let \(g(x)\) be the function described in Problem 2-25 above. Then on the \((-1,1)\) interval, \(f(x) = g(x-1)g(x+1)\). Since \(g\) is infinitely differentiable on that interval, by the Product Rule and induction \(f\) is also infinitely differentiable on \((-1,1)\). For \(x \in (-\infty,-1) \cup (1,\infty)\) \(f\) is constant, so we only need to check that the derivative exists at \(1,-1\). For example, for \(x=1\) we have \begin{align*} \lim_{h\rightarrow 1} \frac{g(h+1)g(h-1)}{h} &= g(2)\lim_{h\rightarrow 0} g(h)\\ &= e^{-1/4}0, \end{align*} by Problem 2-25. That \(f\) is positive follows from the definition of an exponential.
(b) Let \(h(x) = f\bigl(\frac{2}{\epsilon}(x-1)\bigr)\). Then \(h\) is positive on \((0,\epsilon)\) and zero elsewhere, and by the Chain Rule is in \(C^\infty\). Now consider \(H(x) = \int_0^x h/\int_0^\epsilon h\). It's easy to see that \(H(x)=0\) if \(x\leq 0\) and \(H(x)=1\) if \(x\geq\epsilon\). In addition, \(H^({i)}(x) = h^{(i-1)}(x)\) for \(x\geq 1\) by the Fundamental Theorem of Calculus, so \(H\in C^\infty\).
(c)\(g \in C^\infty\) by the Product and Chain rules, and it's easy to see that it's positive only on the specified square and zero elsewhere.
(d)Let \(U_x\) be an open rectangle centered on \(x\in C\), such that \(U_x \subsetneq A\). Then we can pick a finite subcover \(U_{x_1}, ..., U_{x_m}\) of compact \(A\). Let \(f\) be a sum of \(m\) functions of form given in Problem 2-26(c) above, each positive on \(U_{x_m}\) and zero elsewhere. \(g\in C^\infty\) because it's a sum of infinitely differentiable functions (note that we needed \(A\) to be compact, as we can't apply the "sum rule" of differentiation to uncountable sums). In addition, \(g \geq 0\) on \(A\) and is zero outside of \(\bigcup_{i} \bar{U_{x_i}}\) (where we again used the compactness of \(A\)).
(e) We only need to show that \(f(x) \geq \epsilon\) for some \(\epsilon > 0\) for any \(x\in A\). This is true for any continuous function by Problem 1-29. The hint completes the problem.

2-28 In each case, let \(a\) be the value of the component functions as in the book's examples. Then
(a)\begin{align*} D_1F(x,y) &= D_1f(a)k(y)g'(x)+D_2f(a)g'(x)\\ D_2F(x,y) &= D_2f(a)g(x)k'(y)+D_2f(a)k'(y). \end{align*}
(b) \begin{align*} D_1F(x,y,z) &= D_1f(a)g'(x+y)\\ D_2F(x,y,z) &= D_1f(a)g'(x+y) + D_2f(a)h'(y+z)\\ D_3F(x,y,z) &= D_2f(a)h'(y+z). \end{align*}
(c) \begin{align*} D_1F(x,y,z) &= D_1f(a)yx^{y-1}+D_3f(a)x^z\ln{x}\\ D_2F(x,y,z) &= D_1f(a)x^y\ln{y}+D_2f(a)zy^{z-1}\\ D_3F(x,y,z) &= D_2f(a)y^z\ln{z}+D_3f(a)xz^{x-1}. \end{align*}
(d) \begin{align*} D_1F(x,y) &= D_1f(a)+D_2f(a)g'(x)+D_3f(a)D_1h(x,y)\\ D_2F(x,y) &= D_3f(a)D_2h(x,y). \end{align*}

2-29 (a) Follows directly from the definition of a partial derivative.
(b) Let \(D_{tx}f(a) = \lim_{s\rightarrow 0}\frac{f(a+s\:tx)-f(a)}{s}\). We can substitute \(u=st\) to get \[D_{tx}f(a) = \lim_{u\rightarrow 0}\frac{f(a+ux)-f(a)}{(u/t)}=tD_xf(a).\]
(c) Since \(f\) is differentiable, we have \[0 = \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)-Df(a)h}{|h|}\] for any sequence \(h\rightarrow 0\). In particular, for any nonzero \(x\in \mathbf{R}^n\), \begin{align*} 0 &= \lim_{t\rightarrow 0}\frac{f(a+tx)-f(a)-tDf(a)x}{t|x|}\\ &= \frac{1}{|x|}lim_{t\rightarrow 0}\frac{f(a+tx)-f(a)}{t}-Df(a)x, \end{align*} which implies that \(D_xf(a) = Df(a)x\) for any nonzero \(x\), and the same holds for \(x=0\) trivially. Linearity of the directional derivative now follows from linearity of \(Df(a)\).

2-31 This follows directly from Problem 1-26, with \(0=Dg_h(0) = D_hf(0,0)\). Of course, in this case \(f\) is not continuously differentiable, having no derivative at the boundary of \(A\).

2-32 (a) To see that \(f\) is differentiable, compute the derivave directy: \[f'(0) = \lim_{h\rightarrow 0} \frac{f(h)}{h}=\lim_{h\rightarrow 0}h\sin{\frac{1}{h}} =0,\] where the last equality follows as \(|\sin{1/h}|\leq 1\) for any \(h\in \mathbf{R}\). For any \(x\neq 0\), \[f'(x)=2x\sin{\frac{1}{x}}-\cos{\frac{1}{x}}.\] It's easy to see that \(f'\) has no limit as \(x\) goes to zero, hence the derivative is not continuous.
(b) This is entirely analoguous to (a). Note also that we can rewrite \(f\) for \(z\in\mathbf{R}_+\) as \(f(z) = |z|^2\sin{\frac{1}{|z|}}\).

2-33 Note that continuity is used in the proof twice. First, applying the mean-value theorem requires continuity of \(g_i(x):=f(xa^1,...,a^{i-1},x,a^{i+1},...,a^n)\) for each \(i\), which is guaranteed by the existence of partial derivatives. Secondly, the last equality of the proof is based on the continuity of the derivaitves, that is \[\lim_{h\rightarrow 0}\sum_{i=1}^n|D_if(c_i)-D_if(a)|=0\] because \[D_if(c_i)\rightarrow D_if(a)\] when \[c_i\rightarrow a.\] However, applying the mean value theorem only to dimensions \(i\geq 2\), we arrive at \begin{align*} \lim_{h\rightarrow 0}\frac{|f(a+h)-f(a)-\sum_{i=1}^nD_if(a)\cdot h^i|}{|h|} &\leq \lim_{h\rightarrow 0}\frac{1}{|h|}|f(c_1)-f(a)-D_1f(a)\cdot h^1|+\lim_{h\rightarrow 0}\sum_{i=2}^n|D_if(c_i)-D_if(a)|\\ &\leq \lim_{h\rightarrow 0}\biggl|\frac{f(c_1)-f(a)}{h^1}-D_1f(a)\biggr|\\ &\leq \lim_{h\rightarrow 0}|D_1f(a)-D_1f(a)|\\ &= 0. \end{align*}

2-34 For any \(x\in \mathbf{R}^n\), let \(h\colon \mathbf{R}\rightarrow \mathbf{R}^n, h(t)=tx\). Then \(g\colon \mathbf{R}\rightarrow \mathbf{R},g = f\circ h\). From the Chain Rule, \begin{align*} g'(1) &= f'(x)\cdot h'(1)\\ &= \sum_{i=1}^{n}x^i D_if(x). \end{align*} From homogeneity of \(f\) we also have \(g(t)=t^mf(x)\) which from the Product Rule gives directly \(g'(1) = mf(x)\).

2-35 Like in Problem 2-34, the derivative of \(h_x(t)\) is \(h_x'(t) = \sum_{i=1}^nx^iD_if(tx)\). Since noe-dimensional integration is a linear operation (see e.g. Apostol), we have \begin{align*} f(x) &=\int_0^1h_x'(t)dt\\ &= \int_0^1 \sum_{i=1}^n x^iD_if(tx)dt\\ &= \sum_{i=0}^nx^i\int_0^1D_if(tx)dt. \end{align*} Now, take \(g_i(x) = \int_0^1D_if(tx)dt\).

2-36 By the Inverse Function Theorem there exists for any \(y\in f(A)\) an open set \(W_y \subset f(A)\) such that \(y \in W_y\), which proves that \(f(A)\) is open. In addition, and \(f\) has a continuous inverse from some open set \(V_{f^{-1}(y)}\ni f^{-1}(y)\) onto \(W_y\), which is differentiable. We only need to show that if \(W_{y_1}, W_{y_2}\) overlap, the values of the inverse functions agree. Assume \(f\) has a local inverse \(g\colon W_{y_1}\rightarrow V_1\) and a local inverse \(h\colon W_{y_2}\rightarrow V_2\), and that for some \(y\in W_{y_1}\cap W_{y_2}\) \(a=g(y), b=h(y), a\neq b\). Since both \(g,h\) are local inverses, this implies \(f(a)=f(b)=y\), which contradicts the injectivity of \(f\). Thus, \(f\) has a "global" inverse \(f^{-1}\) on \(A\) given by the IFT locally. Since the local inverses agree on all the open subsets of \(f(A)\) their derivatives given by the IFT must also agree, thus \(f^{-1}\) is differentiable.

2-37 (a) Assume for contradiction that \(f\) is injective. Then if \(D_2f(x,y)=0\), there must exist an open set \(A\ni (x,y)\) such that \(D_1f(x,y)\neq 0\text{ if }(x,y)\in A\) - otherwise, \((x,y)\) would be a maximum which contradicts the injectivity of \(f\). Hence, we can assume without loss of generality that \[D_1f(x,y)\neq 0\text{ for }(x,y)\in A, A - \text{ open }.\] Now consider the function \(g(x,y) = (f(x,y),y)\) defined on \(A\). Its derivative has a nonzero determinant, as \[g'(x,y)=\begin{pmatrix}D_1f(x,y) & D_2f(x,y)\\ 0 & 1\end{pmatrix}\]. Thus, \(g\) is of the form discussed in Problem 2-36 and it has a (differentiable) inverse on the open set \(g(A) = \{(f(x,y),y), (x,y)\in A\}\). However, this is impossible: for some \((x,y)\in A\) let \(f(x,y)=b\). Then \(g(x,y) = (b,y)\) so the function \(g^{-1}\) is not defined on any point \((b,z), z\neq y\) as this would imply \(f(x,z)=b\) which contradicts that \(f\) is injective.
(b) Similarily as above, we must have \(D_1f(a^1,...,a^n)\neq 0\) for all \(a:=(a^1,...,a^n)\) in some open set \(A\). Then, consider the function \(g(a^1,...,a^n) = (f(a),a^{n-m},...,a^n).\) Its derivative is \[\begin{pmatrix}D_1f(a) & D_2f(a) & D_3f(a) & \cdots & D_nf(a)\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & & & & \\ 0 & \cdots & & & 1 \end{pmatrix}, \] which has nonzero determinant. The rest of the proof is entirely analoguous to that in (a) above. Note that one can apply this to the case of \(m

2-38 (a) Since \(f\) is differentiable (it has a nonzero derivative everywhere), the Mean Value Theorem applies. Assume \(f(x_1) = f(x_2), x_1\neq x_2\). Then by MVT, for some point \(a\in (x_1,x_2)\) we have \(f'(a)=0\), a contradiction.
(b) The derivative matrix is \[\begin{pmatrix}e^x\cos{x} & -e^x\sin{y}\\e^x\sin{y} & e^x\cos{y}\end{pmatrix},\] whose determinant is \(\det{f'(x,y)}=e^{2x}\) for any \((x,y)\). However, \(f(x,y) = f(x,y+2\pi)\) for any \((x,y)\), by periodicity of the trigonometric functions.

2-39 Checking that \(Df\) is not continuously differentiable at \(0\) proceeds similarily to Problem 2.32. It's easy to see that \(f'(0)=1/2\), so we could apply the IFT at \(0\) had it not required a continuous derivative. However, for any \(x > 0\) we have \(f'(x) = 1/2-\cos{1/x}+2x\sin{1/x}\). The derivative function is continuous at \(x>0\) and for small \(x\) approximates \(1/2-\cos{1/x}\) arbitrarily well: it has value zero at points with negative and positive second derivatives. Thus \(f\) has both local maxima and minima arbitrarily close to zero, and cannot be inverted around the origin.

2-41 NOTE: I'll assume the function is in \(C^\infty\) (enough to assume twice-continuously differentiability).
(a) Consider the function \(D_2f:\mathbf{R}^2\rightarrow \mathbf{R}\). It is continuously differentiable with a zero at \(x,c(x)\), and \(D_2(D_2f)(x,c(x))\neq 0\) by assumption. Thus the Implicit Function Theorem applies directly and guarantees that \(c\) is differentiable. To compute its derivative, differentiate \[0=D_2f(x,c(x))\] on both sides. Using the Chain Rule gives \begin{align*} 0 &= D_1(D_2f)(x,c(x))\\ &= D_{2,1}f(x,c(x))+D_{2,2}f(x,c(x))c'(x)\\ &\implies c'(x)=-\frac{D_{2,1}f(x,c(x))}{D_{2,2}f(x,c(x))}. \end{align*}
(b) Since \(c'(x)=0\), it follows from (a) above that \(D_{2,1}f(x,y)=0\) for \(y=c(x)\), that is \(y\) such that \(D_2f(x,y)=0\).
(c) Do the appropriate differentiations and boundary bookkeeping, remembering that Theorem 2-6 applies thanks to results in (a,b) above.

3-1 Consider the partition \(P=((0,1/2,1),(0,1))\). Then \(L(f,P) = U(f,P) = 0.5\) which implies (through Corollary 3-2) that \(\sup{L(f,P)} = \inf{U(f,P)}\) -- thus the function is integrable and \(\int_{[0,1]\times [0,1]} = 0.5\).

3-2 Let \(x_1,\cdots,x_N\) be all the points such that \(f(x_i)\neq g(x_0)\) and let \(\delta = \max{|f(x_i)-g(x_i)|}\). Let \(P\) be a partition and note that thanks to Lemma 3-1 we can assume that \(P\) is fine enough to enclose each \(x_i\) in a separate rectangle \(S_i\). Finally, let \(\epsilon = \max{v(S_i)}\). Then we have \begin{align*} |L(f,P)-L(g,P)| &\leq N\delta\epsilon\\ |U(f,P)-U(g,P)| &\leq N\delta\epsilon\\ \end{align*} Thus by refining \(P\) so that the \(x_i\) are enclosed in smaller rectangles we can get the lower and upper sums of \(g\) arbitrarily close to the lower and upper sums of \(f\). This, together with integrability of \(f\), implies that \(g\) is integrable on \(A\) and has the same integral.

3-3 Easy, just follow the definitions.

3-4 If \(f|_S)\) is integrable for all \(S\), let \(\Delta(f,P) = \max_S{U(f|_S,S)-L(f|_S,S)}.\) If \(N\) is the number of subrectangles in \(P\) we have \(U(f,P)-L(f,P) \leq N\Delta(f,P).\) Since by refining \(P\) we can get \(\Delta\) to be arbitrarily small, \(f\) is integrable. For the other direction, assume \(f\) is integrable but for some subrectangle \(S\), \(\inf{U(f|_S,S)}-\sup{L(f|_S,S)} = \delta > 0.\) Then for any partition \(P\) we must have \(U(f,P) - L(f,P) \geq \delta\) -- otherwise we could refine the partition \(P\) into \(P'\) by adding the boundaries of \(S\). But this would imply that \(\inf{U(f|_S,S\in P')}-\sup{L(f|_S,S\in P')} < U(f,P) - L(f,P) < \delta,\) a contradiction.

3-5 Since \(f,g\) are integrable, \(\int_A f = \inf{U(f,P)}\) and \(\int_A g = \inf{U(f,P)}.\) But for any partition \(P\), \begin{align*} U(f,P) &= \sum_S M_S(f) v(S)\\ &\leq \sum_S M_S(g) v(S)\\ &= U(g,P). \end{align*}

3-6 Let \(P\) be a partition. Then for any subrectangle \(S\), \[M_S(f) - m_S(f) \geq |M_S(f)|-|m_S(f)| = M_S(|f|) - m_S(|f|).\] This shows that \(U(f,P) - L(f,P) \geq U(|f|,P) - L(|f|,P)\) and so \(|f|\) is integrable. Furthermore, \begin{align*} \left|\int_Af\right| &= \left|\inf{U(f,P)}\right|\\ &= \inf{\left|\sum_S M_S(f)v(S)\right|}\\ &\leq \inf{\sum_S \left|M_S(f)\right|v(S)}\\ &= \inf{U(\left|f\right|,P)}\\ &= \int_A\left|f\right|. \end{align*}

3-7 First note that for any partition \(P\), any \(m_S(f) = 0\). Thus, it is enough to show that for small enough rectangles, \(M_S(f)v(S)\) gets arbitrarily close to zero. For a natural number \(q\), consider the partition \[P = \bigl((0,1/q,2/q,\cdots,(q-1)/q,1),(0,1)\bigr).\] Clearly if \(x \in [p/q, (p-1)/q], p< q\), then \(x = a/b, b\geq q\). Thus, for any square in this partition \[M_S(f)v(S) = 1/q^2.\] Since \(q\) can be chosen arbitrarily large, we have shown that the upper sums of \(f\) are arbitrarily close to the lower sums for appropriate partitions, and thus \(f\) is integrable. Moreover, \begin{align*} \int_{[0,1]\times [0,1]} f &= \inf{U(f,P)}\\ &= \inf_q{q (1/q^2)}\\ &= 0. \end{align*}

3-8 Analoguous to the proof of Theorem 3-5, but with much more hairy notation.

3-9 (a) Let \(U\) be an unbounded subset of \(\mathbf{R}^n\) and suppose it has content zero. Then we can cover it with a finite cover of closed rectangles of arbitrarily small -- that is, finite -- volume. But a closed rectangle of finite volume must be bounded, and a finite sum of bounded sets is bounded. Thus, we now have the unbounded \(U\) as a subset of a bounded set, a contradiction.
(b) Let \(\mathbf{N}\) be the (closed) set of all natural numbers. It can be arranged in a sequence, and thus has measure zero. However, it is unbounded, and thus cannot have content zero by (a) above.

3-10 (a) The boundary of \(C\) belongs to its closure, which by definition means that the boundary belongs to every closed set containing \(C\). If \(S_1,\cdots,S_N\) is any cover of \(C\) by closed sets, it is also a closed set (being a finite sum of closed sets), and it contains \(C\), thus it also covers the boundary.
(b) Let \(C\) be the set of all rational numbers contained in the interval \([0,1]\). We've seen that this set has measure zero. However, its boundary is the whole interval, obviously doesn't have measure zero.

3-11 Assume we can cover the boundary by a cover of closed intervals \([k_1,l_1], [k_2,l_2],\cdots\) such that \(\sum_i l_i-k_i \leq \epsilon\) for any \(\epsilon\). But by Problem 1-18, the boundary of \(A\) is \([0,1]-A\), so \(A = [0,1]-\text{boundary}A\). Hence, we would have \(\sum_i (b_i-a_i) = 1-\epsilon\), a contradiction.

3-12 The result of the Hint follows easily from Problem 1-30. We can use it to show that the points at which \(f\) is discontinuous can be ordered in a sequence. By Theorem 1-10, \(f\) is discontinuous at \(a\) iff its oscillation at \(a\) is nonzero. Thus for each \(i\in 1,2,\cdots\) we can in turn take the finite set of \(x\) such that \(o(f,x) > 1/n\). Since finite sets have measure zero, Theorem 3-4 implies the result.

3-13 (a) Treat each rectangle as a finite sequence \((a_1,b_1,a_2,b_2,\cdots, a_n,b_n)\). Since we can order the rational numbers, there is a bijection between the rectangles and the set of vectors \(k=(k_1,k_2,\cdots,k_{2n})\) with positive integer coefficients -- thus we only need to arrange this set. Now for any \(N_j\in 1,2\cdots\) there is only a finite number of such vectors with \(\sum_i k_i \leq N_j\). By Theorem 3-4, the set of such vectors thus has measure zero.
(b) The Hint solves the problem.

3-14 Let \(B_f\) and \(B_g\) be the sets of points at which \(f\) and \(g\) are discontinuous. Clearly \(f\dot g\) can only be discontinuous at \(x\) such that either \(g\) is dicontinuous at \(x\) or \(f\) is discontinuous at \(g(x)\). Thus \(B_{f\dot g}\subset B_f \cup B_g\) which is a set of measure zero.

3-15 By Problem 3-9(a), \(C\) is a bounded set and thus is a subset of some closed rectangle \(A\). By Problem 3-10(a), the boundary of \(C\) has content zero and thus measure zero, thus \(C\) is Jordan-measurable. Because \(C\) has content zero, for any partition \(P\) we can create a subpartition in which the elements of \(C\) are enclosed in a finite number of subrectangles of joint volume less than \(\epsilon\); this shows that \(\int_A\chi_C = 0.\)

3-16 The subset of the rationals contained in the interval \([0,1]\) (we've already seen that its boundary is the whole interval, which we know doesn't have measure zero).

3-17 Let \(P\) be a partition of \(A\). For any \(\epsilon\), enclose \(C\) in closed rectangles \(S_1,\cdots,S_i\cdots\) such that \(\sum v(S_i) < \epsilon\). Let \(P'\) be a subpartition of \(P\) containing all these rectangles. Then \[L(\chi_C,P') = \sum m_{S_i}(\chi_C)v(S_i) < \epsilon.\] Since \(L(\chi_C,P) \leq L(\chi_C,P')\) for any refinement \(P'\), this implies \(L(\chi_C,P)=0\) and, since the integral exists, \(\int_A \chi_C = 0\).

3-18 We have that \(\inf{U(f,P)} = 0\). But then for any \(P\) there must exist a subpartition such that \(\{x\colon f(x)>1/n\}\) is enclosed in rectangles of arbitrarily small volume; otherwise the set would contribute at least \(\epsilon 1/n\) to the largest lowe bound on \(U\). Thus, \(\{x\colon f(x)>1/n\}\) has content zero. Now, notice that \(C = \{x\colon f(x) \neq 0\} = \bigcup_i \{x\colon f(x) > 1/i\}\) -- that is, \(C\) is a countable set-sum of sets of measure zero, thus is itself of measure zero.

3-19 By Problem 3-11, \(U\) has a boundary of measure not zero. But \[\text{Boundary}(U) - V \subset \text{Boundary}(U-V),\] and thus if \(V\) has measure zero, \(\text{Boundary}(U-V)\) has a subset of measure not zero, and thus cannot itself have measure zero.

3-20 This follows directly from Problem 3-12.

3-21 By definition of the boundary, \(x\in \text{Boundary}(C)\) iff it can be enclosed in an arbitrarily small rectangle \(S_X\in S_1.\) From this, and from the fact that \(S_2\subset S_1\), the result of the problem follows easily.

3-22 Partition the underlying set as in Problem 3-21, so that the boundary of \(A\) is covered by rectangles in \(S_1-S_2\) of volume less than \(\epsilon\). Then \(S_2\) is a closed set (consisting of a finite sum of closed sets), and it's bounded as a subset of bounded \(A\). Thus, \(C\) is compact and \(\int_{A-C}1 < \sum_{S\in S_1-S_2}v(S) < \epsilon\).

3-23 Since \(C\) has content zero, the characteristic function on it is integrable with integral zero per Problem 3-15. Fubini's Theorem then tells us that \[ \int_A \mathbf{\mathcal{L}} = \int_A \mathbf{\mathcal{U}} = 0, \] which by linearity of integration (Problem 3-3) implies \[ \int_A \mathbf{\mathcal{U}}-\mathbf{\mathcal{L}} = 0. \] Now, the integrand is non-negative, so by Problem 3-18 the set at which it is non-zero has measure zero. We will show this set equals \(A'\). Take any such \(x\); then we have \[ \mathcal{U}\int_B\chi_C(x,y)dy > \mathcal{L}\int_B\chi_C(x,y)dy. \] But this implies that \(\{y: (x,y) \in C\}\) is not of content zero (otherwise both the lower and upper sum would have to be equal), and consequently \(x\in A'\).

3-24 First, show that \(C\) has content zero. Take any \(\epsilon > 0\), and let \(S_1 = [0,1]\times[0,\epsilon/2]\). This rectangle covers all the elements \((p/q,y), y\leq 1/q\)of \(C\) with \(1/q \leq \epsilon/2\). The remaining elements are contained in a finite number (say \(N\)) of segments of a line of form \(p/q \times 1/q\), which can easily be covered by \(N\) rectangles of volume less than \(\epsilon/2\). Now, for any \(p/q\) the set \([0,1/q]\) is not of content zero (Theorem 3-5), thus in this case \(A'\) is the set of all rational numbers in \([0,1]\) which has measure, but not content, zero.

3-25 Using Fubini's Theorem and induction it's easy to see that \[ \int_{[a_1,b_1]\times \cdots \times [a_n,b_n]} 1 > 0 \] for any \(n\) -- that is, the set has nonzero volume. However, by Problem 3-15 sets of content zero have zero volume.

3-26 It's easy to see that the boundary of \(A_f\) equals \([0,1]\times \{0\} \cup \{0\} \times [0,f(0)] \cup \{1\}\times [0,f(1)] \cup \text{graph}(f)\). The first three elements of this set sum are easily seen to have content zero. We thus only need to prove that \(\{(x,f(x))\colon x\in [a,b]\}\) has content zero. This is almost immediate from the definition of integrability: \(f\) is integrable so we can partition \([a,b]\) into intervals \([k_i,l_i]\) such that \[\sum_i (l_i-k_i)(M_{[k_i,l_i]}(f)-m_{[k_i,l_i]}(f))\] is arbitrarily small. In other words, we can cover the graph of \(f\) with rectangles \([k_i,l_i]\times [m_{[k_i,l_i]}(f),M_{[k_i,l_i]}(f)]\) of arbitrarily small volume. Now we can apply Fubini's Theorem. First, note that \(f\) must achieve a maximum on the compact interval \([a,b]\). Call this maximum \(B\). Then \begin{align*} \int_{A_f} 1 &= \int_{[a,b]\times[0,B]} \chi_{A_f}\\ &=\int_{[a,b]}\int_{[0,f(x)]}1\,dx\\ &=\int_{[a,b]}f(x)\,dx. \end{align*}

3-27 Let \(C = \{(x,y)\in [a,b]^2\colon x\leq y\}\). By Problem 3-26, \(C\) is Jordan-measurable. Since \(f\) is continuous it is integrable, Fubini's Theorem gives us directly \[ \int_Cf = \int_a^b \int_a^y f(x,y)dx dy = \int_a^b \int_x^b f(x,y)dy dx.\]

3-28 For any \(a\), either \(D_{1,2}f(a) = D_{2,1}f(a)\) or we can assume \(D_{1,2}f(a) - D_{2,1}f(a) > 0\) (if the difference is smaller then zero, we use exactly the same reasoning to arrive at a contradiction). Now, since the derivatives are continuous, \(D_{1,2}f(a) - D_{2,1}f(a) > 0\) implies that there is an open set \(O\) containing a rectangle \(A = [a,b]\times [c,d]\ni a\) such that \(D_{1,2}f - D_{2,1}f > 0\) on \(A\). Now, we can use Fubini's Theorem to get \begin{align*} \int_A D_{1,2}f - D_{2,1}f &= \int_{[a,b]} \int_{[c,d]}D_{1,2}f(x,y) - D_{2,1}f(x,y)\,dy\,dx\\ &=\int_{[a,b]}\bigl(D_1f(x,d)-D_1f(x,c)\bigr)\,dx -\int_{[a,b]}\bigr(D_2f(b,y)-D_2f(a,y)\bigl)\,dy\\ &=f(b,d)-f(b,c)-f(a,d)+f(a,c)-(f(b,d)-f(a,c)-f(a,d)+f(a,c))\\ &=0, \end{align*} a contradiction. Note that above, we've used the theorem to write down the iterated integrals in the first line, and to switch the order of integration of the second integral in the second line.

3-30 ERRATUM: The problem as stated is trivial. We'll show instead that \[\int_{[0,1]}\int_{[0,1]}\chi_C(x,y)\,dx\,dy = \int_{[0,1]}\int_{[0,1]}\chi_C(x,y)\,dy\,dx = 0.\] First, note that \(\int_{[0,1]^2}\chi_C\) does not exist, since its boundary is the whole unit square. However, \[\int_{[0,1]}\chi_C(x,y)dx = 0\] for any \(y\), as by definition of \(C\) there is only one point on the line \([0,1]\times \{y\}\). By similar reasoning, \[\int_{[0,1]}\chi_C(x,y)dy =0.\] From these two facts, the result follows directly as \(\int_{[0,1]} 0 = 0\).

3-31 Define \(g_{x^{-i}}(x^i) := g_{x^1,\cdots,x^{i-1},x^{i+1},\cdots,x^n}(x^i) = f(x^1,\cdots,x^n).\). Also, let \(A_{-i}(x)\) indicate the set \([a_1,x^1]\times \cdots \times [a_{i-1},x^{i-1}] \times [a_{i+1},x^{i+1}] \times \cdots \times [a_n,x^n]\). Since \(f\) is continuous, by Fubini's Theorem we can freely exchange the order of integration to get \[F(x) = \int_A f = \int_{[a_i,x^i]}\int_{A_{-i}(x)} f.\] This yields \begin{align*} D_iF(x) &= D_i\int_{[a_i,x^i]} \int_{A_{-i}(x)} f\\ &= \int_{A_{-i}(x)} f(t^1,\cdots,t^{i-1},x^i,t^{i+1},\cdots,t^n) dt^{-i}, \end{align*}

3-32 Since \(\int_c^yD_2f(x,y)\,dy = f(x,y)-f(x,c)\), we have \[ F(y) = \int_a^bf(x,y)\,dx = \int_a^b\left(\int_c^yD_2f(x,y)\,dy + f(x,c)\right)\,dx. \] Now, \begin{align*} F(y) &= \int_a^b\left(\int_c^yD_2f(x,y)\,dy + f(x,c)\right)\,dx\\ &= \int_a^b\int_c^yD_2f(x,y)\,dy\,dx+\int_a^bf(x,c)\,dx\\ &= \int_c^y\int_a^bD_2f(x,y)\,dx\,dy+\int_a^bf(x,c)\,dx, \end{align*} where we used linearity of the integral to get the second equality, and Fubini's Theorem to get the final equality. From this and the Fundamental Theorem of Calculus it follows directly that \[F'(y) = \int_a^bD_2f(x,y)\,dx.\] Note that the use of Fubini's Theorem here didn't require that \(D_2f\) be continuous. Indeed, the exchange of the derivative ordering required only that both functions \(g(x) = D_2f(x,y)\) and \(g(y) = D_2f(x,y)\) be integrable, so that appropriate lower and upper integrals from the theorem are equal.

3-33 (a) From the Fundamental Theorem of Calculus, \(D_1F(x,y) = f(x,y)\). By Problem 3-32, \(D_2F(x,y) = \int_a^xD_2f(t,y)\,dt.\)
(b) We can write \(G(x) = F(g(x),x).\) Using the Chain Rule, we have \begin{align*} DG(x) &= D_1F(g(x),x)Dg(x)+D_2F(g(x),x)D id\\ &= f(g(x),x)g'(x)+\int_a^{g(x)}D_2f(t,x)\,dt. \end{align*}

3-34 Keeping in mind Leibnitz's rule, compute \begin{align*} D_1f(x,y) &= g_1(x,0) + \int_0^y D_1g_2(x,t)\,dt\\ &= g_1(x,0)+\int_0^yD_2g_1(x,t)\,dt\\ &= g_1(x,y). \end{align*}

3-36 This follows directly from Fubini's Theorem.

3-37 (a) Suppose first that \(\int_{(0,1)}f\) exists. That is, there is an admissible open cover \(\mathcal{O}\) of \((0,1)\) and a partition of unity \(\Phi\) subordinate to \(\mathcal{O}\) such that \[ \sum_{\phi\in\Phi}\int_{(0,1)}\phi f \] converges, assuming \(f\) is bounded in some open set around each point of \((0,1)\). By Theorems 3-11 and 3-12 we can choose \[\mathcal{O} = \{(\epsilon, 1-\epsilon)\}\] for any sequence \(\epsilon\rightarrow 0.\) Furthermore, by Theorem 3-11(4), we can arrange \(\phi\in\Phi\) in a sequence such that whenever \(i\leq j\), \(\phi_i\) is zero outside \((\epsilon_i,1-\epsilon_i)\) and \(\phi_j\) is zero outside \((\epsilon_j,1-\epsilon_j)\) with \((\epsilon_j\leq \epsilon_i)\). Then each partial sum \(\sum_{i=1}^k\int_{(0,1)}\phi_i f\) equals \(\int_{\epsilon_k}^{1-\epsilon_k}f\) for some \(\epsilon_k, \) with \(\epsilon_k\rightarrow 0.\) Thus, the extended integral exists if and only if the limit exists.
(b) This question is broken -- see errata e.g. at http://www.jirka.org/spivak-errata.html.

3-38 Define \(\mathcal{A}_n=\int_{A_n}f.\) Since the series \((\mathcal{A}_n)\) converges conditionally it can be rearranged to converge to arbitrary real values. Let \((\alpha_n)_{n=1}^\infty\) be one such rearrangement that converges to \(\alpha\), and \((\beta_n)_{n=1}^\infty\) another rearrangement, converging to \(\beta.\) We can now choose \(i_1\) such that \(|\sum_{n=1}^{i_1}\alpha_n - \alpha| \leq 1/4\). Next, choose \(i_2\) such that \(|\sum_{n=1}^{i_2}\alpha_n - \alpha| \leq 1/8\). In this way, we can pick an index sequence \((i_1,i_2,\cdots)\) that partitions the original two sequences in a way such that \(|\sum_{n=i_k}^{i_{k+1}}\alpha_n| \leq 1/2^k\). If \(a_k=\sum_{n=i_k}^{i_{k+1}}\alpha_n,\) the sequence \((a_i)_{i=1}^\infty\) converges absolutely to \(\alpha.\) Of course we can partition \((\beta_i)\) similarily to get a sequence of sums that converges absolutely to \(\beta.\) Now consider the following partition of unity: \[ \phi_k(x) = \begin{cases} 1 & \text{if } x\in \alpha_i, i\in [i_k,i_{k+1})\\ 0 & \text{otherwise} \end{cases}, \] where with some abuse of notation we write \(x\in \alpha_i\) if \(\alpha_i = \int_{A_n}f\) and \(x\in A_n.\) Then by construction, \(\sum_{\phi\in\Phi}|\int\phi\cdot f| = \alpha.\) We can construct a partition of unity \(\Phi\) such that the sum converges to \(\beta\) in the same way.

3-39 Let \(B\) be the set of critical points (i.e. \(\det g'(x) = 0\) if \(x\in B.\). By Sard's Theorem and Problem 3-4, \begin{align*} \int_{g(A)}f &= \int_{g(A)-g(B)}f + \int_{g(B)} f\\ &= \int_{g(A)-g(B)}f. \end{align*} Since the determinant is nonzero on \(g(A)-g(B)\), we can apply Theorem 3-13 as follows: \begin{align*} \int_{g(A)}\chi_{g(A)-g(B)}\cdot f &= \int_{A} (\chi_{g(A)-g(B)}\cdot f \circ g)|\det g'|\\ &= \int_{g(A)} (\chi_{g(A)-g(B)}\circ g)\cdot (f\circ g) \cdot |\det g'|\\ &= \int_{g(A)} (f\circ g) \cdot |\det g'|, \end{align*} where the last equality follows as \(\chi_{g(A)-g(B)} = 0\) if and only if \(|\det g'| = 0\) by construction of \(g(B).\)

3-41 (a) If \(r\cos{\theta} = s\cos{\phi}\) and \(r\sin{\theta} = s\sin{\phi}\), we can square and add together the two equations to see that \(r=s.\) But then \(\sin{\theta}=\sin{\phi}\) and sine is \(1-1\) on \((0,2\pi)\), so \(f\) is injective. We can compute the derivative directly: \[ f'(r,\theta) = \begin{pmatrix} \cos{\theta} & -r\sin{\theta}\\ \sin{\theta} & r\cos{\theta} \end{pmatrix}. \] It's equally easy to check that \[det f'(r,\theta) = r > 0,\] and the last part of the subproblem also follows easily (the half-line \(y=0, x>0\) is excluded because the second dimension of the domain doesn't contain an integer multiple of \(2\pi\)).
(b) Checking that this is the inverse is easy. For the derivative in case \(x\neq 0\), compute directly \[ P'(x,y) = \begin{pmatrix} \frac{x}{\sqrt{x^2+y^2}} & -\frac{y}{\sqrt{x^2+y^2}}\\ \frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} \end{pmatrix}. \] For \(x=0\) use L'Hospital's rule to get the same result.
(c) The first part follows directly from Theorem 3-13 and the fact that \(|det P'(x,y)| = \sqrt{x^2+y^2} = r(x,y).\) The set \(B_r\) is just the ball of radius \(r\), which is the set between the circle of radius \(0\) and the circle of radius \(r\), so the result also follows.
(d) First, note that if \(h(x,y) = e^{-(x^2+y^2)}\) and \(g(r,\theta) = e^{-r^2}\) then \(h(x,y) = g(r(x,y),\theta(x,y)).\) Then it follows from (c) that \begin{align*} \int_{B_r} e^{-(x^2+y^2)} &= \int_0^r\,\int_0^{2\pi}r e^{-r^2} d\theta dr\\ &= 2\pi \int_0^r re^{-r^2} dr\\ &= 2\pi \bigl(-\frac{1}{2}e^{-r^2}\bigr)_0^r\\ &= \pi(1-e^{-r^2}), \end{align*} where we used Fubini's Theorem to arrive at the second equality. For the second equality, use Fubini's Theorem again: \begin{align*} \int_{C_r}e^{-(x^2+y^2)}dx\,dy &= \int_{-r}^r\int_{-r}^r e^{-(x^2+y^2)}dx\,dy\\ &= \int_{-r}^r\int_{-r}^r e^{-x^2}e^{-y^2}dx\,dy\\ &= \left(\int_{-r}^r e^{-x^2}dx\right)^2. \end{align*}
(e) \(C_r\) is the square of side \(r\), which contains \(B_r\) and is contained in \(B_{r\sqrt{2}}.\) By (d), we know that \(\lim_{R\rightarrow\infty} \int_{B_R} e^{-(x^2+y^2)}dx\,dy = \pi\), which thus implies that \(\lim_{R\rightarrow\infty} \int_{C_R} e^{-(x^2+y^2)}dx\,dy = \pi.\) From this and the second equation in (d) it finally follows that \(\int_{-r}^r e^{-x^2}\,dx = \sqrt{\pi}.\)

Suppose X and Y Are Two Bounded Subsets of R and Write X+y, Prove Sup(X+y)

Source: http://www.vision.caltech.edu/kchalupk/spivak.html

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