Advanced Calculus for Economics and Finance

Consider a function \(f: X \to Y\) and let \(A,B \subseteq X\). Check that \(f(A \cup B)=f(A) \cup f(B)\), \(f(A \cap B)=f(A) \cap f(B)\), if \(A \subseteq B\), \(f(A) \subseteq f(B)\). If the function is bijective, \(f(A^c)=(f(A))^c\) and \(f(A \setminus B)= f(A) \setminus f(B)\).

Consider a function \(f: X \to Y\) and let \(A,B \subseteq Y\). Check that \(f^{-1}(A \cup B)=f^{-1}(A) \cup f^{-1}(B)\), \(f^{-1}(A \cap B)=f^{-1}(A) \cap f^{-1}(B)\), if \(A \subseteq B\), \(f^{-1}(A) \subseteq f^{-1}(B)\), \(f^{-1}(A^c)=(f^{-1}(A))^c\), \(f^{-1}(A \setminus B)= f^{-1}(A) \setminus f^{-1}(B)\).

Continuity condition Prove that the function \(f: X \to Y\) between two topological spaces is continuous if and only if \(\forall B \subseteq Y\), \(f^{-1}(\text{int} B) \subseteq \text{int} f^{-1}(B)\).

Continuity condition Prove that the function \(f: X \to Y\) between two topological spaces is continuous if and only if \(\forall A \subseteq X\), \(f(\bar{A}) \subseteq \overline{f(A)}\).

Maximum and minimum of semicontinuous functions With reference to Example 2.20, let \(f\) and \(g\) be two functions from a topological space \((X,T)\) to \(\mathbb{R}\). Define \(M(x)=\max \{f(x),g(x)\}\) and \(m(x)=\min \{f(x),g(x)\}\). Prove that if \(f\) and \(g\) are upper (lower) semicontinuous, then \(M\) and \(m\) are upper (lower) semicontinuous. Hint: \(M(x) < z\) if both \(f(x) < z\) and \(g(x) < z\), while \(m(x) < z\) if either \(f(x) < z\) or \(g(x) < z\).

Subbase Consider a set \(X\) and a collection of subsets \(\mathbb{A}=\{A_\alpha\} \subseteq 2^X\). The number of elements in \(\mathbb{A}\) can be finite or infinite. In general \(\mathbb{A}\) might not be the base of a topology. Now consider all finite intersections of the elements of \(\mathbb{A}\), together with the entire set and the empty set, \(\mathbb{B}=\{ \cap_{i=1}^n A_i \mid A_i \in \mathbb{A}, n\in \mathbb{N} \}\cup X \cup \emptyset\). Prove that \(\mathbb{B}\) is the base of a topology \(T_\mathbb{A}\) on X. The set \(\mathbb{A}\) is said a subbase of the topology \(T_\mathbb{A}\).

Subbase of the Euclidean topology With reference to the previous exercise, prove that the set \(\mathbb{A}=\{(-\infty,a) \mid a \in \mathbb{R}\} \cup \{(b,+\infty) \mid b \in \mathbb{R}\}\) is a subbase of the Euclidean topology in \(\mathbb{R}\).

Dense subsets. Let \((X,d)\) be a metric space, and \(A \subset X\). Prove that \(A\) is dense in \(X\) (Definition 2.2, point 8) if and only if \(\forall x \in X\) and \(\forall \epsilon >0\), \(B(x,\epsilon) \cap A \neq \emptyset\).

Let \(B([0,1])\) be the set of real bounded functions defined over the closed interval \([0,1]\) and \(x_1,x_2 \in [0,1]\). Consider the operator \(h: f \to (f(x_1),f(x_2))\) from \((B([0,1]),\|.\|_\infty)\) to \((\mathbb{R}^2,\|.\|_2)\). Prove that it is a linear operator and compute its norm.

Let \(B([0,1])\) be the set of real bounded functions defined over the closed interval \([0,1]\) and consider the operator \(h: (a,b) \to a + b x^2\) from \((\mathbb{R}^2,\|.\|_2)\) to \((B([0,1]),\|.\|_\infty)\). Prove that it is a linear operator and compute its norm.

Discontinuous linear function. Let \((V,\rho)\) be an infinite dimensional normed space and \((\mathbf{e}_n)\) a sequence of linearly independent elements of \(V\) with norm equal to one. Define the function \(F: V \to \mathbb{R}\) as \(F(\mathbf{x})=\sum_{n=1}^\infty n c_n\) if \(\mathbf{x} = \sum_{n=1}^\infty c_n \mathbf{e}_n\) and zero otherwise. That is, the function \(F\) is zero on all elements of \(V\) that do not belong to the subspace spanned by the sequence \((\mathbf{e}_n)\). Prove that the function \(F\) is linear and, according to Definition 4.3, unbounded. Thus, by Theorem 4.6, \(F\) is discontinuous.

Consider a converging sequence \(\sum_{n=1}^\infty x_n\) and define \(\tilde{s}_n=\sum_{k=n}^\infty x_k\). Prove that \(\lim_{n \to \infty} \tilde{s}_n=0\). Hint: Use the Cauchy property of the sequence of partial sums.

With reference to Example 2.20, let \((f_n)\) be a sequence functions from a topological space to \(\mathbb{R}\). Prove that if the functions are upper semicontinuous, then \(m(x)=\inf_n f_n(x)\) is upper semicontinuous, while if they are lower semicontinuous, then \(M(x)=\sup_n f_n(x)\) is lower semicontinuous.

Limit of monotone sequences of continuous functions. Let \((f_n)\) be a pointwise converging sequence \(f_n \to f\) of continuous functions from a topological space to \(\mathbb{R}\). Prove that if the sequence is nonincreasing, \(f_n(x) \leq f_{n-1}(x)\), then f is upper semicontinuous, while if the sequence is nondecreasing, \(f_n(x) \geq f_{n-1}(x)\), \(f\) is lower semicontinuous. Hint: If the sequence is nonincreasing, \(\left\{x | f(x) < z \right\}\) is equal to \(\cup_n \left\{ x | f_n (x) < z \right\}\), \(\forall z\).

Prove that if two polynomials \(P(x)=\sum_{h=1}^n p_h x^h\) and \(Q(x)=\sum_{h=1}^m q_h x^h\) are equal in an interval \((a,b) \subset \mathbb{R}\), then they are the same polynomial.

Level sets Consider a scalar function \(f: \mathbb{R}^n \to \mathbb{R}\), a real number \(a\), and the associated level set, \(L(a)=\{ \mathbf{x} \in \mathbb{R}^n \mid f(\mathbf{x}) = a \}\), upper level set, \(L(a)^+=\{ \mathbf{x} \in \mathbb{R}^n \mid f(\mathbf{x}) \geq a \}\), and lower level set, \(L(a)^-=\{ \mathbf{x} \in \mathbb{R}^n \mid f(\mathbf{x}) \leq a \}\). Prove that if \(f\) is continuous, \(L(a)\), \(L(a)^+\), and \(L(a)^-\) are closed.

Integral of uniformly convergent sequences. Let \((f_n)\) be a sequence of continuous functions defined on \([a,b]\) converging uniformly to \(f\) (see Definition 5.19). Prove that \(\lim_{n \to \infty } \int_a^b d x f_n(x) = \int_a^b d x f(x)\).

Around p.7.

Lemma The union of a countable number of countable sets is countable.

Proof Imagine to stack the countable sets one below the other. In each row, arrange the elements of a set in increasing fashion, using the order relation induced by their one-to-one map with the natural numbers. Then follow the counting procedure illustrated in the proof of Theorem 1.6 to build a one-to-one relationship with \(\mathbb{N}\).

At the bottom of p.42, before "Using this notions…" add the following sentence: \(T_Y\) is said the subspace topology or relative topology induced by the topology \(T\) on the subset \(Y\).

After theorem 5.3.

We can easily prove that also the inverse is true in first-countable topologies. Footnote: The following statement can be extended to any topological space if instead of using sequences we used the more general notion of net. Nets are not particularly difficult mathematical objects, but they rest outside the limited scope of this textbook.

Theorem In a first-countable topological space, if any converging sequence has one limit, then the space is Hausdorff.

Proof We show that if the space is not Hausdorff, then there is a converging sequence with two limits. Assume that the space is not Hausdorff and let \(x,y \in X\), \(x \neq y\), be such that they do not have disjoint neighborhoods. Since the space is first-countable, there exists a countable base \(\{B_n(x)\}\) of the neighborhoods of \(x\) and a countable base \(\{B_n(y)\}\) of the neighborhoods of \(y\). Consider the decreasing bases \(\{B'_n(x)\}\) and \(\{B'_n(y)\}\) defined as \(B'_n(x)=\cap_{i=1}^n B_i(x)\) and \(B'_n(y)=\cap_{i=1}^n B_i(y)\). Note that if \(n \leq m\) than \(B'_m (x) \subseteq B'_n(x)\) and \(B'_m(y) \subseteq B'_n(y)\). Because \(x\) and \(y\) do not have disjoint neighborhoods, \(B'_n(x) \cap B'_n(y) \neq \emptyset\). Let \(x_n \in B'_n(x) \cap B'_n(y)\). The sequence \((x_n)\) converges to both \(x\) and \(y\). To see it, note that \(\forall N(x)\) and \(\forall N(y)\), \(\exists B_n(x) \subseteq N(x)\) and \(\exists B_m(x) \subseteq N(y)\). Let \(k =\max\{n,m\}\). Then, \(\forall h >k\), \(x_h \in B'_k(x) \subseteq N(x)\) and \(B'_k(x) \subseteq N(y)\).

Example (Banach space of bounded sequences) Let \(p \geq 1\). Consider a real sequence \((x_n)\) and, with some abuse of notation (see Sect. 4.2.2) define \(\| (x_n) \|_p = ( \sum_{n=1}^\infty |x_n|^p)^{1/p}\). Let \(l_p\) be the set of all bounded sequences, \(l_p=\{ (x_n) \mid \| (x_n) \|_p < +\infty\}\). The set of sequences is a linear space under the natural operation \((x_n) + (y_n) = (x_n + y_n)\) and \(a (x_n) = (a x_n)\), \(\forall a \in \mathbb{R}\). The zero element is the constant sequence zero, \((0)\). Applying the Minknowski inequality (Theorem 4.5) to any truncated sequence and taking the limit, it is easy to see that \(\| (x_n) + (y_n) \|_p \leq \| (x_n) \|_p + \| (y_n) \|_p\). This implies that the set \(l_p\) is closed with respect to the addition of its elements, and that \((l_p, \| . \|_p )\) is a normed space. The associated metric space \((l_p,d_p )\) is defined through the distance function \(d_p((x_n), (y_n))= \|(x_n) - (y_n)\|_p\). We want to prove that this space is complete.

Let \((x_n)^m\) be a Cauchy sequence in \(m\) of elements of \(l_p\). That is, \(\forall \epsilon>0\), \(\exists m_\epsilon\) such that, \(\forall m,l > m_\epsilon\),

\[ d_p((x_n)^m, (x_n)^l) = \|(x_n)^m - (x_n)^l\|_p = \left( \sum_{n=1}^\infty |x_n^m-x_n^l|^p \right)^{1/p} < \epsilon. \]

This implies that \(\forall n\), \((x_n^m)\) is a Cauchy, and hence converging, sequence in \(m\). Let \(x^*_n=\lim_{m \to \infty} x_n^m\). For the continuity of the distance function, if \(m> m_\epsilon\),

\[ \lim_{l \to \infty} d_p((x_n)^m, (x_n)^l)= d_p((x_n)^m, \lim_{l \to \infty} (x_n)^l) = d_p((x_n)^m, (x_n^*)) \leq \epsilon. \]

This implies that \(\lim_{m \to \infty} d_p((x_n)^m, (x_n^*))=0\). That is, the sequence \((x_n)^m\) of elements of \(l_p\) converges to \((x_n^*)\) in the distance \(d_p\). It remains to show that \((x_n^*) \in l_p\). By the triangular inequality, \(\forall m\), \(d_p((x_n^*),(0)) \leq d_p((x_n^*),(x_n)^m)+ d_p((x_n)^m,(0))\). Fix \(\epsilon>0\) and pick an \(m\) such that \(d_p((x_n^*),(x_n)^m)<\epsilon\). As \((x_n)^m \in l_p\), \(\exists M>0\) such that \(d_p((x_n)^m,(0))=\| (x_n)^m \|_p < M\). Thus \(d_p((x_n^*),(0)) = \| (x_n^*)\|_p < M+\epsilon\).

At the end of section 7.4

Example (Level set, tangent hyperplane, and differential) Consider a function \(f:E \subseteq \mathbb{R}^n \to \mathbb{R}\), \(f \in C^1(E)\) and has the second-order partial derivatives, \(\mathbf{x}_0 \in \text{int} E\), and \(df(\mathbf{x}_0) \neq\mathbf{0}\). We are interested in the level set of the function \(f\) through \(\mathbf{x}_0\), that is the locus of points \(L=\{ \mathbf{x} \in \mathbb{R}^n \mid f(\mathbf{x}) = f(\mathbf{x}_0) \}\) (this set will be more thoroughly discussed in Section~7.6). Using the Taylor polynomial,

\[ f(\mathbf{x})=f(\mathbf{x}_0)+df(\mathbf{x}_0) \cdot (\mathbf{x}-\mathbf{x}_0) + o(\| \mathbf{x}-\mathbf{x}_0 \|). \]

Thus, in a sufficiently small neighborhood of \(\mathbf{x}_0\), the set \(L\) is well approximated by the hyperplane

\[ df(\mathbf{x}_0) \cdot (\mathbf{x}-\mathbf{x}_0)= \sum_{i=1}^m \partial_i f(\mathbf{x}_0) (x_i-x_{0 i})=0. \]

This hyperplane is a line if \(n=2\) and a plane if \(n=3\). Locally, it has a single point in common with the set \(L\), namely \(\mathbf{x}_0\). Hence, it can be considered the tangent hyperplane to \(L\). It is generated by all the vectors in \(\mathbb{R}^n\) orthogonal to \(df(\mathbf{x}_0)\). In this sense we can say that the differential \(df(\mathbf{x}_0)\) is orthogonal to the level set of the function \(f\) passing through \(\mathbf{x}_0\). See the example in the picture above.

After Theorem 7.23, on page 204.

Example (Second-order conditions) We want to maximize the function \(f(x,y)=y-\beta x^2\) with the equality constraint \(h(x,y)=y-\alpha x^2=0\). The origin of the axis \(\mathbf{0}=(0,0)\) is a candidate solution with multiplier \(\mu=1\), \(df(\mathbf{0})=dh(\mathbf{0})=(0,1)\). The value of the function in this point is zero. The second-order conditions derived in Theorem 7.23 suggest to investigate the expression \(\mathbf{u}^\intercal H_L \mathbf{u}\), where

\[ H_L = H_f-H_h= \begin{pmatrix} 2(\alpha-\beta) &0 \\ 0 &0 \end{pmatrix}, \]

for the nonzero vectors orthogonal to the constraints in \(\mathbf{0}\), that is the vectors of the type \(\mathbf{u}=(u,0)\), \(u \in \mathbb{R} \setminus \{0\}\). Thus we have to study the quantity \(2 u^2 (\alpha-\beta)\). If \(\alpha>\beta\) this quantity is strictly positive, suggesting that the candidate solution cannot be a maximum. In this case (see the picture above) the curve \(h(\mathbf{x})=0\) passes above the contour line \(f(\mathbf{x})=0\), that is the locus of points in which the value of the function \(f\) is zero, so that, along the curve \(h(\mathbf{x})=0\), the function \(f\) takes positive values in any neighborhood of the origin. If instead \(\alpha<\beta\), than the curve \(h(\mathbf{x})=0\) passes through a region of the domain in which the function is negative, or zero in \(\mathbf{0}\). In this case, the condition is sufficient to guarantee that the candidate solution is a strict local maximum. Finally, if \(\alpha=\beta\), than the value of the function is zero when restricted to the points \(\mathbf{x}\) such that \(h(\mathbf{x})=0\). In this case the candidate solution is a local maximum, but not a strict maximum.

As the next example shows, monotonicity is a strong property when integration is concerned.

Example (Pointwise converging sequence of monotone functions) Let \((f_n)\) be a sequence of monotone real valued functions, all increasing or all decreasing, defined on \([a,b]\). Assume that the sequence is pointwise converging to a function \(f\), that is \(\forall x \in [a,b]\), \(f_n(x) \to f(x)\). It is immediate to see that \(f\) must be itself monotone. Thus, by Theorem 8.7, \(\forall n\), \(f_n \in \mathcal{R}([a,b])\), and \(f \in \mathcal{R}([a,b])\). We can actually prove that \[ \lim_{n \to \infty } \int_a^b d x f_n(x) = \int_a^b d x \lim_{n \to \infty } f_n(x). \] Assume for definiteness that the functions are all increasing. Consider \(\forall \epsilon>0\) and let \(P=\{x_0=a,x_1,\ldots,x_{N-1},x_N=b\}\) be a partition of the interval \([a,b]\). Then, \[ \left| U_P(f) - U_P(f_n)\right| \leq \sum_{i=1}^N (x_{i}-x_{i-1}) |f(x_i)-f_n(x_i)|, \text{ and } \left| L_P(f) - L_P(f_n)\right| \leq \sum_{i=1}^N (x_{i}-x_{i-1}) |f(x_{i-1})-f_n(x_{i-1})|, \]

where we have used the monotonic behaviour of the functions. Because the sequence of functions is monotonically convergent, \(\exists n_P\) such that \(\forall n> n_P\), \(\forall i\), \(|f(x_i)-f_n(x_i)|< 1/2 \epsilon/(b-a)\). Then, substituting in the previous equation, \(\forall n > n_P\), \(\left| U_P(f) - U_P(f_n)\right|< \epsilon/2\) and \(\left| L_P(f) - L_P(f_n)\right|< \epsilon/2\). This implies that definitely in \(n\), \(L_P(f)-\epsilon/2 < \int_a^b d x f_n(x) < U_P(f)+\epsilon/2\). Note that the index \(n_P\) will generally depends on the partition \(P\). However, as \(f \in \mathcal{R}([a,b])\), there exists a partition \(P\) for which \(U_{P}(f)-L_{P}(f)<\epsilon/2\), thus, \(\forall n > n_{P}\), \(\left|\int_a^b d x f(x) - \int_a^b d x f_n(x)\right| < \epsilon\), which proves the assertion.

This result differs from the more power monotone convergence, Theorem 9.11, and dominated converge, Theorem 9.14, available when considering Lebesgue integrals. In fact, the latter do not assume that the functions in the sequence are all monotone, albeit Theorem 9.11 requires the convergence itself being monotone. Similar results for the Riemann integral are not generally available for continuous functions. In general, we have to assume that the sequence of functions converges uniformly (see the Exercise above)

After Theorem 9.2. A third useful property of the measure is how it interacts with nested sequences of sets.

Theorem Let \((A_n)\) be an nondecreasing nested sequence of sets, \(A_n \subseteq A_{n+1}\), \(\forall n\). Then \(\lim_{n \to \infty} \mu(A_n) = \mu( \cup_{n=1}^\infty A_n )\). Analogously, if \((B_n)\) is a nonincreasing nested sequence of sets, \(B_{n+1} \subseteq B_{n}\), \(\forall n\), \(\lim_{n \to \infty} \mu(B_n) = \mu(\cap_{n=1}^\infty B_n)\).

Proof First, note that the sequence \((\mu(A_n))\) is monotone nondecreasing and bounded from above by \(\mu(\cup_{n=1}^\infty A_n)\), so that it has a limit. Note also that \(\cup_{n=1}^\infty A_n=\cup_{n=1}^\infty (A_n\setminus A_{n-1})\) and, for different \(n\), the sets \(A_n\setminus A_{n-1}\) are disjoint. Thus, for the additive property of the measure,

\[ \mu(\cup_{n=1}^\infty A_n)=\mu( \cup_{n=1}^\infty (A_n\setminus A_{n-1}) ) = \sum_{n=1}^\infty \mu ( A_n\setminus A_{n-1} )= \lim_{n \to \infty} \sum_{k=1}^n \mu ( A_k\setminus A_{k-1} )= \lim_{n \to \infty} \mu ( \cup_{k=1}^n A_k\setminus A_{k-1} )= \lim_{n \to \infty} \mu ( A_n). \]

To prove the second part of the statement, consider a measurable set \(C\) that contains all the sets of the sequence, \(B_1 \subseteq C\). Define the sequence \((C_n = C \setminus B_n)\). This is a nondecreasing nested sequence of sets, so that, according to the first statement of the theorem, \(\lim_{n \to \infty} \mu(C_n) = \mu( \cup_{n=1}^\infty C_n )\). But \(\mu(C_n) = \mu(C)-\mu(B_n)\) and \(\mu( \cup_{n=1}^\infty C_n )=\mu(C \setminus \cap_{n=1}^\infty B_n )= \mu(C)-\mu(\cap_{n=1}^\infty B_n)\), whence the assertion.

In the previous theorem, \(\cup_{k=1}^n A_k = A_n\) and \(\cap_{k=1}^n B_k = B_n\). Thus, with some abuse of notation, when considering nested sequences, one can move the limit inside the measure, and write \(\lim_{n \to \infty} \mu(A_n) = \mu( \lim_{n \to \infty} A_n )\) or \(\lim_{n \to \infty} \mu(B_n) = \mu( \lim_{n \to \infty} B_n )\).