Exercise set C

Please, see the general comment on the tutorial exercises

Question C.1

Consider two convergent sequences in \(\mathbb{R}^n\), \(\{{\bf x}_i\}_{i=1}^\infty\) and \(\{{\bf y}_i\}_{i=1}^\infty\) such that

\[ \lim_{i \to \infty} {\bf x}_i = {\bf x} \in \mathbb{R}^n, \quad \lim_{i \to \infty} {\bf y}_i = {\bf y} \in \mathbb{R}^n \]

Prove the following properties of the limits:

• \(\lim_{i \to \infty} ({\bf x}_i+{\bf y}_i) = {\bf x} + {\bf y}\)

• \(\lim_{i \to \infty} ({\bf x}_i'{\bf y}_i) = {\bf x}'{\bf y}\)

• \({\bf x}_i \le {\bf y}_i\) for \(\forall i\) component-wise \(\implies {\bf x} \le {\bf y}\)

Definition

The scalar product \({\bf x}'{\bf y}\) of two vector is \(\mathbb{R}\) is defined by \({\bf x}'{\bf y} = \sum_{j=1}^n x_j y_j\)

Component-wise comparison of the vectors is defined as \({\bf x} \le {\bf y} \iff x_j \le y_j\) for all \(j\in\{1,\dots,N\}\)

Tip

Definition of the limit in \(\mathbb{R}^n\) is essentially unchanged from the definition of the limit in \(\mathbb{R}\)

Question C.2

Show that the Cobb-Douglas production function \(f(k,l) = k^\alpha l^\beta\) from \([0,\infty) \times [0,\infty)\) to \(\mathbb{R}\) is continuous everywhere in its domain.

Tip

You can use the fact that, for any \(a \in \mathbb{R}\) the function \(g(x) = x^a\) is continuous at any \(x \in [0,\infty)\). This was mentioned towards the end of lecture 4. Also, remember that norm convergence implies element by element convergence.

Question C.3

Let \(\beta \in (0,1)\). Show that the utility function \(u(c_1,c_2) = \sqrt{c_1} + \beta \sqrt{c_2}\) from \([0,\infty) \times [0,\infty)\) to \(\mathbb{R}\) to \(\mathbb{R}\) is continuous everywhere in its domain.

Question C.4

Let \(A\) be the set of all consumption pairs \((c_1,c_2)\) such that \(c_1 \ge 0\), \(c_2 \ge 0\) and \(p_1 c_1 + p_2 c_2 \le M\) Here \(p_1\), \(p_2\) and \(M\) are positive constants. Show that \(A\) is a closed subset of \(\mathbb{R}^2\).

Tip

Weak inequalities are preserved under limits

Question C.5

Let \({\bf x}, {\bf y} \in \mathbb{R}^N\) and \(\| {\bf x} \| \) denote the Euclidean norm. Verify the Parallelogram Equality given by

\[ \| {\bf x} + {\bf y} \|^2 + \| {\bf x} - {\bf y} \|^2 = 2 \big( \| {\bf x} \|^2 + \| {\bf y} \|^2 \big) \]

Solutions

Question C.1

Let \(\{\mathbf{x}_i\}_{i=1}^\infty\) and \(\{\mathbf{y}_i\}_{i=1}^\infty\) be two converget sequences in \(\mathbb{R}^n\) such that \(\mathbf{x}_i \to \mathbf{x} \in \mathbb{R}^n\) and \(\mathbf{y}_i \to y \in \mathbb{R}^n\) as \(i \to \infty\). Recall that \(\mathbf{x}_i \to \mathbf{x}\) if and only if \(\|\mathbf{x}_i - \mathbf{x}\| \to 0\).

Proof of \(\lim_{i \to \infty} (\mathbf{x}_i + \mathbf{y}_i) = \mathbf{x} + \mathbf{y}\)

Fix \(\varepsilon > 0\). Since \(\mathbf{x}_i \to \mathbf{x}\) and \(\mathbf{y}_i \to \mathbf{y}\), there exists an \(N \in \mathbb{N}\) such that

\[ \|\mathbf{x}_i - \mathbf{x}\| < \frac{\varepsilon}{2} \qquad \text{and} \qquad \|\mathbf{y}_i - \mathbf{y}\| < \frac{\varepsilon}{2} \]

for all \(i \geq N\). Hence, the triangle inequality yields

\[ \|(\mathbf{x}_i + \mathbf{y}_i) - (\mathbf{x} + \mathbf{y})\| = \|\mathbf{x}_i - \mathbf{x} + \mathbf{y}_i - \mathbf{y}\| \leq \|\mathbf{x}_i - \mathbf{x}\| + \|\mathbf{y}_i - \mathbf{y}\|< \varepsilon \]

for all \(i \geq N\). Therefore, since \(\varepsilon\) is arbitrary, we have \(\mathbf{x}_i + \mathbf{y}_i \to \mathbf{x} + \mathbf{y}\) as \(i \to \infty\).

Proof of \(\lim_{i \to \infty} (\mathbf{x}_i' \mathbf{y}_i) = \mathbf{x}' \mathbf{y}\)

Since \(\{\mathbf{x}_i\}_{i=1}^\infty\) and \(\{\mathbf{y}_i\}_{i=1}^\infty\) are convergent sequences, they are bounded (why?) by constants \(M_x > 0\) and \(M_y >0\), respectively. That is, we have \(\|\mathbf{x}_i\|\leq M_x\) and \(\|\mathbf{y}_i\|\leq M_y\) for all \(i\in\mathbb{N}\).

Let \(M:= \max\{M_x, M_y, \|\mathbf{x}\|, \|\mathbf{y}\|\}\). Fix \(\varepsilon > 0\). Again, since these sequences are convergent, there is an \(N \in \mathbb{N}\) such that

\[ \|\mathbf{x}_i - \mathbf{x}\| < \frac{\varepsilon}{2M}, \qquad \text{and} \qquad \|\mathbf{y}_i - \mathbf{y}\| < \frac{\varepsilon}{2M} \]

for all \(i \geq N\). The tirangle inequality and Cauchy-Schwarz inequality imply

\[\begin{split} |\mathbf{x}_i' \mathbf{y}_i - \mathbf{x}'\mathbf{y}| = |\mathbf{x}_i' \mathbf{y}_i - \mathbf{x}' \mathbf{y}_i + \mathbf{x}' \mathbf{y}_i - \mathbf{x}'\mathbf{y}| \\ \leq |\mathbf{x}_i' \mathbf{y}_i - \mathbf{x}' \mathbf{y}_i| + |\mathbf{x}' \mathbf{y}_i - \mathbf{x}'\mathbf{y}| \\ = |(\mathbf{x}_i - \mathbf{x})' \mathbf{y}_i| + |\mathbf{x}'(\mathbf{y}_i-\mathbf{y})| \\ \leq \|\mathbf{y}_i\| \|\mathbf{x}_i-\mathbf{x}\| + \|\mathbf{x}\|\|\mathbf{y}_i - \mathbf{y}\| \\ \leq M \|\mathbf{x}_i - \mathbf{x}\| + M \|\mathbf{y}_i - \mathbf{y}\| \\ < M \frac{\varepsilon}{2M}+ M \frac{\varepsilon}{2M}< \varepsilon \end{split}\]

for all \(i \geq N\). Therefore, since \(\varepsilon\) is arbitrary, we have \(\mathbf{x}_i'\mathbf{y}_i \to \mathbf{x}'\mathbf{y}\) as \(i \to \infty\).

Proof of “\(\mathbf{x}_i \leq \mathbf{y}_i\) for all \(i\) component-wise \(\Longrightarrow \mathbf{x} \leq \mathbf{y}\)”

Assume that \(\mathbf{x}_i \leq \mathbf{y}_i\) for all \(i\). Toward contradiction, suppose that \(\mathbf{x} \nleq \mathbf{y}\). Then, there is \(k\in\{1, 2, \dots, n\}\) such that \(\mathbf{x}(k) > \mathbf{y}(k)\), where \(\mathbf{x}(k)\) denotes the \(k\)-th component of \(\mathbf{x} \in \mathbb{R}^n\). Let \(\varepsilon = \mathbf{x}(k) - \mathbf{y}(k) > 0\). Since the convergences of \(\{\mathbf{x}_i\}_{i=1}^\infty\) and \(\{\mathbf{y}_i\}_{i=1}^\infty\) imply the convergences of \(\{\mathbf{x}_i(k)\}_{i=1}^\infty\) and \(\{\mathbf{y}_i(k)\}_{i=1}^\infty\) (why?), there is an \(N\in\mathbb{N}\) such that

\[\begin{split} |\mathbf{x}_i(k) - \mathbf{x}(k)| < \frac{\varepsilon}{4} \qquad \text{and} \qquad |\mathbf{y}_i(k) - \mathbf{y}(k)| < \frac{\varepsilon}{4} \\ \Rightarrow \mathbf{x}_i(k) > \mathbf{x}(k) - \frac{\varepsilon}{4} \qquad \text{and} \qquad \mathbf{y}_i(k) < \mathbf{y}(k) + \frac{\varepsilon}{4} \end{split}\]

for all \(i \geq N\). This implies that \(\mathbf{x}_i(k) > \mathbf{y}_i(k)\) for all \(i \geq N\). To see this, observe that

\[ \mathbf{x}_i(k) - \mathbf{y}_i(k) > \mathbf{x}(k) - \frac{\varepsilon}{4} - \Big(\mathbf{y}(k) +\frac{\varepsilon}{4}\Big)> \mathbf{x}(k) - \mathbf{y}(k) -\frac{\varepsilon}{2} = \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}> 0 \]

for all \(i \geq N\). Since \(\mathbf{x}_i(k)\leq \mathbf{y}_i(k)\) for all \(i\) by assumption, we obtain a contraction, which implies that it must be \(\mathbf{x} \leq \mathbf{y}\).

Question C.2

Let \((k, \ell)\) be any point in \(A\), and let \(\{(k_n, \ell_n)\}\) be any sequence converging to \((k, \ell)\) in the sense of convergence in \(\mathbb{R}^2\). We wish to show that

\[ f(k_n, \ell_n) \to f(k, \ell) \]

Since \((k_n, \ell_n) \to (k, \ell)\) in \(\mathbb{R}^2\), we know from the facts on convergence in norm that the individual components converge in \(\mathbb{R}\). That is,

\[ k_n \to k \quad \text{and} \quad \ell_n \to \ell \]

We also know from the facts that, for any \(a\), the function \(g(x) = x^a\) is continuous at \(x\). It follows from the definition of continuity and the convergence in \(\mathbb{R}\) above that \(k_n^\alpha \to k^{\alpha}\) and \(\ell^{\beta}_n \to \ell^\beta\).

Moreover, we know that, for any sequences \(\{y_n\}\) and \(\{z_n\}\), if \(y_n \to y\) and \(z_n \to z\), then \(y_n z_n \to yz\). Hence

\[ k_n^\alpha \ell^{\beta}_n \to k^{\alpha}\ell^\beta \]

That is, \(f(k_n, \ell_n) \to f(k, \ell)\). Hence \(f\) satisfies the definition of continity.

Question C.3

The proof is analogous to the proof of continuity of the Cobb-Douglas production function given above.

Question C.4

To show that \(A\) is closed, we need to show that the limit of any sequence contained in \(A\) is also in \(A\). To this end, let \(\{{\bf x}_n\}\) be an arbitrary sequence in \(A\) coverging to a point \({\bf x} \in \mathbb{R}^2\). Since \({\bf x}_n \in A\) for all \(n\) we have \({\bf x}_n \geq {\bf 0}\) in the sense of the component-vise vector inequality and \({\bf x}_n' {\bf p} \leq m\), where \({\bf p} = (p_1, p_2)\). We need to show that the same is true for \({\bf x}\).

Since \({\bf x}_n \to {\bf x}\), we have \({\bf x}_n' {\bf p} \to {\bf x}' {\bf p}\). Since limits preserve weak inequalities and \({\bf x}_n' {\bf p} \leq m\) for all \(n\), we have \({\bf x}' {\bf p} \leq m\). Hence it remains only to show that \({\bf x} \geq 0\). Again using the fact that weak inequalities are preserved under limits, combined with \({\bf x}_n \geq {\bf 0}\) for all \(n\), gives \({\bf x} \geq {\bf 0}\) as required.

Question C.5

From the definition of Euclidean norm, we have \(\|{\bf x}\|^2 = \sum_{i=1}^{N}x_i^2\) where \(x_i\) are components of \({\bf x}\). Therefore:

\[\begin{split} \| {\bf x} + {\bf y} \|^2 + \| {\bf x} - {\bf y} \|^2 = \sum_{i=1}^{N}(x_i+y_i)^2 + \sum_{i=1}^{N}(x_i-y_i)^2 = \\ = \sum_{i=1}^{N} \Big( 2 x_i^2 + 2 x_i y_i - 2 x_i y_i + 2 y_i^2 \Big) = 2 \sum_{i=1}^{N}x_i^2 + 2 \sum_{i=1}^{N}y_i^2 = 2 \big( \| {\bf x} \|^2 + \| {\bf y} \|^2 \big) \end{split}\]