Exercise set E#

Please, see the general comment on the tutorial exercises

Question E.1#

Show that the function \(f(x) = - |x|\) from \(\mathbb{R}\) to \(\mathbb{R}\) is concave.

Question E.2#

Let \({\bf A}\) be the \(1 \times 1\) matrix \((a)\). Give a necessary and sufficient condition on \(a\) (that is, an “if and only if” condition on \(a\)) under which \({\bf A}\) is nonsingular.

Question E.3#

Consider the function \(f\) from \(\mathbb{R}\) to \(\mathbb{R}\) defined by

\[ % f(x) = (c x)^2 + z % \]

Give a necessary and sufficient (if and only if) condition on \(c\) under which \(f\) has a unique minimizer.

Question E.4#

Let \({\bf C}\) be an \(N \times K\) matrix, let \(z \in \mathbb{R}\) and consider the function \(f\) from \(\mathbb{R}^K\) to \(\mathbb{R}\) defined by

\[ % f({\bf x}) = {\bf x}' {\bf C}' {\bf C} {\bf x} + z % \]

Show that \(f\) has a unique minimizer on \(\mathbb{R}^K\) if and only if \({\bf C}\) has linearly independent columns.

Tip

Obviously, you should draw intuition from the preceding question. Also, what does linear independence of the columns of \({\bf C}\) say about the vector \({\bf C} {\bf x}\) for different choices of \({\bf x}\)?

Question E.5#

Consider the maximization problem

\[ % \max_{c_1, c_2} ( \sqrt c_1 + \beta \sqrt{c_2}) % \]

subject to \(c_1, c_2 \geq 0\) and \(p_1 c_1 + p_2 c_2 \leq m\). Here \(p_1, p_2\) and \(m\) are nonnegative constants, and \(\beta \in (0, 1)\).

Show that this problem has a solution if and only if \(p_1\) and \(p_2\) are both strictly positive.

Question E.6#

Let \(A\) be a nonempty bounded set and let \(B := \{ b \in \mathbb{R} \colon b = 2a \text{ for some } a \in A\}\). Obtain \(\sup B\) in terms of \(\sup A\). Justify your answer.

Solutions

Question E.1

Pick any \(x, y \in \mathbb{R}\) and any \(\lambda \in [0, 1]\). By the triangle inequality, we have

\[ % |\lambda x + (1 - \lambda) y| \leq |\lambda x| + |(1 - \lambda) y| % \]

and hence

\[ % - |\lambda x + (1 - \lambda) y| \geq - |\lambda x| - |(1 - \lambda) y| = - \lambda |x| - (1 - \lambda) |y| % \]

That is, \(f(\lambda x + (1 - \lambda) y) \geq \lambda f(x) + (1 - \lambda)f(y)\). Hence \(f\) is concave as claimed.

Question E.2

The condition for nonsingularity of \((a)\) is \(a \ne 0\). There are many ways we could show this. One is that \({\bf A}\) is nonsingular when \({\bf A} {\bf x} = {\bf 0} \implies {\bf x} = {\bf 0}\). Here this translates to \(ax = 0 \implies x = 0\). The question then becomes, for what \(a\) is this implication true? It is true exactly when \(a \ne 0\), for if \(a \ne 0\) and \(ax=0\), the only possibility is that \(x=0\).

Question E.3

The function \(f\) has a unique minimizer at \(x^* = 0\) if and only if \(c \ne 0\). Here’s one proof: If \(c \ne 0\) then the function is strictly convex. Moreover, it is stationary at \(x^* = 0\). Hence, by our facts on minimization under convexity, \(x^*\) is the unique minimizer. The condition is necessary and sufficient because if \(c = 0\), then \(f\) is a constant function, which clearly does not have a unique minimizer.

Here’s a second (more direct) proof that the correct condition is \(c \ne 0\). Suppose first that \(c \ne0\) and pick any \(x \in \mathbb{R}\). We have

\[ % f(x) = (cx)^2 + z \geq z = f(0) % \]

This tells us that \(x^* = 0\) is a minimizer. Moreover,

\[ % f(x) = (cx)^2 + z > z = f(0) \quad \text{whenever} \quad x \ne x^* % \]

Hence \(x^* = 0\) is the unique minimizer.

Suppose next that \(x^* = 0\) is the unique minimizer. Then it must be that \(c \ne0\), for if \(c=0\) then \(f(x) = f(x^*)\) for every \(x \in \mathbb{R}\).

Question E.4

Suppose first that \({\bf C}\) has linearly independent columns. We claim that \({\bf x} = {\bf 0}\) is the unique minimizer of \(f\) on \(\mathbb{R}^K\). To see this observe that if \({\bf x} = {\bf 0}\) then \(f({\bf x}) = z\). On the other hand, if \({\bf x} \ne {\bf 0}\), then, by linear independence, \({\bf C}{\bf x}\) is not the origin, and hence \(\| {\bf C}{\bf x} \| > 0\). Therefore

\[ % f({\bf x}) = {\bf x}' {\bf C}' {\bf C} {\bf x} + z = ({\bf C} {\bf x} )' {\bf C} {\bf x} + z = \| {\bf C} {\bf x} \|^2 + z > z % \]

Thus \({\bf x} = {\bf 0}\) is the unique minimizer of \(f\) on \(\mathbb{R}^K\) as claimed.

Since this is an “if and only if” proof we also need to show that when \(f\) has a unique minimizer on \(\mathbb{R}^K\), it must be that \({\bf C}\) has linearly independent columns. Suppose to the contrary that the columns of \({\bf C}\) are not linearly independent. We will show that multiple minimizers exist.

Since \(f({\bf x}) = \| {\bf C} {\bf x} \|^2 + z\) it is clear that \(f({\bf x}) \geq z\), and hence \({\bf x} = {\bf 0}\) is one minimizer. (At this point, \(f\) evaluates to \(z\).) Since the columns of \({\bf C}\) are not linearly independent, there exists a nonzero vector \({\bf y}\) such that \({\bf C} {\bf y} = {\bf 0}\). At this vector we clearly have \(f({\bf y}) = z\). Hence \({\bf y}\) is another minimizer.

Question E.5

First, show that \(U(c_1, c_2) = \sqrt c_1 + \beta \sqrt{c_2}\) is continuous and that

\[ B := \{ (c_1, c_2) \colon c_i \geq 0 \text{ and } p_1 c_1 + p_2 c_2 \leq m\} \]

is closed. Hence, by the Weierstrass extreme value theorem, a maximizer will exist whenever \(B\) is bounded. If \(p_1\) and \(p_2\) are strictly positive then \(B\) is bounded. This is intuitive but we can also show it formally by observing that \((c_1, c_2) \in B\) implies \(c_i \leq m / p_i\) for \(i =1,2\). Hence

\[ % {\bf c} := (c_1, c_2) \in B \implies \| {\bf c} \| \leq M := \sqrt{ \left(\frac{m}{p_1}\right)^2 + \left(\frac{m}{p_2}\right)^2 } % \]

We also need to show that if one price is zero then no maximizer exists. Suppose to the contrary that \(p_1 = 0\). Intuitively, no maximizer exists because we can always consumer more of good one, thereby increasing our utility. To formalize this we can suppose that a maximizer exists and derive a contradiction. To this end, suppose that \({\bf c}^* = (c_1^*, c_2^*)\) is a maximizer of \(U\) over \(B\). Since \(p_1 = 0\), the fact that \((c_1^*, c_2^*) \in B\) implies \({\bf c}^{**} := (c_1^* + 1, c_2^*) \in B\). Since \(U\) is strictly increasing in its first argument, we also have \(U({\bf c}^{**}) > U({\bf c}^*)\). This contradicts the statement that \({\bf c}^*\) is a maximizer of \(U\) over \(B\).

Question E.6

Let \(A\) and \(B\) be as stated in the question. We claim that \(\sup B = \bar b\) where \(\bar b := 2 \sup A\). According to the definition of the supremum, to prove this we need to show that

  1. \(b \leq \bar b\) for all \(b \in B\)

  2. \(\bar b \leq u\) for all \(u \in U(B)\)

Regarding part 1, pick any \(b \in B\). By definition, \(b = 2a\) for some \(a \in A\). We know that \(a \leq \sup A\) and hence \(2 a \leq 2 \sup A\). Therefore \(b = 2a \leq \bar b\), as was to be shown.

Regarding part 2, take any \(u \in U(B)\). For any \(a \in A\) we have \(2a \in B\) and hence \(2a \leq u\). Therefore \(a \leq u/2\) for all \(a \in A\), and hence \(u/2\) is an upper bound of \(A\). Therefore \(\sup A \leq u/2\). Rearranging gives \(\bar b \leq u\), as claimed.