# π¬ Tutorial problems *iota*#

## \(\iota\).1#

Find the largest domain \(S \subset \mathbb{R}^2\) on which

is concave.

How about strictly concave?

It is useful to review the Hessian based conditions for concavity and the conditions for definiteness of a Hessian of \(\mathbb{R}^2 \to \mathbb{R}\) functions.

Combining the fact on definiteness of the Hessian of two-variate functions from the lecture notes with the Hessian based conditions for convexity of functions, we can formulate the following fact.

Fact: the sufficient conditions for concavity/convexity in 2D

Let \(z = f(x,y)\) be a twice continuously differentiable function defined for all \((x, y) \in R^2\).

Then it holds:

\(f \text{ is convex } \iff f''_{1,1} \ge 0, \; f''_{2,2} \ge 0 , \text{ and } f''_{1,1} f''_{2,2} - (f''_{1,2})^2 \ge 0\)

\(f \text{ is concave } \iff f''_{1,1} \le 0, \; f''_{2,2} \le 0 , \text{ and } f''_{1,1} f''_{2,2} - (f''_{1,2})^2 \ge 0\)

\(f''_{1,1} > 0 \text{ and } f''_{1,1} f''_{2,2} \implies f \text{ is strictly convex}\)

\(f''_{1,1} < 0 \text{ and } f''_{1,1} f''_{2,2} \implies f \text{ is strictly concave}\)

The proof is a simple combination of the known facts named above.

Now we have

\(f^{\prime}_{1}(x, y) = 2x - y -3x^2\)

\(f^{\prime}_{2}(x, y) = -2y - x\)

\(f^{\prime\prime}_{1, 1}(x, y) = 2 - 6x\)

\(f^{\prime\prime}_{2, 2}(x, y) = -2\)

\(f^{\prime\prime}_{1, 2}(x, y) = -1\)

Thus, \(S = \{(x, y) \in \mathbb{R^2}: x \geq \frac{5}{12}\}= [\frac{5}{12}, +\infty) \times \mathbb{R}\).

For the strict concavity we just replace all the inequalities to strict inequalities, and we get \(S = \{(x, y) \in \mathbb{R^2}: x > \frac{5}{12}\}= (\frac{5}{12}, +\infty) \times \mathbb{R}\).

## \(\iota\).2#

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

Because the function is not differentiable everywhere in its domain, using the definition of concavity could be an easier way.

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

and hence

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

## \(\iota\).3#

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

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

The function \(f\) has a unique minimizer at \(x^* = 0\) if and only if \(c \ne 0\).

Hereβs 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.

Hereβs 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

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

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 \(x \in \mathbb{R}\).

## \(\iota\).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

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

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 \({\bf x}\)?

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

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

Since this is an βif and only ifβ 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.

## \(\iota\).5#

Consider the maximization problem

subject to \(c_1 \geq 0\), \(c_2 \geq 0\) 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.

Solve the problem by substitution and using the tangency (relative slope) condition. Discuss, which solution approach is easier.

To answer the first part of the question, review facts of existence of optima.

First, note that \(U(c_1, c_2) = \sqrt c_1 + \beta \sqrt{c_2}\) is continuous as a composition of continuous functions. Then, observe that the admissible set

is closed, all inequalities are weak.

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

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 \(B\).

Now, to solve the problem explicitly, review the argument that the inequality \(p_1 c_1 + p_2 c_2 \leq m\) can be replaced by an equality with no effect on the set of maximizers. Indeed, if \(p_1 c_1 + p_2 c_2 < m\) then we can increase \(c_1\) or \(c_2\) until the equality is satisfied, and because the criterion function is strictly increasing in each argument, the interior point is not a maximizer.

Inverting \(p_1 c_1 + p_2 c_2 = m\) with respect to \(c_2\) gives \(c_2 = (m - p_1 c_1) / p_2\).

First, solve by substitution. The first order condition is

It is convenient to check the second order conditions right away for the one dimensional problem. Simple derivation (left for exercise) yields that the second derivative is negative independent of \(c_1\), and so the function \(\sqrt{c_1} + \beta \sqrt{\tfrac{m}{p_2}-\tfrac{p_1}{p_2}c_1}\) is strictly concave. Therefore, any point that satisfies the first order condition is a unique maximizer.

It is only left to find maximizing \(c_2\) from the constraint.

To solve the problem using the tangency condition, recall that the maximizer is characterized by

where \(f(x_1,x_2)\) is the criterion and \(g(x_1,x_2)=0\) is the constraint, and the equality condition itself.

In our case \(x_1=c_1\), \(x_2=c_2\), and we have

Hence, the optimum is the solution of

Clearly, this system leads to the same equation as before after the substitution \(c_2 = (m - p_1 c_1) / p_2\) is made.