Mathematical methods for economic theory

Martin J. Osborne
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3.1 Exercises on concave and convex functions of a single variable

  1. The functions f and g are both concave functions of a single variable. Neither function is necessarily differentiable.
    1. Is the function h defined by h(x) = f(x) + g(x) necessarily concave, necessarily convex, or not necessarily either?

    2. Is the function h defined by h(x) = −f(x) necessarily concave, necessarily convex, or not necessarily either?

    3. Is the function h(x) = f(x)g(x) necessarily concave, necessarily convex, or not necessarily either?

    [In questions like this, it is probably helpful to first draw some diagrams to get an idea of whether each claim is true or false. If you think a claim is false, try to find an example that is inconsistent with it. For instance, if you think that the first claim in the first part of this question is false---that is, you think that h is not necessarily concave---then try to find two functions whose sum is not concave. If you think a claim is true, try to prove it. To start a proof, write down precise statements of what you know---in this case, the precise conditions that f and g satisfy, given they are concave. Then write down on a separate piece of paper, or at the bottom of the page, a precise statement of the conclusion you want to reach---in this case, the precise condition under which h is concave. Now you need to figure out a way to get from the first set of conditions to the second set of conditions in logical steps.]

    Solution

    1. We have
      h((1 − λ)x + λy)  = f((1 − λ)x + λy) + g((1 − λ)x + λy)
      ≥ (1 − λ)f(x) + λf(y) + (1 − λ)g(x) + λg(y)
           (using the concavity of f and of g)
      = (1 − λ)(f(x) + g(x)) + λ(f(y) + g(y))
      = (1 − λ)h(x) + λh(y).
      Thus h is necessarily concave.
    2. Since f is concave, we have
      f((1 − λ)x + λy) ≥ (1 − λ)f(x) + λf(y) for all x, y, and λ.
      Hence
      −f((1 − λ)x + λy) ≤ (1 − λ)(−f(x)) + λ(−f(y)) for all x, y, and λ,
      so that −f is convex.
    3. The function h is neither necessarily concave nor necessarily convex. If f(x) = x and g(x) = x then both f and g are concave, but h is convex and not concave. Thus h is not necessarily concave. If f(x) = x and g(x) = −x then both f and g are concave, and h is strictly concave, and hence not convex. Thus h is not necessarily convex.

  2. The function f(x) is concave, but not necessarily differentiable. Find the values of the constants a and b for which the function af(x) + b is concave. (Give a complete argument; no credit for an argument that applies only if f is differentiable.)

    Solution

    Let g(x) = af(x) + b. Because f is concave we have
    f((1 − λ)x + λy) ≥ (1 − λ)f(x) + λf(y) for all x, y, and λ ∈ (0,1).
    Now,
    g((1 − λ)x + λy) = af((1 − λ)x + λy) + b
    and
    (1 − λ)g(x) + λg(y) = a[(1 − λ)f(x) + λf(y)] + b.
    Thus
    g((1 − λ)x + λy) ≥ (1 − λ)g(x) + λg(y) for all x, y, and λ ∈ (0,1)
    if and only if
    af((1 − λ)x + λy) ≥ a[(1 − λ)f(x) + λf(y)],
    or if and only if a ≥ 0 (using the concavity of f).

  3. The function g of a single variable is defined by g(x) = f(ax + b), where f is a concave function of a single variable that is not necessarily differentiable, and a and b are constants with a ≠ 0. (These constants may be positive or negative.) Either show that the function g is concave, or show that it is not necessarily concave. [Your argument must apply to the case in which f is is not necessarily differentiable.]

    Solution

    We have
    g((1 − λ)x1 + λx2)  = f(a((1 − λ)x1 + λx2) + b)
    = f((1 − λ)(ax1 + b) + λ(ax2 + b))
    ≥ (1 − λ)f(ax1 + b) + λf(ax2 + b)
           (by the concavity of f)
    = (1 − λ)g(x1) + λg(x2).
    Thus g is concave.

  4. Determine the concavity/convexity of f(x) = −(1/3)x2 + 8x − 3.

    Solution

    The function is twice-differentiable, because it is a polynomial. We have f'(x) = −2x/3 + 8 and f"(x) = −2/3 < 0 for all x, so f is strictly concave.

  5. Let f(x) = Axα, where A > 0 and α are parameters. For what values of α is f (which is twice differentiable) nondecreasing and concave on the interval [0, ∞)?

    Solution

    We have f'(x) = αAxα−1 and f"(x) = α(α − 1)Axα−2. For any value of β we have xβ ≥ 0 for all x ≥ 0, so for f to be nondecreasing and concave we need α ≥ 0 and α(α − 1) ≤ 0, or equivalently 0 ≤ α ≤ 1.

  6. Find numbers a and b such that the graph of the function f(x) = ax3 + bx2 passes through (−1, 1) and has an inflection point at x = 1/2.

    Solution

    For the graph of the function to pass through (−1,1) we need f(−1) = 1, which implies that −a + b = 1. Now, we have f'(x) = 3ax2 + 2bx and f"(x) = 6ax + 2b, so for f to have an inflection point at 1/2 we need f"(1/2) = 0, which yields 3a + 2b = 0. Solving these two equations in a and b yields a = −2/5, b = 3/5.

  7. A competitive firm receives the price p > 0 for each unit of its output, and pays the price w > 0 for each unit of its single input. Its output from using x units of the variable input is f(x) = x1/4. Is this production function concave? Is the firm's profit concave in x?

    Solution

    The function f is twice-differentiable for x > 0. We have f'(x) = (1/4)x−3/4 and f"(x) = −(3/16)x−7/4 < 0 for all x, so f is concave for x > 0. It is continuous, so it is concave for all x ≥ 0. The firm's profit, pf(x) − wx, is thus the sum of two concave functions, and is hence concave.

  8. A firm has the continuous production function f, sells output at the price p, and pays wi per unit for input i, i = 1, ..., L. Consider its profit maximization problem
    maxzpf(z) − w·z subject to z ≥ 0.
    Fix the value of the vector w. Assume that for each value of p the problem has a solution z*(p) with z*(p) > 0. Let
    π(p) = pf(z*(p)) − w·z*(p),
    the maximal profit of the firm when the price is p. Show that π is a convex function of p. [Hint: Let p and p' be arbitrary values of the parameter p, and let p" = (1 − λ)p + λp' (for convenience). We have π(p") = p"f(z*(p")) − w·z*(p"). How are pf(z*(p")) − w·z*(p") and p'f(z*(p")) − w·z*(p") related to π(p) and π(p')?]

    Solution

    We need to show that π(p'') ≤ (1 − λ)π(p) + λπ(p') for all p, p' and all λ ∈ (0,1). (Remember that p'' = (1 − λ)p + λp'.) Now,
    π(p'')  =  p''f(z*(p'')) − w·z*(p'')
    =  (1 − λ)[pf(z*(p'')) − w·z*(p'')] + λ[p'f(z*(p'')) − w·z*(p'')].
    Further, π(p) ≥ pf(z) − w·z for every value of z, by the definition of π(p), which is the maximal profit of the firm when the price is p. Thus, choosing z = z*(p''), we have π(p) ≥ pf(z*(p'')) − w·z*(p''). Similarly, π(p') ≥ p'f(z*(p'')) − w·z*(p''). Hence π(p'') ≤ (1 − λ)π(p) + λπ(p').