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I'm studying the Solow growth model from the Acemoglu's book.

Consider the following standard assumptions:

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The fact that $F()$ exhibits constant returns to scale means that F() is linear homogenous. Why the book says that $F()$ is concave? Why it is not strictly concave?

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    $\begingroup$ I'm not certain that I've understood your question fully. If the book assumes that $F$ is concave, this is a weaker condition than assuming it is strictly concave. Therefore, the analysis conducted in the book, which holds for concave functions, will also hold for strictly concave functions. $\endgroup$
    – Amit
    Commented Apr 1, 2023 at 0:49
  • $\begingroup$ Thanks for your answer. The book says : "Prove that Assumption 1 implies that F(A,K, L) is concave in K and L but not strictly so." $\endgroup$
    – John M.
    Commented Apr 1, 2023 at 13:35
  • $\begingroup$ @JohnM. Are you asking if there any concave functions that are not strictly concave, or what exactly is your question here? $\endgroup$
    – Giskard
    Commented Apr 1, 2023 at 18:45
  • $\begingroup$ @JohnM. Let me ask you a follow-up question. Can you think of any production function $F$ that satisfy assumption $1$? $\endgroup$
    – Amit
    Commented Apr 1, 2023 at 21:52
  • $\begingroup$ Thanks for your replies. Assumption 1, in my understanding implies that $F$ is strictly concave. Strict concavity implies concavity. Why the book says that $F$ is concave only (and not strictly concave as well)? $\endgroup$
    – John M.
    Commented Apr 2, 2023 at 8:38

1 Answer 1

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If you consider the production function $f(K, L) = F(K, L, A = 1)= K^{\frac{1}{2}}L^{\frac{1}{2}}$, this function meets all the conditions specified in assumption 1 within the interior, but not on the axes due to its lack of differentiability on the axes. Typically, this detail is not explicitly mentioned. Check that $f(K, L)= K^{\frac{1}{2}}L^{\frac{1}{2}}$ is concave and not strictly concave. To show that this is not strictly concave, consider $(K_1, L_1)=(0,0)$, $(K_2, L_2)=(2,2)$ and $\lambda = \frac{1}{2}$,

$f(\lambda (K_1, L_1) +(1-\lambda) (K_2, L_2)) = f(1,1) =1 =\frac{1}{2}(0) + \frac{1}{2}(2) = \lambda f(K_1, L_1) + (1-\lambda) f(K_2, L_2) $

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  • $\begingroup$ I understand the definition of concave function as you wrote above, and makes sense. But I also know that if the second (partial) derivatives are strictly negative, then $F$ is strictly concave. Why this is not the case? $\endgroup$
    – John M.
    Commented Apr 2, 2023 at 10:46
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    $\begingroup$ @JohnM. The example example is a counter-example to your claim "if the second (partial) derivatives are strictly negative, then $F$ is strictly concave". Your claim is wrong which is why you have a counter-example. $\endgroup$
    – user43302
    Commented Apr 2, 2023 at 11:02
  • $\begingroup$ Thanks. I see. Ok, my statement was simply wrong $\endgroup$
    – John M.
    Commented Apr 2, 2023 at 11:06

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