Concave functions of two variables
While we will not provide a proof here, the following three definitions are equivalent if the
function f is differentiable.1
Definition 1: Concave Function
The function f is concave on X 
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if for any x0 , x1  X and any   (0,1)
f ( x )  (1   ) f ( x0 )   f ( x1 )
Definition 2: Concave function
The differentiable function f is concave on X if for any x0 , x1  X and any   (0,1)
f ( x1 )  f ( x 0 ) 
f 0 1 0
f 0 1
( x )( x1  x1 ) 
( x )( x2  x20 )
x1
x2
Definition 3: Concave function
A twice continuously differentiable function f is concave if and only if
2 f
2 f
2 f
2 f
2 f
(i)
( x)  0, i  1, 2 and (ii)
( x)
( x) 
( x)
( x)  0
xi xi
x1x1
x2x2
x1x2
x2x1
In the one variable case a function is concave if the derivative of the function is decreasing. We
can use this result and the following proposition to define a class of concave function in higher
dimensions.
Proposition 1: The sum of concave functions is concave
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See Module 4 for a proof of the equivalence of D1 and D2.
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Exercise: Complete the following argument to prove Proposition 1.
Suppose that f1 ( x) and f 2 ( x) are both concave. Then for any x 0 , x1 and convex combination
x ,
f1 ( x )  (1   ) f1 ( x0 )   f1 ( x1 ) and f 2 ( x )  (1   ) f 2 ( x0 )   f 2 ( x1 )
Define g ( x)  f1 ( x)  f 2 ( x) . You need to show that g ( x )  (1   ) g ( x0 )  g ( x1 ) .
Remark: If follows that if u1 ( x1 ) and u2 ( x2 ) are both concave then U ( x)  u1 ( x1 )  u2 ( x2 ) is
concave.
In economics we very often assume that a set is convex. Consider for example a
consumer whose preferences are represented by the utility function U ( x1 , x2 ) . For any x̂ , the
set of points having utility equal to that at x̂ is called a contour set.
C  {x | U ( x)  U ( x0 )}
Economists call this an indifference curve. The set of points that are preferred to x̂ is called an
upper contour set.
CU  {x | U ( x)  U ( xˆ)} .
Consider the figure below showing three contour sets of an increasing function. The upper
contour set is the shaded region.
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Economists typically assume that upper contour sets are strictly convex. The following
result establishes that the is property holds for all convex function
Proposition 2: The upper contour sets of a concave function are convex
Exercise: Complete the following argument to prove Proposition 2.
The upper contour set CU  {x | U ( x)  U ( xˆ )} is convex if for any x 0 , x1 in this set each convex
combination x  lies in the set as well. Thus it must be shown that if U ( x0 )  U ( xˆ ) and
U ( x1 )  U ( xˆ) then U ( x )  U ( xˆ) . Now appeal to Definition 1 for a concave function.
Example: Concave function
Suppose f ( x)  1 ln x1   2 ln x2 . Note that
d
1
ln( x j ) 
. This is a decreasing
dx j
xj
function of x j . Therefore  j ln x j is concave and so, by Proposition 1, f ( x) is concave.
By Proposition 2 it follows that the upper contour sets of f are convex.
Proposition 3: Sufficient condition for the upper contour sets of a function to be convex
Suppose that g () is a strictly increasing function and that h( x)  g ( f ( x) is concave. Then the
upper contour sets of f are convex.
Exercise: Proof of Proposition 3
Complete the following argument to prove Proposition 3
Proof: Appealing to Proposition 2, the upper contour sets of h are convex. That is, for any x̂
the set CU  {x | h( x)  h( xˆ)} is convex. That is, for any x̂ the set CU  {x | g ( f ( x))  g ( f ( xˆ))}
is convex. Then argue that g ( f ( x))  g ( f ( xˆ )) if and only if f ( x)  f ( xˆ) .
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Example: f ( x)  x11 x22 where x 
2

Define h( x)  ln f ( x)  1 ln x1   2 ln x2 . As argued above, this is concave and hence
has convex upper contour sets. Since ln() is strictly increasing it follows from
proposition 3 that the upper contour sets of f are convex.
Just as in the one variable case we now show that if f is concave, the FOC for a
maximum are both necessary and sufficient.
Proposition 4: Necessary and sufficient conditions for a maximum
If f is a differentiable concave function then the following conditions are both necessary and
sufficient for f to take on its maximum value at x 0 .
f 0
f 0
(x ) 
(x )  0 .
x1
x2
FOC
Proof: If f takes on its maximum at x 0 consider changes only in x j , j  1, 2 . This reduces the
problem to a one variable problem so we know that the FOC must hold.
f 0
f 0
(x ) 
(x )  0 .
x1
x2
From Definition 2, for any x1  x0 ,
f ( x1 )  f ( x 0 ) 
f 0 1 0
f 0 1
( x )( x1  x1 ) 
( x )( x2  x20 )  f ( x 0 )
x1
x2
Thus f takes on its maximum at x 0 .
QED
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