[PDF] Concave and Convex Functions - Department of Mathematics





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17 déc. 2009 Définition 2. Soit E un evn et C une partie convexe de E. Soit f : C ? R. f est dite convexe (resp. strictement convexe resp. concave ...



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Concave and Convex Functions - Department of Mathematics

f is both concave and convex i for any a;b2RN and any 2(0;1) f( a+ (1 )b) = f(a) + (1 )f(b) A function fis a ne i there is a 1 Nmatrix Aand a number y 2R such that for all x2C f(x) = Ax+ y fis linear if it is a ne with y = 0 Theorem 2 f: RN!R is a ne i it is both concave and convex Proof 1



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What is the difference between concave and Convexe?

Donc la fonction est concave, toujours située en dessous de ses tangentes, avec une pente de plus en plus faible. et . Comme 1 < x nous avons g’’(x) strictement positive, g’(x) strictement croissante. Donc g est convexe, située au-dessus de toutes ses tangentes, avec une pente de plus en plus forte.

What does it mean when a function is convexed?

Rappel : on considère que toute fonction (définie) dérivable sur un intervalle est continue sur cet intervalle. • Une fonction f, définie, dérivable (donc continue) sur un intervalle I est convexe sur I si sa représentation graphique est entièrement située au-dessus de chacune de ses tangentes.

Is convex a word?

Convex has the word vex in it (because it is vexing that this word is hard to remember), and means “curved or rounded outward.” Neither word is particularly recent; concave has been in English since the 15th century, and convex since the 16th.

Is the second derivative of a function concave or convex?

This is consistent withthe fact that the second derivative of any ane function is the zero matrix.Showing that other functions are concave or convex typically requires work. For = 1, Theorem7can be used to show that many standard functions are concave,strictly concave, and so on.

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Concave and Convex Functions - Department of Mathematics

John Nachbar

Washington University

March 27, 2018

Concave and Convex Functions

1

1 Basic Denitions.

Denition 1.LetCRNbe non-empty and convex and letf:C!R. 1. (a) fisconcavei for anya;b2Cand any2[0;1], f( a+ (1)b)f(a) + (1)f(b); (b)fisstrictly concavei for anya;b2Cand any2(0;1), the above inequality is strict. 2. (a) fisconvexi for anya;b

2Cand any2[0;1],

f(a+ (1)b)f(a) + (1)f(b); (b)fisstrictly convexi for anya;b2Cand any2(0;1), the above inequality is strict. The following equivalence is immediate from the denitions. Theorem 1.LetCRNbe non-empty and convex and letf:C!R.fis convex ifis concave.fis strictly convex ifis strictly concave. fis both concave and convex i for anya;b2RNand any2(0;1),f(a+ (1)b) =f(a) + (1)f(b). A functionfisanei there is a 1Nmatrix

Aand a numbery2Rsuch that for allx2C,f

(x) =Ax+y.fislinearif it is ane withy= 0. Theorem 2.f:RN!Ris ane i it is both concave and convex.

Proof.

1.). For anya;b2Cand any2(0;1),f(a+(1)b) =A(a+(1)b)+y=

(Aa+y)+(1)(Ab+y) =f(a)+(1)f(b), so thatfis both concave and convex.

2.(. Lety=f(0) and letg(x) =f(x)y, so thatg(0) = 0. Sincefis both

concave and convex, so isg.1 . This work is licensed under the Creative Commons Attribution-NonCommercial-

ShareAlike 4.0 License.

1

Claim: for anya2RN, for any

0,g( a) = g(a).

The claim is trivially true for

equal to either 0 or 1. Suppose

2(0;1).

Theng(

a) =g( a+ (1 )0) = g(a) + (1 )g(0) = g(a).

On the other hand, if

>1, then 1=

2(0;1) and henceg(a) =

g((1= a+ (11= )0) = (1= )g( a) + (11= )g(0) = (1= )g( a).

Multiplying through by

gives g(a) =g( a).

Claim:for anya;b2RN,g(a+b) =g(a) +g(b).

g(a+b) =g((1=2)(2a) + (1=2)(2b)) = (1=2)g(2a) + (1=2)g(2b) =g(a) + g(b), where the last equality comes from the previous claim. ConstructAby settingan=g(en), whereen= (0;:::;0;1;0;:::;0) is the coordinatenunit vector. Sincex=P nxnen, induction on the second claim above givesg(x) =g(P nxnen) =P ng(xnen) =P nxng(en) =P nxnan=

Ax. Finally,f(x) =g(x) +y=Ax+y.

ForN= 1, the next result says that a function is concave i, informally, its slope is weakly decreasing. If the function is dierentiable then the implication is that the derivative is weakly decreasing.

Theorem 3.LetCRbe an open interval.

1.f:C!Ris concave i for anya;b;c2C, witha < b < c,

f(b)f(a)baf(c)f(b)cb; and, f(b)f(a)baf(c)f(a)ca: For strict concavity, the inequalities are strict.

2.f:C!Ris convex i for anya;b;c2C, witha < b < c,

f(b)f(a)baf(c)f(b)cb; and, f(b)f(a)baf(c)f(a)ca: For strictly convexity, the inequalities are strict. 2 Proof.Take anya;b;c2C,a < b < c. Sincebaandcb >0, the rst inequality under (1), holds i [f(b)f(a)](cb)[f(c)f(b)](ba); which holds i (collecting terms inf(b)), f(b)(ca)f(a)(cb) +f(c)(ba); which (sinceca >0) holds i f(b)cbca f(a) +baca f(c): Take= (cb)=(ca)2(0;1) and verify that, indeed,b=a+ (1)c. Then the last inequality holds sincefis concave. Conversely, the preceding argument shows that if the rst inequality in (1) holds thenfis concave (take anya < c, any

2(0;1), and letb=a+ (1)c). The proofs of the other claims are similar.

It is also possible to characterize concavity or convexity of functions in terms of the convexity of particular sets. Given the graph of a function, thehypographoff, written hypf, is the set of points that lies on or below the graph off, while the epigraphoff, written epif, is the set of points that lies on or above the graph of f.2Formally, epif=f(x;y)2RN+1:yf(x)g; hypf=f(x;y)2RN+1:yf(x)g:

Theorem 4.LetCRNbe convex and letf:C!R.

1.fis concave ihypfis convex.

2.fis convex iepifis convex.

Proof.Suppose thatfis concave. I will show that hypfis convex. Take any z

1;z22hypfand any2[0;1]. Then there is ana;b2Candy1;y22R, such

thatz1= (a;y1),z2= (b;y2), withf(a)y1,f(b)y2. By concavity off, f(a+(1)b)f(a)+(1)f(b). Hencef(a+(1)b)y1+(1)y2. Thequotesdbs_dbs7.pdfusesText_5
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