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The one dimensional results Pascal Gourdel1 Université Paris 11

October 18, 2008

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

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Outline

1

Existence results Weierstrass’s Theorem coercivity

2

First order condition Necessary condition Alternative formulation in terms of multipliers qualification

3

Convex and concave functions Definitions

Pascal Gourdel (Université Paris 1)

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Weierstrass’s Theorem The main starting point of existence results is Weierstrass’s Theorem and its corollaries.

Theorem 1 (Weierstrass) If the domain is compact, and if f is continuous then f is bounded and the extremum are reached.

The proof is based on the notion of maximizing sequence (respectively minimizing). When the domain is not bounded, numerous problems can be solved using the coercivity condition.

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Weierstrass’s Theorem The main starting point of existence results is Weierstrass’s Theorem and its corollaries.

Theorem 1 (Weierstrass) If the domain is compact, and if f is continuous then f is bounded and the extremum are reached.

The proof is based on the notion of maximizing sequence (respectively minimizing). When the domain is not bounded, numerous problems can be solved using the coercivity condition.

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coercivity Let f : R → R be a continuous function, and let us consider two optimisation problems.  (P1 )

min f (x) x ∈ [a, b]

 (P2 )

min f (x) x ∈R

We can apply Weierstrass’s Theorem in order to prove that Sol(P1 ) is nonempty but what about Sol(P2 ) ?

Let us consider by example f (x) = x 8 − x 7 + x 6 + x 3 (1 + sin x) + 4x It is obvious that maxa∈R f (x) has no solutions. Pascal Gourdel (Université Paris 1)

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coercivity Let f : R → R be a continuous function, and let us consider two optimisation problems.  (P1 )

min f (x) x ∈ [a, b]

 (P2 )

min f (x) x ∈R

We can apply Weierstrass’s Theorem in order to prove that Sol(P1 ) is nonempty but what about Sol(P2 ) ?

Let us consider by example f (x) = x 8 − x 7 + x 6 + x 3 (1 + sin x) + 4x It is obvious that maxa∈R f (x) has no solutions. Pascal Gourdel (Université Paris 1)

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coercivity Definition 2 The function f : R → R is said coercive if lim f (x) = lim f (x) = +∞.

x→−∞

x→+∞

The previous example f (x) = x 8 − x 7 + x 6 + x 3 (1 + sin x) + 4x, corresponds to a coercive function.

Exercise 1 Let us assume that f : R → R is coercive and continuous, show that f has a lower bound and that there exists at least a minimum.

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coercivity Definition 2 The function f : R → R is said coercive if lim f (x) = lim f (x) = +∞.

x→−∞

x→+∞

The previous example f (x) = x 8 − x 7 + x 6 + x 3 (1 + sin x) + 4x, corresponds to a coercive function.

Exercise 1 Let us assume that f : R → R is coercive and continuous, show that f has a lower bound and that there exists at least a minimum.

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By extension, we can adapt this definition for example for functions defined on [0, +∞[ by requiring that lim f (x) = +∞.

x→+∞

We can also use a change of variable in order to study a problem defined on ]1, +∞[.  (P1 )

min f (x) x ∈ ]1, +∞[

 (P2 )

min g(y ) = f (1 + exp(y )) y ∈R

x ∈ Sol(P1 ) if and only if ln(x − 1) ∈ Sol(P2 ) y ∈ Sol(P2 ) if and only if ey + 1 ∈ Sol(P1 )

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By extension, we can adapt this definition for example for functions defined on [0, +∞[ by requiring that lim f (x) = +∞.

x→+∞

We can also use a change of variable in order to study a problem defined on ]1, +∞[.  (P1 )

min f (x) x ∈ ]1, +∞[

 (P2 )

min g(y ) = f (1 + exp(y )) y ∈R

x ∈ Sol(P1 ) if and only if ln(x − 1) ∈ Sol(P2 ) y ∈ Sol(P2 ) if and only if ey + 1 ∈ Sol(P1 )

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

6 / 26

By extension, we can adapt this definition for example for functions defined on [0, +∞[ by requiring that lim f (x) = +∞.

x→+∞

We can also use a change of variable in order to study a problem defined on ]1, +∞[.  (P1 )

min f (x) x ∈ ]1, +∞[

 (P2 )

min g(y ) = f (1 + exp(y )) y ∈R

x ∈ Sol(P1 ) if and only if ln(x − 1) ∈ Sol(P2 ) y ∈ Sol(P2 ) if and only if ey + 1 ∈ Sol(P1 )

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

6 / 26

By extension, we can adapt this definition for example for functions defined on [0, +∞[ by requiring that lim f (x) = +∞.

x→+∞

We can also use a change of variable in order to study a problem defined on ]1, +∞[.  (P1 )

min f (x) x ∈ ]1, +∞[

 (P2 )

min g(y ) = f (1 + exp(y )) y ∈R

x ∈ Sol(P1 ) if and only if ln(x − 1) ∈ Sol(P2 ) y ∈ Sol(P2 ) if and only if ey + 1 ∈ Sol(P1 )

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

6 / 26

Outline

1

Existence results Weierstrass’s Theorem coercivity

2

First order condition Necessary condition Alternative formulation in terms of multipliers qualification

3

Convex and concave functions Definitions

Pascal Gourdel (Université Paris 1)

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Necessary condition Proposition 1 Let −∞ ≤ a < b ≤ ∞, we let I = [a, b] ∩ R and we consider f : I → R differentiable on I, if we assume that the problem maxx∈I f (x) has a local solution x ∈ R, then 1

if x ∈ ]a, b[, then f 0 (x) = 0,

2

if x = a, then f 0 (x) ≤ 0,

3

if x = b, then f 0 (x) ≥ 0.

Exercise 2 1) Prove Proposition 1. 2) Let us fix some real number a ∈ R. We assume that f (a) exists. Show that a is a solution (even a global solution) of maxx∈[a,a] f (x), can we deduce a condition on f 0 (a) ?

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Necessary condition Proposition 1 Let −∞ ≤ a < b ≤ ∞, we let I = [a, b] ∩ R and we consider f : I → R differentiable on I, if we assume that the problem maxx∈I f (x) has a local solution x ∈ R, then 1

if x ∈ ]a, b[, then f 0 (x) = 0,

2

if x = a, then f 0 (x) ≤ 0,

3

if x = b, then f 0 (x) ≥ 0.

Exercise 2 1) Prove Proposition 1. 2) Let us fix some real number a ∈ R. We assume that f (a) exists. Show that a is a solution (even a global solution) of maxx∈[a,a] f (x), can we deduce a condition on f 0 (a) ?

Pascal Gourdel (Université Paris 1)

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October 18, 2008

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Necessary condition Proposition 1 Let −∞ ≤ a < b ≤ ∞, we let I = [a, b] ∩ R and we consider f : I → R differentiable on I, if we assume that the problem maxx∈I f (x) has a local solution x ∈ R, then 1

if x ∈ ]a, b[, then f 0 (x) = 0,

2

if x = a, then f 0 (x) ≤ 0,

3

if x = b, then f 0 (x) ≥ 0.

Exercise 2 1) Prove Proposition 1. 2) Let us fix some real number a ∈ R. We assume that f (a) exists. Show that a is a solution (even a global solution) of maxx∈[a,a] f (x), can we deduce a condition on f 0 (a) ?

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

8 / 26

Necessary condition Proposition 1 Let −∞ ≤ a < b ≤ ∞, we let I = [a, b] ∩ R and we consider f : I → R differentiable on I, if we assume that the problem maxx∈I f (x) has a local solution x ∈ R, then 1

if x ∈ ]a, b[, then f 0 (x) = 0,

2

if x = a, then f 0 (x) ≤ 0,

3

if x = b, then f 0 (x) ≥ 0.

Exercise 2 1) Prove Proposition 1. 2) Let us fix some real number a ∈ R. We assume that f (a) exists. Show that a is a solution (even a global solution) of maxx∈[a,a] f (x), can we deduce a condition on f 0 (a) ?

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

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Alternative formulation in terms of multipliers

Let us focus on the previous problem when the domain is bounded (a and b finite). The problem maxx∈[a,b] f (x) can be formulated as    max f (x)  max f (x) x −b ≤0 g (x) ≤ 0 i.e (P1 ) (P)  1  g2 (x) ≤ 0 a−x ≤0 with the notations g1 (x) = x − b g2 (x) = a − x

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We can state Proposition 1 under the following version.

Proposition 2 Let −∞ < a < b < ∞, we let I = [a, b] and we consider f : I → R differentiable on I, we assume that the problem (P1 ) has a local solution x, then there exist multipliers λ1 ≥ 0 and λ2 ≥ 0 such that f 0 (x) = λ1 g10 (x) + λ2 g20 (x) and λ1 g1 (x) = λ2 g2 (x) = 0.

Exercise 3 Deduce Proposition 2 from Proposition 1.

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Be careful, if we consider the optimization problem,  max 3x (P) g1 (x) = x 2 ≤ 0

(1)

It is easy to prove that x = 0 is the unique solution of (P) but nevertherless the system  0  f (x) = λ1 g10 (x) λ ≥0  1 λ1 g1 (x) = 0 has no solution.

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Definition 3 Let us consider the following domain, the set O is open and gi are continuous on O.  x ∈O    g1 (x) ≤ 0 ...    gp (x) ≤ 0 W say that the Slater condition is satisfied for this list of constraints if there exists a feasible point xˆ such that xˆ ∈ O, and for each i = 1, . . . , p,   either gi is an affine function and gi (xˆ ) ≤ 0 or  gi is a convex function and gi (xˆ ) < 0

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Exercise 4 Let f : [a, b] → R, let us consider the optimization problems   max f (x) ∈ R x −b ≤0 (P1 )  a−x ≤0

  max f (x) (P2 ) (x − b)3 ≤ 0  a−x ≤0

1) Compare the two problems. 2) Show that the Slater condition is satisfied for (P1 ). 3) Show that the Slater condition is NOT satisfied for (P2 ).

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Exercise 4 Let f : [a, b] → R, let us consider the optimization problems   max f (x) ∈ R x −b ≤0 (P1 )  a−x ≤0

  max f (x) (P2 ) (x − b)3 ≤ 0  a−x ≤0

1) Compare the two problems. 2) Show that the Slater condition is satisfied for (P1 ). 3) Show that the Slater condition is NOT satisfied for (P2 ).

Pascal Gourdel (Université Paris 1)

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Exercise 4 Let f : [a, b] → R, let us consider the optimization problems   max f (x) ∈ R x −b ≤0 (P1 )  a−x ≤0

  max f (x) (P2 ) (x − b)3 ≤ 0  a−x ≤0

1) Compare the two problems. 2) Show that the Slater condition is satisfied for (P1 ). 3) Show that the Slater condition is NOT satisfied for (P2 ).

Pascal Gourdel (Université Paris 1)

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Exercise 4 Let f : [a, b] → R, let us consider the optimization problems   max f (x) ∈ R x −b ≤0 (P1 )  a−x ≤0

  max f (x) (P2 ) (x − b)3 ≤ 0  a−x ≤0

1) Compare the two problems. 2) Show that the Slater condition is satisfied for (P1 ). 3) Show that the Slater condition is NOT satisfied for (P2 ).

Pascal Gourdel (Université Paris 1)

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Exercise 5 Let us consider again the problem  max 3x (P) g1 (x) = x 2 ≤ 0 Show that this system does not satisfy the Slater condition.

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Proposition 3 Let x be a local solution of the optimization problem:  max f (x)      x ∈O g1 (x) ≤ 0 (P)   ...    gp (x) ≤ 0 Let us assume that the Slater condition is satisfied and that each function is differerentiable on O, then, there exists (λi )pi=1 ∈ Rp+ such that  0 f (x) = λ1 g10 (x) + . . . + λp gp0 (x) λi gi (x) = 0 for all i = 1, . . . , p

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Outline

1

Existence results Weierstrass’s Theorem coercivity

2

First order condition Necessary condition Alternative formulation in terms of multipliers qualification

3

Convex and concave functions Definitions

Pascal Gourdel (Université Paris 1)

The one dimensional results

October 18, 2008

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In this part, f is a real-valued function defined on U convex subset (interval) of R.

Definition 4 The function f is convex (resp. concave) if for all (x, y ) ∈ U × U and for all λ ∈ [0, 1], f (λx + (1 − λ)y ) ≤ λf (x) + (1 − λ)f (y ). (respectively ) f (λx + (1 − λ)y ) ≥ λf (x) + (1 − λ)f (y ). The function f is strictly convex if for all (x, y ) ∈ U × U such that x 6= y and for all λ ∈ ]0, 1[, f (λx + (1 − λ)y ) < λf (x) + (1 − λ)f (y ). A function f is convex if and only if −f is concave. Consequently, the results obtained for convex functions can be translated in terms of concave functions. These notions can be defined locally.

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In this part, f is a real-valued function defined on U convex subset (interval) of R.

Definition 4 The function f is convex (resp. concave) if for all (x, y ) ∈ U × U and for all λ ∈ [0, 1], f (λx + (1 − λ)y ) ≤ λf (x) + (1 − λ)f (y ). (respectively ) f (λx + (1 − λ)y ) ≥ λf (x) + (1 − λ)f (y ). The function f is strictly convex if for all (x, y ) ∈ U × U such that x 6= y and for all λ ∈ ]0, 1[, f (λx + (1 − λ)y ) < λf (x) + (1 − λ)f (y ). A function f is convex if and only if −f is concave. Consequently, the results obtained for convex functions can be translated in terms of concave functions. These notions can be defined locally.

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In this part, f is a real-valued function defined on U convex subset (interval) of R.

Definition 4 The function f is convex (resp. concave) if for all (x, y ) ∈ U × U and for all λ ∈ [0, 1], f (λx + (1 − λ)y ) ≤ λf (x) + (1 − λ)f (y ). (respectively ) f (λx + (1 − λ)y ) ≥ λf (x) + (1 − λ)f (y ). The function f is strictly convex if for all (x, y ) ∈ U × U such that x 6= y and for all λ ∈ ]0, 1[, f (λx + (1 − λ)y ) < λf (x) + (1 − λ)f (y ). A function f is convex if and only if −f is concave. Consequently, the results obtained for convex functions can be translated in terms of concave functions. These notions can be defined locally.

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Definition 5 The epigraph and the hypograph (denoted by epi(f ) and hypo(f )) of a function f are the sets (see figure 1) defined by epi(f ) = {(x, t) ∈ U × R | t ≥ f (x)}. hypo(f ) = {(x, t) ∈ U × R | t ≤ f (x)}.

E(f )

Cf

H(f )

Figure: graph, epigraph and hypograph

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Theorem 6 The three following properties are equivalent: (i) f is convex (resp. concave); (ii)Pfor all k ≥ 2, (xi ) ∈ U k P and λ ∈ S k , k f ( i=1 λi xi ) ≤ (resp. ≥) ki=1 λi f (xi ); (iii) The epigraph (resp. the hypograph) of f is a convex subset of R2 .

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Examples: 1

Let us recall that an affine function is a function R → R, such that f (x) = αx + β. Any affine function is both convex and concave, but not strictly. It can be easily proved that if f is convex and concave on U, then it is the restriction on U of an affine function.

2

|.| is a convex function.

3

If C is a nonempty convex convex subset of R, the distance to C defined by dC (x) = inf{|x − c| such that c ∈ C} is convex.

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Examples: 1

Let us recall that an affine function is a function R → R, such that f (x) = αx + β. Any affine function is both convex and concave, but not strictly. It can be easily proved that if f is convex and concave on U, then it is the restriction on U of an affine function.

2

|.| is a convex function.

3

If C is a nonempty convex convex subset of R, the distance to C defined by dC (x) = inf{|x − c| such that c ∈ C} is convex.

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Examples: 1

Let us recall that an affine function is a function R → R, such that f (x) = αx + β. Any affine function is both convex and concave, but not strictly. It can be easily proved that if f is convex and concave on U, then it is the restriction on U of an affine function.

2

|.| is a convex function.

3

If C is a nonempty convex convex subset of R, the distance to C defined by dC (x) = inf{|x − c| such that c ∈ C} is convex.

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Proposition 4 if f is convex (resp. concave) and if x < x 0 < x 00 are elements of I, then (figure 2), f (x) − f (x 00 ) f (x 00 ) − f (x 0 ) f (x) − f (x 0 ) ≤ ≤ x − x0 x − x 00 x 00 − x 0 . b

b

b

O

x

x′

x′′

Figure: Monotonicity of the slope of a convex function Pascal Gourdel (Université Paris 1)

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Corollary 7 Let f be a convex function on some interval U, 1) if x1 < x2 < x3 < x4 < x5 are elements of U, then f (x2 ) − f (x3 ) f (x3 ) − f (x4 ) f (x4 ) − f (x5 ) f (x1 ) − f (x2 ) ≤ ≤ ≤ x1 − x2 x2 − x3 x3 − x4 x4 − x5 2) The right and left derivatives exist on the interior of U and moreover if x1 < x2 are interior points, then fr0 (x1 ) ≤

f (x2 ) − f (x1 ) ≤ f`0 (x2 ). x2 − x1

3) The right and left derivatives are non decreasing on the interior of U, and more over if x is an interior point, f`0 (x1 ) ≤ fr0 (x1 ).

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Corollary 7 Let f be a convex function on some interval U, 1) if x1 < x2 < x3 < x4 < x5 are elements of U, then f (x2 ) − f (x3 ) f (x3 ) − f (x4 ) f (x4 ) − f (x5 ) f (x1 ) − f (x2 ) ≤ ≤ ≤ x1 − x2 x2 − x3 x3 − x4 x4 − x5 2) The right and left derivatives exist on the interior of U and moreover if x1 < x2 are interior points, then fr0 (x1 ) ≤

f (x2 ) − f (x1 ) ≤ f`0 (x2 ). x2 − x1

3) The right and left derivatives are non decreasing on the interior of U, and more over if x is an interior point, f`0 (x1 ) ≤ fr0 (x1 ).

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Corollary 7 Let f be a convex function on some interval U, 1) if x1 < x2 < x3 < x4 < x5 are elements of U, then f (x2 ) − f (x3 ) f (x3 ) − f (x4 ) f (x4 ) − f (x5 ) f (x1 ) − f (x2 ) ≤ ≤ ≤ x1 − x2 x2 − x3 x3 − x4 x4 − x5 2) The right and left derivatives exist on the interior of U and moreover if x1 < x2 are interior points, then fr0 (x1 ) ≤

f (x2 ) − f (x1 ) ≤ f`0 (x2 ). x2 − x1

3) The right and left derivatives are non decreasing on the interior of U, and more over if x is an interior point, f`0 (x1 ) ≤ fr0 (x1 ).

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Corollary 8 Let f be a convex and differentiable function on an open interval containing a and b. If x1 < x2 are elements from ]a, b[, then f 0 (a) ≤

f (x2 ) − f (x1 ) ≤ f 0 (b). x2 − x1

In particular f is lipschitz on [a, b]

Corollary 9 Let f be a differentiable function on the interval U, f is convex on U if and only if f 0 is non decrasing on U.

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Corollary 8 Let f be a convex and differentiable function on an open interval containing a and b. If x1 < x2 are elements from ]a, b[, then f 0 (a) ≤

f (x2 ) − f (x1 ) ≤ f 0 (b). x2 − x1

In particular f is lipschitz on [a, b]

Corollary 9 Let f be a differentiable function on the interval U, f is convex on U if and only if f 0 is non decrasing on U.

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Corollary 10 Let f be a twice differentiable function on the interval U, f is convex if and only if for all x ∈ U, f 00 (x) ≥ 0. Be careful, f may be discontinuous on the boundary of the domain. For example,  1 if x = 0 f (x) = 0 if x > 0 is convex on R+ .

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Corollary 10 Let f be a twice differentiable function on the interval U, f is convex if and only if for all x ∈ U, f 00 (x) ≥ 0. Be careful, f may be discontinuous on the boundary of the domain. For example,  1 if x = 0 f (x) = 0 if x > 0 is convex on R+ .

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Exercise 6 1) Show that the function f defined by f (x) = x 4 is strictly convex on R but nevertheless the second derivative is not always positive on R since f 00 (0) = 0. 2) Let f be a twice differentiable function on the interval U. Show that if for all x ∈ U, f 00 (x) > 0, then f is strictly convex on U. 3) Let f be a C 2 function on U a neighborhood of x, and assume that f 00 (x) > 0. Show that there exists V a (possibly smaller) neighborhood of x such that f is strictly convex on V .

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Exercise 6 1) Show that the function f defined by f (x) = x 4 is strictly convex on R but nevertheless the second derivative is not always positive on R since f 00 (0) = 0. 2) Let f be a twice differentiable function on the interval U. Show that if for all x ∈ U, f 00 (x) > 0, then f is strictly convex on U. 3) Let f be a C 2 function on U a neighborhood of x, and assume that f 00 (x) > 0. Show that there exists V a (possibly smaller) neighborhood of x such that f is strictly convex on V .

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Exercise 6 1) Show that the function f defined by f (x) = x 4 is strictly convex on R but nevertheless the second derivative is not always positive on R since f 00 (0) = 0. 2) Let f be a twice differentiable function on the interval U. Show that if for all x ∈ U, f 00 (x) > 0, then f is strictly convex on U. 3) Let f be a C 2 function on U a neighborhood of x, and assume that f 00 (x) > 0. Show that there exists V a (possibly smaller) neighborhood of x such that f is strictly convex on V .

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In the following picture, f 00 (0) = 0.

0,16

0,12

0,08

0,04

-0,12

-0,08

-0,04

0

0,04

0,08

0,12

Figure: f (x) = x 2 (1 + sin(1/x)), g(x) = 2x 2

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