HW1 Solution - IE 8534

Hamidreza Badri

February 16, 2014

Problem 1.1

Image of convex set under perspective mapping remains convex:

P (x, t) = x/t, x ∈ Rn, t > 0

Let S be a convex set, therefore we have to show that if (x1, t1), (x2, t2) ∈ S, then λP (x1, t1) +(1− λ)P (x2, t2) ∈ P (S) for all λ ∈ [0, 1].

λP (x1, t1) + (1− λ)P (x2, t2)

= λx1t1

+ (1− λ)x2t2

=λt2x1 + (1− λ)t1x2

t1t2

=

λt2λt2+(1−λ)t1x1 + (1−λ)t1

λt2+(1−λ)t1x2t1t2

λt2+(1−λ)t1

By defining λ′

= λt2λt2+(1−λ)t1 :

λP (x1, t1) + (1− λ)P (x2, t2) =λ

′x1 + (1− λ′

)x2λ′t1 + (1− λ′)t2

Since S is a convex set, therefore λ′x1 + (1− λ′

)x2 ∈ S and 0 ≤ λ′= λt2

λt2+(1−λ)t1 ≤ 1.

Inverse image of convex set under perspective mapping remains convex:

The inverse function can be defined as follow

P−1(S) = {(x, t)|x/t ∈ S, t > 0}

Therefore for (x1, t1), (x2, t2) ∈ P−1(S), we must show λ(x1, t1) + (1− λ)(x2, t2) ∈ P−1(S). Wehave

1

λx1 + (1− λ)x2λt1 + (1− λ)t2

=λt1

x1t1

+ (1− λ)t2x2t2

λt1 + (1− λ)t2

=λt1

λt1 + (1− λ)t2

x1t1

+(1− λ)t2

λt1 + (1− λ)t2

x2t2

By defining λ′

= λt1λt1+(1−λ)t2 ∈ [0, 1], it shows that λ

′(x1, t1) + (1− λ′

)(x2, t2) ∈ P−1(S).

Problem 1.2

Image of convex sets under a linear-fractional mapping remains convex:

Linear-fractional mapping function is defined by F (x) = Ax+bcT x+d

, where cTx+ d > 0. Let S be aconvex set, therefore we have to show for every x1, x2 ∈ S, λF (x1) + (1− λ)F (x2) ∈ F (S).

λF (x1) + (1− λ)F (x2)

λF (x1) + (1− λ)F (x2) = λAx1 + b

cTx1 + d+ (1− λ)

Ax2 + b

cTx2 + d

If we pick

λ =λ

′(cTx1 + d)

λ′(cTx1 + d) + (1− λ′)(cTx2 + d), λ, λ

′ ∈ [0, 1]

We can get

λF (x1) + (1− λ)F (x2) =

λ′(cTx1 + d)

λ′(cTx1 + d) + (1− λ′)(cTx2 + d)

Ax1 + b

cTx1 + d+ (1− λ

′(cTx1 + d)

λ′(cTx1 + d) + (1− λ′)(cTx2 + d))Ax2 + b

cTx2 + d

=A(λ

′x1 + (1− λ′

)x2) + b

cT (λ′x1 + (1− λ′)x2) + d

Since S is a convex set and λ′ ∈ [0, 1], λ

′x1 +(1−λ′

)x2 ∈ S. Therefore λF (x1)+(1−λ)F (x2) ∈F (S).

Inverse image of convex sets under a Linear-fractional mapping remains convex:

The inverse image of thew convex set S under a Linear-fractional mapping function is F−1(S) ={x| Ax+b

cT x+d∈ S, cTx+d > S}. We need to show that λx1+(1−λ)x2 ∈ F−1(S) for all x1, x2 ∈ F−1(S),

0 ≤ λ ≤ 1.

2

A(λx1 + (1− λ)x2) + b

cT (λx1 + (1− λ)x2) + d

=λ(Ax1 + b) + (1− λ)(Ax2 + b)

λ(cTx1 + d) + (1− λ)(cTx2 + d)

=λ(cTx1 + d) (Ax1+b)

(cT x1+d)+ (1− λ)(cTx2 + d) (Ax2+b)

(cT x2+d)

λ(cTx1 + d) + (1− λ)(cTx2 + d)

=λ(cTx1 + d) (Ax1+b)

(cT x1+d)+ (1− λ)(cTx2 + d) (Ax2+b)

(cT x2+d)

λ(cTx1 + d) + (1− λ)(cTx2 + d)

=λ(cTx1 + d)

λ(cTx1 + d) + (1− λ)(cTx2 + d)

(Ax1 + b)

(cTx1 + d)+

(1− λ)(cTx2 + d)

λ(cTx1 + d) + (1− λ)(cTx2 + d)

(Ax2 + b)

(cTx2 + d)

Bt defining λ′

= λ(cT x1+d)λ(cT x1+d)+(1−λ)(cT x2+d) , 0 ≤ λ′ ≤ 1, we have

A(λx1 + (1− λ)x2) + b

cT (λx1 + (1− λ)x2) + d= λ

′F (x1) + (1− λ′

)F (x2)

Since S is convex, λx1 + (1− λ)x2 ∈ S and it completes the proof.

Problem 2.1

epi(f) = {(x, t)|f(x) ≤ t, x ∈ dom(f)}

Suppose f be a convex function, then for (x1, t1), (x2, t2) ∈ epi(f) we have

f(λx1 + (1− λ)x2) ≤ λf(x1) + (1− λ)f(x2)

and since f(x1) ≤ t1 and f(x2) ≤ t2, we can get

λf(x1) + (1− λ)f(x2) ≤ λt1 + (1− λ)t2

Therefore {λx1 + (1 − λ)x2, λt1 + (1 − λ)t2} ∈ epi(f) and it shows that epi(f) is a convex setfor convex functions. Conversely assume epi(f) is a convex set, then if (x1, t1), (x2, t2) ∈ epi(f) wehave (λx1 + (1− λ)x2, λt1 + (1− λ)t2) ∈ epi(f), which means:

f(λx1 + (1− λ)x2) ≤ λt1 + (1− λ)t2

3

We can always pick t1 = f(x1) and t2 = f(x2) which imply the definition of convex functions.

Problem 2.2

Assume x1, x2 ∈ dom(f), then

g(λx1 + (a− λ)x2) = inf(f(λx1 + (1− λ)x2), y)

≤ f(λx1 + (1− λ)x2, λy1 + (1− λ)y2)

≤ λf(x1, y1) + (1− λ)f(x2, y2)

≤ λg(x1) + (1− λ)g(x2) + ε

By taking the limit of the right hand side of the last inequity when ε → 0, proof completes.Also the last inequality comes from the fact that for ε ≥ 0, there are y1 and y2 such that

f(x1, y1) ≤ g(x1) + ε

f(x2, y2) ≤ g(x2) + ε

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