R n → R with domf, f∗(y) = sup {yT x

f :Rⁿ→Rwith domf, f^∗(y) = sup

x∈domf

{y^Tx−f(x)},

where domf^∗={y|y^Tx−f(x)<+∞,∀x∈domf} is called the conjugate function off.

∀x∈domf,∀y∈domf^∗, f^∗(y)>y^Tx−f(x) ⇐⇒ y^Tx6f(x) +f^∗(y) e.g. f(x) = 1

2x^TQx⇒f^∗(y) = 1

2y^TQ⁻¹y, whereQ∈Sⁿ++

∀x, y, y^Tx61

2x^TQx+1

2y^TQ⁻¹y

Observation. Forf^∗(y) =1

2y^TQ⁻¹y, f^∗∗(z) =1

2z^T(Q⁻¹)⁻¹z= 1 2z^TQz.

Then, is the statement,f^∗∗ =f, true in general?

The answer is yes iff is a convex and closed function.

Note. f is a closed function if epif is a closed set.

When f is closed, f can be represented as the following:

f(x) = sup{h(x)|hare affine functions such that h(x)6f(x),∀x∈domf}

Proposition. f^∗∗=f iff is a convex and closed function.

Proof.

1. f^∗∗>f

Any affine function minorizingf ofxcan be represented asy^Tx+c, wherec is a constant.

⇒y^Tx+c6f(x),∀x∈domf

⇒y^Tx−f(x)6−c,∀x∈domf

⇒y^Tx−f(x) is bounded above over domf

⇒the supremum ofy^Tx−f(x) exists, andy∈domf^∗

⇒y∈domf^∗, f^∗(y)6−c

⇒f(x) = sup

y^Tx+c6f(x)

∀x∈domf

{y^Tx+c}= sup

y∈domf^∗ f^∗(y)6−c

{y^Tx+c}6 sup

y∈domf^∗

{y^Tx−f∗(y)}=f^∗∗(x)

2. f^∗∗6f

Sincef^∗(y) = sup

x∈domf

{y^Tx−f(x)}, x^Ty−f^∗(y)6f(x),∀x∈domf,∀y∈domf^∗.

⇒ sup

y∈domf^∗

{x^Ty−f^∗(y)}6f(x),∀x∈domf

The left-hand-side of the inequality above can be defined asf^∗∗(x) if domf^∗∗= domf. Sincex^Ty−f^∗(y)<+∞,∀x∈domf,∀y∈domf^∗, domf^∗∗⊇domf.

If there existsx^o∈/ domf butx^o∈domf^∗∗, thenf^∗∗(x)6f(x) = +∞.

It leads to contradiction, so domf^∗∗⊆domf. Hence, domf^∗∗= domf, and sup

y∈domf^∗

{x^Ty−f^∗(y)}=f^∗∗(x).

Therefore,f^∗∗(x)6f(x),∀x∈domf = domf^∗∗

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Legendre transformation

Letf be a convex and differentiable function with domf =Rⁿ.

Ify^Tx−f(x) be maximized atx^∗, theny=5f(x^∗)⇒f^∗(y) =Of(x^∗)^Tx^∗−f(x^∗)

Quasiconvex functions

f :Rⁿ→Rwith domf is quasiconvex(unimodal) function

if its every sublevel set is convex, that is,Sα={x|f(x)6α}is convex, ∀α∈R. f is quasiconcave function if−f is quasiconvex.

f is quasilinear function iff is both quasiconvex and quasiconvex.

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